[{"id":"math-functions-domain-001","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":1,"stem":"What is the domain of $f(x) = \\sqrt{2x - 1}$?","options":[{"id":"A","text":"$x > 1/2$"},{"id":"B","text":"$x \\geq 1/2$"},{"id":"C","text":"$x \\leq 1/2$"},{"id":"D","text":"All real numbers"}],"answer":"B","explanation":"For an even-root radical, the radicand must be $\\geq 0$, so $2x - 1 \\geq 0$.","knowledgePoint":"domain","source":{"exam":"CSCA Mathematics","note":"Adapted from past paper style"},"tags":["domain","radical"],"status":"published"},{"id":"math-functions-domain-002","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":1,"stem":"What is the domain of $f(x) = \\dfrac{1}{x - 2}$?","options":[{"id":"A","text":"$x \\neq 2$"},{"id":"B","text":"$x > 2$"},{"id":"C","text":"$x \\geq 2$"},{"id":"D","text":"All real numbers"}],"answer":"A","explanation":"The denominator cannot be zero, so $x - 2 \\neq 0$.","knowledgePoint":"domain","tags":["domain","rational"],"status":"published"},{"id":"math-functions-domain-003","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":2,"stem":"What is the domain of $f(x) = \\log(x + 3)$?","options":[{"id":"A","text":"$x \\geq -3$"},{"id":"B","text":"$x > -3$"},{"id":"C","text":"$x < -3$"},{"id":"D","text":"All real numbers"}],"answer":"B","explanation":"The argument of a logarithm must be strictly greater than 0.","knowledgePoint":"domain","tags":["domain","logarithm"],"status":"published"},{"id":"math-functions-domain-004","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":2,"stem":"What is the domain of $f(x) = \\dfrac{\\sqrt{x + 1}}{x - 3}$?","options":[{"id":"A","text":"$x \\geq -1$"},{"id":"B","text":"$x \\neq 3$"},{"id":"C","text":"$x \\geq -1$ and $x \\neq 3$"},{"id":"D","text":"$x > -1$ and $x \\neq 3$"}],"answer":"C","explanation":"Both conditions hold: the radical needs $x + 1 \\geq 0$ and the denominator needs $x \\neq 3$.","knowledgePoint":"domain","tags":["domain","composite"],"status":"published"},{"id":"math-functions-domain-005","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":2,"stem":"The domain of $f(x) = \\sqrt{9 - x^2}$ is the interval $[-a, a]$. Find a.","options":[{"id":"A","text":"9"},{"id":"B","text":"-3"},{"id":"C","text":"6"},{"id":"D","text":"3"}],"answer":"D","explanation":"$9 - x^2 \\geq 0$ gives $x^2 \\leq 9$, so $-3 \\leq x \\leq 3$ and $a = 3$.","knowledgePoint":"domain","tags":["domain","interval"],"status":"published"},{"id":"math-functions-range-001","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":2,"stem":"What is the range of $f(x) = x^2 + 1$?","options":[{"id":"A","text":"$[1, \\infty)$"},{"id":"B","text":"$(1, \\infty)$"},{"id":"C","text":"$[0, \\infty)$"},{"id":"D","text":"All real numbers"}],"answer":"A","explanation":"$x^2 \\geq 0$ for every real x, so $x^2 + 1 \\geq 1$, with minimum 1 at $x = 0$.","knowledgePoint":"range","tags":["range","quadratic"],"status":"published"},{"id":"math-functions-monotonicity-001","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":2,"stem":"On which interval is $f(x) = 2x - 5$ strictly increasing?","options":[{"id":"A","text":"$(0, \\infty)$"},{"id":"B","text":"$(-\\infty, \\infty)$"},{"id":"C","text":"$(-\\infty, 0)$"},{"id":"D","text":"Nowhere"}],"answer":"B","explanation":"A linear function with positive slope is strictly increasing everywhere.","knowledgePoint":"monotonicity","tags":["monotonicity","linear"],"status":"published"},{"id":"math-functions-parity-001","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":2,"stem":"Which statement shows $f(x) = x^3$ is an odd function?","options":[{"id":"A","text":"$f(-x) = f(x)$"},{"id":"B","text":"$f(-x) = -f(x)$"},{"id":"C","text":"$f(-x) = 0$"},{"id":"D","text":"$f(x) = f(x)$"}],"answer":"B","explanation":"$(-x)^3 = -x^3 = -f(x)$, the defining property of an odd function.","knowledgePoint":"parity","tags":["parity","odd"],"status":"published"},{"id":"math-functions-inverse-001","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":3,"stem":"If $f(x) = 3x - 2$, the inverse function is $f^{-1}(x) = \\dfrac{x + a}{3}$. Find a.","options":[{"id":"A","text":"-2"},{"id":"B","text":"3"},{"id":"C","text":"2"},{"id":"D","text":"$\\dfrac{1}{3}$"}],"answer":"C","explanation":"Solve $y = 3x - 2$ for $x: x = (y + 2)/3$, so $f^{-1}(x) = (x + 2)/3$.","knowledgePoint":"inverse","tags":["inverse","linear"],"status":"published"},{"id":"math-functions-power-001","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":1,"stem":"Simplify $2^3 \\cdot 2^4$.","options":[{"id":"A","text":"$2^7$"},{"id":"B","text":"$2^{12}$"},{"id":"C","text":"$4^7$"},{"id":"D","text":"$2^6$"}],"answer":"A","explanation":"Same base: add $exponents, 2^{3+4} = 2^7$.","tags":["power","exponent"],"status":"published"},{"id":"math-functions-exponential-001","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":2,"stem":"Solve $2^x = 8$ for x.","options":[{"id":"A","text":"2"},{"id":"B","text":"3"},{"id":"C","text":"4"},{"id":"D","text":"6"}],"answer":"B","explanation":"$8 = 2^3$, so $2^x = 2^3$ and $x = 3$.","tags":["exponential"],"status":"published"},{"id":"math-functions-logarithmic-001","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":2,"stem":"$\\log_{2}(32) =$ ?","options":[{"id":"A","text":"4"},{"id":"B","text":"5"},{"id":"C","text":"6"},{"id":"D","text":"16"}],"answer":"B","explanation":"$2^5 = 32$, so $\\log_{2}(32) = 5$.","tags":["logarithm"],"status":"published"},{"id":"math-functions-trig-001","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":2,"stem":"What is $\\sin(\\pi/6)$?","options":[{"id":"A","text":"1/2"},{"id":"B","text":"$\\sqrt{3}/2$"},{"id":"C","text":"1"},{"id":"D","text":"0"}],"answer":"A","explanation":"$\\sin(30^\\circ) = 1/2$.","tags":["trigonometry","sine"],"status":"published"},{"id":"math-functions-calculus-001","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":3,"stem":"The derivative of $f(x) = x^2$ at $x = 3$ is ?","options":[{"id":"A","text":"6"},{"id":"B","text":"3"},{"id":"C","text":"9"},{"id":"D","text":"2"}],"answer":"A","explanation":"$f'(x) = 2x$, so $f'(3) = 6$.","knowledgePoint":"derivatives","tags":["calculus","derivative"],"status":"published"},{"id":"math-functions-domain-006","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":2,"stem":"Find the domain of $f(x) = \\dfrac{\\sqrt{2x - 1}}{x - 3}$.","stemZh":"函数 $f(x) = \\dfrac{\\sqrt{2x - 1}}{x - 3}$ 的定义域是（　　）","options":[{"id":"A","text":"$[1/2, +\\infty)$"},{"id":"B","text":"$(1/2, 3) \\cup (3, +\\infty)$"},{"id":"C","text":"$[1/2, 3) \\cup (3, +\\infty)$"},{"id":"D","text":"$(-\\infty, 3) \\cup (3, +\\infty)$"}],"answer":"C","explanation":"Two conditions hold at once: the radical needs $2x - 1 \\geq 0$ and the denominator needs $x \\neq 3$, giving $[1/2, 3) \\cup (3, +\\infty)$.","explanationZh":"需满足 $2x - 1 \\geq 0 且 x \\neq 3，即 x \\geq 1/2 且 x \\neq 3。$","knowledgePoint":"domain","tags":["domain","radical","rational"],"status":"published"},{"id":"math-functions-domain-007","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":2,"stem":"Find the domain of $f(x) = \\dfrac{1}{\\sqrt{5 - 2x}}$.","stemZh":"函数 $f(x) = \\dfrac{1}{\\sqrt{5 - 2x}}$ 的定义域是（　　）","options":[{"id":"A","text":"$(-\\infty, 5/2)$"},{"id":"B","text":"$(-\\infty, 5/2]$"},{"id":"C","text":"$(5/2, +\\infty)$"},{"id":"D","text":"$[5/2, +\\infty)$"}],"answer":"A","explanation":"The radical sits in the denominator, so its radicand must be strictly $positive: 5 - 2x > 0$.","explanationZh":"$根式在分母中，被开方数必须严格大于零：5 - 2x > 0。$","knowledgePoint":"domain","tags":["domain","radical","rational"],"status":"published"},{"id":"math-functions-domain-008","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":2,"stem":"Find the domain of $f(x) = \\ln(x + 2) + \\dfrac{1}{x - 1}$.","stemZh":"函数 $f(x) = \\ln(x + 2) + \\dfrac{1}{x - 1}$ 的定义域是（　　）","options":[{"id":"A","text":"$(-2, +\\infty)$"},{"id":"B","text":"$(-2, 1) \\cup (1, +\\infty)$"},{"id":"C","text":"$[-2, 1) \\cup (1, +\\infty)$"},{"id":"D","text":"$\\mathbb{R} \\setminus \\{-2, 1\\}$"}],"answer":"B","explanation":"The logarithm needs $x + 2 > 0 (strictly,\\text{ not merely }x + 2 \\neq 0)$ and the fraction needs $x - 1 \\neq 0$.","explanationZh":"需同时满足 $x + 2 > 0（真数为正，不是 x + 2 \\neq 0）与 x - 1 \\neq 0；C$ 错在把真数条件写成 $\\geq 0，D$ 是把对数当成分母处理的典型错解。","knowledgePoint":"domain","tags":["domain","logarithm","rational"],"status":"published"},{"id":"math-functions-domain-009","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":3,"stem":"Find the domain of $f(x) = \\sqrt{\\dfrac{x - 1}{x + 2}}$.","stemZh":"函数 $f(x) = \\sqrt{\\dfrac{x - 1}{x + 2}}$ 的定义域是（　　）","options":[{"id":"A","text":"$(-\\infty, -2] \\cup [1, +\\infty)$"},{"id":"B","text":"$(-2, 1]$"},{"id":"C","text":"$[-2, 1]$"},{"id":"D","text":"$(-\\infty, -2) \\cup [1, +\\infty)$"}],"answer":"D","explanation":"Solve $\\dfrac{x - 1}{x + 2} \\geq 0$ with $x \\neq -2$. The critical points are -2 and 1, giving $(-\\infty, -2) \\cup [1, +\\infty)$. The equality sign can only go to the zero of the numerator, never to $x = -2$.","explanationZh":"$解 \\dfrac{x - 1}{x + 2} \\geq 0 且 x \\neq -2，临界点为 -2 和 1。等号只能给分子零点 x = 1，不能给使分母为零的 x = -2。$","knowledgePoint":"domain","tags":["domain","radical","rational-inequality"],"status":"published"},{"id":"math-functions-domain-010","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":3,"stem":"Find the domain of $f(x) = \\log_{2}(3 - x) + \\sqrt{x + 1}$.","stemZh":"函数 $f(x) = \\log_{2}(3 - x) + \\sqrt{x + 1}$ 的定义域是（　　）","options":[{"id":"A","text":"$[-1, 3)$"},{"id":"B","text":"$(-1, 3)$"},{"id":"C","text":"$[-1, 3]$"},{"id":"D","text":"$(-\\infty, 3)$"}],"answer":"A","explanation":"Both $3 - x > 0$ and $x + 1 \\geq 0$ must hold, so $-1 \\leq x < 3$.","explanationZh":"需同时满足 $3 - x > 0 与 x + 1 \\geq 0，即 -1 \\leq x < 3。$","knowledgePoint":"domain","tags":["domain","logarithm","radical"],"status":"published"},{"id":"math-functions-domain-011","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":3,"stem":"Find the domain of $f(x) = \\dfrac{1}{\\sqrt{x^2 - 5x + 6}}$.","stemZh":"函数 $f(x) = \\dfrac{1}{\\sqrt{x^2 - 5x + 6}}$ 的定义域是（　　）","options":[{"id":"A","text":"$(-\\infty, 2] \\cup [3, +\\infty)$"},{"id":"B","text":"$(2, 3)$"},{"id":"C","text":"$(-\\infty, 2) \\cup (3, +\\infty)$"},{"id":"D","text":"$\\mathbb{R} \\setminus \\{2, 3\\}$"}],"answer":"C","explanation":"The radical is in the denominator, so $x^2 - 5x + 6 > 0$, i.e. $(x - 2)(x - 3) > 0$. Option A wrongly uses $\\geq 0$; option D only removes the zeros.","explanationZh":"根式在分母中，需 $x^2 - 5x + 6 > 0，即 (x - 2)(x - 3) > 0。A$ 错用 $\\geq 0，B$ 取了两根之间，D 只排除零点。","knowledgePoint":"domain","tags":["domain","radical","quadratic"],"status":"published"},{"id":"math-functions-domain-012","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":3,"stem":"The domain of f is $[-2, 3)$. Find the domain of $g(x) = f(2x - 1) + \\dfrac{1}{\\sqrt{1 - x}}$.","stemZh":"已知函数 f(x) 的定义域是 $[-2, 3)，则函数 g(x) = f(2x - 1) + 1/\\sqrt{1 - x}$ 的定义域是（　　）","options":[{"id":"A","text":"$[-1/2, 2)$"},{"id":"B","text":"$[-1/2, 1)$"},{"id":"C","text":"$(-1/2, 1)$"},{"id":"D","text":"$[-1, 1)$"}],"answer":"B","explanation":"$f(2x - 1)$ needs $-2 \\leq 2x - 1 < 3$, so $-\\tfrac{1}{2} \\leq x < 2$. The term $\\dfrac{1}{\\sqrt{1 - x}}$ needs $1 - x > 0$, so $x < 1$. Intersecting gives $[-\\tfrac{1}{2}, 1)$.","explanationZh":"$f(2x - 1)$ 要求 $-2 \\leq 2x - 1 < 3，即 -1/2 \\leq x < 2；1/\\sqrt{1 - x}$ 要求 $1 - x > 0，即 x < 1。取交集得 [-1/2, 1)。A$ 遗漏第二项，C 左端点开闭错误，D 是移项出错。","knowledgePoint":"domain","tags":["domain","composite","radical"],"status":"published"},{"id":"math-functions-domain-013","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":4,"stem":"The domain of f is $(-1, 2]$. Find the domain of $g(x) = f(x^2 - 2)$.","stemZh":"已知函数 f(x) 的定义域是 $(-1, 2]，则函数 g(x) = f(x^2 - 2)$ 的定义域是（　　）","options":[{"id":"A","text":"$[-2, 2]$"},{"id":"B","text":"$(-2, -1) \\cup (1, 2)$"},{"id":"C","text":"$[-2, -1] \\cup [1, 2]$"},{"id":"D","text":"$[-2, -1) \\cup (1, 2]$"}],"answer":"D","explanation":"We need $-1 < x^2 - 2 \\leq 2$, i.e. $1 < x^2 \\leq 4$, so $|x| > 1$ and $|x| \\leq 2$. The strict left inequality excludes $x = \\pm 1$; the non-strict right one keeps $x = \\pm 2$.","explanationZh":"$需 -1 < x^2 - 2 \\leq 2，即 1 < x^2 \\leq 4，亦即 |x| > 1 且 |x| \\leq 2。左端严格不等式排除了 x = \\pm 1，右端非严格不等式保留了 x = \\pm 2。A$ 只解了 $x^2 \\leq 4。$","knowledgePoint":"domain","tags":["domain","composite","quadratic"],"status":"published"},{"id":"math-functions-domain-014","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":4,"stem":"Find the domain of $f(x) = \\dfrac{1}{\\ln(x^2 - 4x + 3)}$.","stemZh":"函数 $f(x) = \\dfrac{1}{\\ln(x^2 - 4x + 3)}$ 的定义域是（　　）","options":[{"id":"A","text":"$(-\\infty, 1) \\cup (3, +\\infty)$, with $x \\neq 2 \\pm \\sqrt{2}$"},{"id":"B","text":"$(-\\infty, 1) \\cup (3, +\\infty)$"},{"id":"C","text":"$(-\\infty, 2 - \\sqrt{2}) \\cup (2 + \\sqrt{2}, +\\infty)$"},{"id":"D","text":"$(-\\infty, 1] \\cup [3, +\\infty)$, with $x \\neq 2 \\pm \\sqrt{2}$"}],"answer":"A","explanation":"The argument needs $x^2 - 4x + 3 > 0$, so $x < 1$ or $x > 3$. The denominator also needs $\\ln(x^2 - 4x + 3) \\neq 0$, i.e. $x^2 - 4x + 3 \\neq 1$, giving $x \\neq 2 \\pm \\sqrt{2}$. Both excluded points fall inside the valid $intervals: 2 - \\sqrt{2} \\approx 0.59 < 1$ and $2 + \\sqrt{2} \\approx 3.41 > 3$.","explanationZh":"真数需 $x^2 - 4x + 3 > 0，即 x < 1 或 x > 3；分母还需 \\ln(x^2 - 4x + 3) \\neq 0，即 x^2 - 4x + 3 \\neq 1，解得 x \\neq 2 \\pm \\sqrt{2}。两个排除点都落在有效区间内（2 - \\sqrt{2} \\approx 0.59 < 1，2 + \\sqrt{2} \\approx 3.41 > 3）。B$ 遗漏分母不为零，C 遗漏真数大于零，D 错用 $\\geq 0。$","knowledgePoint":"domain","tags":["domain","logarithm","quadratic"],"status":"published"},{"id":"math-functions-domain-015","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":4,"stem":"Find the domain of $f(x) = \\sqrt{\\dfrac{x - 1}{x + 2}} + \\dfrac{1}{\\ln(3 - x)}$.","stemZh":"函数 $f(x) = \\sqrt{\\dfrac{x - 1}{x + 2}} + \\dfrac{1}{\\ln(3 - x)}$ 的定义域是（　　）","options":[{"id":"A","text":"$(-\\infty, -2) \\cup [1, 3)$"},{"id":"B","text":"$(-\\infty, -2] \\cup [1, 2) \\cup (2, 3)$"},{"id":"C","text":"$(-\\infty, -2) \\cup [1, 2) \\cup (2, 3)$"},{"id":"D","text":"$[1, 2) \\cup (2, 3)$"}],"answer":"C","explanation":"Intersect three $conditions: \\dfrac{x - 1}{x + 2} \\geq 0$ gives $(-\\infty, -2) \\cup [1, +\\infty); 3 - x > 0$ gives $x < 3$; and $\\ln(3 - x) \\neq 0$ gives $x \\neq 2$. The result is $(-\\infty, -2) \\cup [1, 2) \\cup (2, 3)$.","explanationZh":"$三个条件取交集：\\dfrac{x - 1}{x + 2} \\geq 0 得 (-\\infty, -2) \\cup [1, +\\infty)；3 - x > 0 得 x < 3；\\ln(3 - x) \\neq 0 得 x \\neq 2。A$ 遗漏 $x = 2，B$ 错把 $x = -2$ 留下，D 遗漏 $(-\\infty, -2)$ 这一段。","knowledgePoint":"domain","tags":["domain","radical","logarithm","composite"],"status":"published"},{"id":"math-sets-rep-001","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":1,"stem":"Which notation represents the set {2, 4, 6, 8}?","options":[{"id":"A","text":"${x | x < 8}$"},{"id":"B","text":"${x ∈ ℕ⁺ | x\\text{ is even and }x ≤ 8}$"},{"id":"C","text":"$(2, 8)$"},{"id":"D","text":"$[2, 8]$"}],"answer":"B","explanation":"It lists the positive even integers up to 8.","knowledgePoint":"set","tags":["set","notation"],"status":"published"},{"id":"math-sets-rel-001","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":1,"stem":"If $A = {1, 2}$ and $B = {1, 2, 3}$, which is true?","options":[{"id":"A","text":"$A ⊂ B$"},{"id":"B","text":"$B ⊂ A$"},{"id":"C","text":"$A = B$"},{"id":"D","text":"$A ∩ B = ∅$"}],"answer":"A","explanation":"Every element of A is in B, and B has an extra element, so A is a proper subset of B.","knowledgePoint":"subset","tags":["set","subset"],"status":"published"},{"id":"math-sets-op-001","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":2,"stem":"If $A = {1, 2, 3}$ and $B = {3, 4}$, then $|A ∪ B| =$ ?","options":[{"id":"A","text":"3"},{"id":"B","text":"5"},{"id":"C","text":"4"},{"id":"D","text":"6"}],"answer":"C","explanation":"$A ∪ B = {1, 2, 3, 4}$, which has 4 elements.","knowledgePoint":"union","tags":["set","union"],"status":"published"},{"id":"math-ineq-quad-001","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":2,"stem":"Solve $x^{2} − 4 < 0$.","options":[{"id":"A","text":"$x < 2$"},{"id":"B","text":"$−2 < x < 2$"},{"id":"C","text":"$x > 2$ or $x < −2$"},{"id":"D","text":"$x ≠ 2$"}],"answer":"B","explanation":"$(x − 2)(x + 2) < 0$, so x lies strictly between −2 and 2.","tags":["inequality","quadratic"],"status":"published"},{"id":"math-ineq-quad-002","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"The solution set of $x^{2} − 5x + 6 ≤ 0$ is the interval $[a, b]$ with $a <$ b. Find a + b.","options":[{"id":"A","text":"1"},{"id":"B","text":"5"},{"id":"C","text":"6"},{"id":"D","text":"−5"}],"answer":"B","explanation":"$(x − 2)(x − 3) ≤ 0$ gives $[2, 3]$, so $a + b = 5$.","tags":["inequality","quadratic"],"status":"published"},{"id":"math-ineq-rational-001","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $1 / (x − 1) > 0$.","options":[{"id":"A","text":"$x > 1$"},{"id":"B","text":"$x < 1$"},{"id":"C","text":"$x ≠ 1$"},{"id":"D","text":"$x > 0$"}],"answer":"A","explanation":"A positive fraction needs a positive denominator, so $x − 1 > 0$.","tags":["inequality","rational"],"status":"published"},{"id":"math-ineq-abs-001","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":2,"stem":"Solve $|x| < 3$.","options":[{"id":"A","text":"$−3 < x < 3$"},{"id":"B","text":"$x < 3$"},{"id":"C","text":"$x > −3$"},{"id":"D","text":"$x ≤ 3$"}],"answer":"A","explanation":"$|x| < 3$ means x is within 3 units of 0.","tags":["inequality","absolute"],"status":"published"},{"id":"math-ineq-abs-002","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"The solution set of $|x − 2| ≤ 5$ is $[a, 7]$. Find a.","options":[{"id":"A","text":"−3"},{"id":"B","text":"3"},{"id":"C","text":"−7"},{"id":"D","text":"7"}],"answer":"A","explanation":"$−5 ≤ x − 2 ≤ 5$ gives $−3 ≤ x ≤ 7$, so $a = −3$.","tags":["inequality","absolute"],"status":"published"},{"id":"math-set-representation-001","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":1,"stem":"Which of the following lists the set of odd positive integers less than 8?","stemZh":"下列哪个选项用列举法表示了小于8的正奇数集？","options":[{"id":"A","text":"{1, 3, 5, 7}"},{"id":"B","text":"{1, 3, 5, 7, 9}"},{"id":"C","text":"{0, 2, 4, 6}"},{"id":"D","text":"{2, 4, 6}"}],"answer":"A","explanation":"The odd positive integers below 8 are 1, 3, 5 and 7. 9 is not less than 8, and the even numbers do not qualify.","explanationZh":"小于8的正奇数为1、3、5、7。9不小于8，偶数也不符合要求。","tags":["easy","set","listing"],"status":"published"},{"id":"math-set-representation-002","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":1,"stem":"Write the set of real numbers x with $−1 < x ≤ 3$ as an interval.","stemZh":"用区间表示满足 $−1 < x ≤ 3$ 的实数集合。","options":[{"id":"A","text":"$(−1, 3)$"},{"id":"B","text":"$[−1, 3]$"},{"id":"C","text":"$(−1, 3]$"},{"id":"D","text":"$[−1, 3)$"}],"answer":"C","explanation":"The strict inequality $−1 < x$ excludes −1, so the left end is $open; x ≤ 3$ includes 3, so the right end is closed. Hence $(−1, 3]$.","explanationZh":"$−1 < x$ 不含 $−1，左端为开；x ≤ 3 含 3，右端为闭。故为 (−1, 3]。$","tags":["easy","set","interval"],"status":"published"},{"id":"math-set-representation-003","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":1,"stem":"Which symbol is used to say that an object is an element of a set?","stemZh":"表示“某个对象是一个集合的元素”使用哪个符号？","options":[{"id":"A","text":"$⊆$"},{"id":"B","text":"$∈$"},{"id":"C","text":"$∪$"},{"id":"D","text":"$∩$"}],"answer":"B","explanation":"$∈$ relates an element to a set. $⊆$ relates a set to another $set, ∪$ is union and $∩$ is intersection.","explanationZh":"$∈ 表示元素与集合的关系；⊆ 表示集合与集合的包含关系，∪ 是并集，∩$ 是交集。","tags":["easy","set","notation"],"status":"published"},{"id":"math-set-representation-004","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":2,"stem":"Which description represents the set {2, 4, 6, 8}?","stemZh":"下列哪个描述法表示了集合 {2, 4, 6, 8}？","options":[{"id":"A","text":"${x | x\\text{ is }even}$"},{"id":"B","text":"${x | x < 9}$"},{"id":"C","text":"${x | x = 2n, n ∈ ℝ}$"},{"id":"D","text":"${x | x = 2n, n ∈ ℕ⁺, n ≤ 4}$"}],"answer":"D","explanation":"The set contains exactly 2, 4, 6, 8, which is 2n for the positive integers $n = 1, 2, 3, 4$. Option A is all even numbers and option C is all real numbers of the form 2n.","explanationZh":"$集合恰好含2、4、6、8，即 n = 1、2、3、4$ 时的 2n。A 是全体偶数，C 是形如 2n 的一切实数。","tags":["easy","set","set-builder"],"status":"published"},{"id":"math-set-representation-005","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":2,"stem":"Which of the following is NOT the empty set?","stemZh":"下列哪个集合不是空集？","options":[{"id":"A","text":"${x ∈ ℝ | x^{2} + 1 = 0}$"},{"id":"B","text":"${x ∈ ℝ | x > 0\\text{ and }x < 0}$"},{"id":"C","text":"{0}"},{"id":"D","text":"$∅$"}],"answer":"C","explanation":"{0} contains one element, namely the number 0, so it is not empty. $x^{2} + 1 = 0$ has no real solution, and no number is both positive and negative, so A, B and D are all empty.","explanationZh":"{0} 含有一个元素 $0，因此不是空集。x^{2} + 1 = 0 无实数解，也没有数既大于0又小于0，故$ A、B、D 都是空集。","tags":["easy","set","empty-set"],"status":"published"},{"id":"math-set-representation-006","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":2,"stem":"List the set ${x ∈ ℤ | −2 ≤ x < 2}$.","stemZh":"用列举法表示集合 ${x ∈ ℤ | −2 ≤ x < 2}。$","options":[{"id":"A","text":"{−2, −1, 0, 1, 2}"},{"id":"B","text":"{−2, −1, 0, 1}"},{"id":"C","text":"{−1, 0, 1}"},{"id":"D","text":"{−2, −1, 1}"}],"answer":"B","explanation":"$−2 ≤ x$ includes −2, while $x < 2$ excludes 2, so the integers are −2, −1, 0, 1.","explanationZh":"$−2 ≤ x 含 −2，而 x < 2$ 不含 2，故整数为 −2、−1、0、1。","tags":["easy","set","listing"],"status":"published"},{"id":"math-set-representation-007","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":1,"stem":"Write the solution set of $x ≥ 5$ as an interval.","stemZh":"用区间表示 $x ≥ 5$ 的解集。","options":[{"id":"A","text":"$(5, +∞)$"},{"id":"B","text":"$(−∞, 5]$"},{"id":"C","text":"$(−∞, 5)$"},{"id":"D","text":"$[5, +∞)$"}],"answer":"D","explanation":"$x ≥ 5$ includes 5 and extends without bound to the right, so the interval is $[5, +∞)$.","explanationZh":"$x ≥ 5$ 包含 $5，且向右无限延伸，故为 [5, +∞)。$","tags":["easy","set","interval"],"status":"published"},{"id":"math-set-representation-008","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":2,"stem":"Which of the following sets is infinite?","stemZh":"下列哪个集合是无限集？","options":[{"id":"A","text":"${x ∈ ℝ | 0 < x < 1}$"},{"id":"B","text":"${x ∈ ℤ | −3 ≤ x ≤ 3}$"},{"id":"C","text":"{1, 2, 3}"},{"id":"D","text":"${x ∈ ℕ⁺ | x < 10}$"}],"answer":"A","explanation":"Between any two real numbers there are infinitely many reals, so $0 < x < 1$ is infinite. The other three contain 7, 3 and 9 elements respectively.","explanationZh":"任意两个实数之间都有无穷多个实数，故 $0 < x < 1$ 是无限集。其余三个分别含 7、3、9 个元素。","tags":["easy","set","finite-infinite"],"status":"published"},{"id":"math-set-representation-009","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":1,"stem":"Solve the system $x + y = 3, x − y = 1$ and write its solution set by listing.","stemZh":"解方程组 $x + y = 3，x − y = 1，并用列举法写出其解集。$","options":[{"id":"A","text":"{2, 1}"},{"id":"B","text":"${(2, 1)}$"},{"id":"C","text":"$(2, 1)$"},{"id":"D","text":"${x = 2, y = 1}$"}],"answer":"B","explanation":"Adding gives $2x = 4$, so $x = 2$ and $y = 1. A$ solution of a system in two unknowns is an ordered pair, so the solution set is the singleton ${(2, 1)}$.","explanationZh":"两式相加得 $2x = 4，故 x = 2，y = 1。二元方程组的解是有序数对，所以解集为单元素集 {(2, 1)}。$","tags":["easy","set","listing"],"status":"published"},{"id":"math-set-representation-010","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":2,"stem":"List the set $A = {x | x^{2} − 4 = 0}$.","stemZh":"用列举法表示集合 $A = {x | x^{2} − 4 = 0}。$","options":[{"id":"A","text":"{2}"},{"id":"B","text":"{4}"},{"id":"C","text":"{−2, 2}"},{"id":"D","text":"{−2}"}],"answer":"C","explanation":"$x^{2} − 4 = 0$ gives $x^{2} = 4$, so $x = 2$ or $x = −2$. Both roots must be listed.","explanationZh":"$由 x^{2} − 4 = 0 得 x^{2} = 4，故 x = 2 或 x = −2，两个根都要列出。$","tags":["easy","set","listing"],"status":"published"},{"id":"math-set-representation-011","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"List the set $A = {x ∈ ℝ | x^{2} − 3x + 2 = 0}$.","stemZh":"用列举法表示集合 $A = {x ∈ ℝ | x^{2} − 3x + 2 = 0}。$","options":[{"id":"A","text":"{1}"},{"id":"B","text":"{1, 2}"},{"id":"C","text":"{2}"},{"id":"D","text":"{−1, −2}"}],"answer":"B","explanation":"$x^{2} − 3x + 2 = (x − 1)(x − 2) = 0$, so $x = 1$ or $x = 2$.","explanationZh":"$x^{2} − 3x + 2 = (x − 1)(x − 2) = 0，故 x = 1 或 x = 2。$","tags":["medium","set","listing"],"status":"published"},{"id":"math-set-representation-012","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"The set ${x | x = 2k + 1, k ∈ ℤ}$ is","stemZh":"集合 ${x | x = 2k + 1, k ∈ ℤ} 是$","options":[{"id":"A","text":"the set of odd numbers"},{"id":"B","text":"the set of even numbers"},{"id":"C","text":"the set of positive integers"},{"id":"D","text":"the set of integers"}],"answer":"A","explanation":"2k is even for every integer k, so 2k + 1 is always odd, and every odd integer can be written this way.","explanationZh":"对任意整数 k，2k 是偶数，因此 2k + 1 恒为奇数，且每个奇数都能写成这种形式。","tags":["medium","set","set-builder"],"status":"published"},{"id":"math-set-representation-013","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"Which set-builder notation describes the points in the second quadrant?","stemZh":"下列哪个描述法表示第二象限内的点集？","options":[{"id":"A","text":"${(x, y) | x > 0, y > 0}$"},{"id":"B","text":"${(x, y) | x < 0, y < 0}$"},{"id":"C","text":"${(x, y) | x < 0, y > 0}$"},{"id":"D","text":"${(x, y) | x > 0, y < 0}$"}],"answer":"C","explanation":"In the second quadrant the x-coordinate is negative and the y-coordinate is positive, so $x < 0$ and $y > 0$.","explanationZh":"第二象限内点的横坐标为负、纵坐标为正，即 $x < 0 且 y > 0。$","tags":["medium","set","coordinates"],"status":"published"},{"id":"math-set-representation-014","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"How many subsets does the set $A = {1, 2}$ have?","stemZh":"集合 $A = {1, 2}$ 共有多少个子集？","options":[{"id":"A","text":"2"},{"id":"B","text":"3"},{"id":"C","text":"8"},{"id":"D","text":"4"}],"answer":"D","explanation":"A set with n elements has $2^{n}$ subsets. Here $n = 2$, so there are $2^{2} = 4 subsets: ∅, {1}, {2}, {1, 2}$.","explanationZh":"含 n 个元素的集合有 $2^{n}$ 个子集。此处 $n = 2，故有 2^{2} = 4 个：∅、{1}、{2}、{1, 2}。$","tags":["medium","set","subsets"],"status":"published"},{"id":"math-set-representation-015","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"If the set $A = {x |\\text{ ax }+ 1 = 0}$ is empty, then $a =$","stemZh":"若集合 $A = {x |$ ax $+ 1 = 0}$ 是空集，则 $a =$","options":[{"id":"A","text":"0"},{"id":"B","text":"1"},{"id":"C","text":"−1"},{"id":"D","text":"any real number"}],"answer":"A","explanation":"When $a = 0$ the equation becomes $1 = 0$, which has no solution, so the set is empty. For $a ≠ 0$ the equation has the solution $x = −1/a$.","explanationZh":"$当 a = 0$ 时方程变为 $1 = 0，无解，集合为空集。当 a ≠ 0$ 时方程有解 $x = −1/a。$","tags":["medium","set","empty-set"],"status":"published"},{"id":"math-set-representation-016","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"List the set ${x ∈ ℕ | |x| < 3}$.","stemZh":"用列举法表示集合 ${x ∈ ℕ | |x| < 3}。$","options":[{"id":"A","text":"{1, 2}"},{"id":"B","text":"{0, 1, 2}"},{"id":"C","text":"{−2, −1, 0, 1, 2}"},{"id":"D","text":"{0, 1, 2, 3}"}],"answer":"B","explanation":"$|x| < 3$ gives $−3 < x < 3$; restricting to the natural numbers (which include $0)$ leaves 0, 1, 2.","explanationZh":"$由 |x| < 3 得 −3 < x < 3；限制在自然数集（含 0）内，剩下 0、1、2。$","tags":["medium","set","listing"],"status":"published"},{"id":"math-set-representation-017","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"The set $A = {x ∈ ℝ | x^{2} + 2x + a = 0}$ has exactly one element. Find a.","stemZh":"集合 $A = {x ∈ ℝ | x^{2} + 2x + a = 0}$ 恰有一个元素，求 a。","options":[{"id":"A","text":"−1"},{"id":"B","text":"0"},{"id":"C","text":"1"},{"id":"D","text":"2"}],"answer":"C","explanation":"Exactly one element means the quadratic has a double root, so the discriminant is $0: Δ = 4 − 4a = 0$, giving $a = 1$.","explanationZh":"恰有一个元素说明二次方程有两个相等实根，判别式为 $0：Δ = 4 − 4a = 0，得 a = 1。$","tags":["medium","set","discriminant"],"status":"published"},{"id":"math-set-representation-018","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"List the set $A = {(x, y) | x + y = 4, x ∈ ℕ, y ∈ ℕ}$.","stemZh":"用列举法表示集合 $A = {(x, y) | x + y = 4, x ∈ ℕ, y ∈ ℕ}。$","options":[{"id":"A","text":"${(1, 3), (2, 2), (3, 1)}$"},{"id":"B","text":"${(0, 4), (1, 3), (2, 2), (3, 1), (4, 0)}$"},{"id":"C","text":"{0, 1, 2, 3, 4}"},{"id":"D","text":"${(4, 0), (3, 1), (2, 2), (1, 3)}$"}],"answer":"B","explanation":"The natural numbers include 0, so the pairs are $(0, 4), (1, 3), (2, 2), (3, 1), (4, 0)$. Option A forgets the two pairs involving 0.","explanationZh":"自然数集含 $0，故数对为 (0, 4)、(1, 3)、(2, 2)、(3, 1)、(4, 0)。A$ 漏掉了含 0 的两组。","tags":["medium","set","listing"],"status":"published"},{"id":"math-set-representation-019","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"Write ${x | x ≠ 2}$ as a union of intervals.","stemZh":"用区间的并表示集合 ${x | x ≠ 2}。$","options":[{"id":"A","text":"$(−∞, 2]$"},{"id":"B","text":"$[2, +∞)$"},{"id":"C","text":"$(−∞, +∞)$"},{"id":"D","text":"$(−∞, 2) ∪ (2, +∞)$"}],"answer":"D","explanation":"Excluding the single point $x = 2$ splits the real line into two open intervals: everything below 2 and everything above 2.","explanationZh":"去掉唯一点 $x = 2$ 后，实数轴被分成两个开区间：小于 2 的部分和大于 2 的部分。","tags":["medium","set","interval"],"status":"published"},{"id":"math-set-representation-020","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"How many sets A satisfy ${1, 2} ⊆ A ⊆ {1, 2, 3, 4, 5}$?","stemZh":"满足 ${1, 2} ⊆ A ⊆ {1, 2, 3, 4, 5}$ 的集合 A 有多少个？","options":[{"id":"A","text":"4"},{"id":"B","text":"6"},{"id":"C","text":"8"},{"id":"D","text":"7"}],"answer":"C","explanation":"A must contain 1 and 2, and each of 3, 4, 5 may be either included or excluded, giving $2^{3} = 8$ possibilities.","explanationZh":"A 必须含 1 和 2，而 3、4、5 每个可选可不选，共 $2^{3} = 8$ 种。","tags":["medium","set","counting"],"status":"published"},{"id":"math-set-representation-021","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":4,"stem":"The set $A = {x | x^{2} + (a − 1)x + 1 = 0}$ has exactly one element. Find a.","stemZh":"集合 $A = {x | x^{2} + (a − 1)x + 1 = 0}$ 恰有一个元素，求 a。","options":[{"id":"A","text":"3"},{"id":"B","text":"−1"},{"id":"C","text":"3 or −1"},{"id":"D","text":"1"}],"answer":"C","explanation":"A single element requires $Δ = 0: (a − 1)^{2} − 4 = 0$, so $(a − 1)^{2} = 4$ and $a − 1 = \\pm2$, giving $a = 3$ or $a = −1$. Both values must be given.","explanationZh":"恰有一个元素需 $Δ = 0：(a − 1)^{2} − 4 = 0，即 (a − 1)^{2} = 4，a − 1 = \\pm2，得 a = 3 或 a = −1，两个值都要给出。$","tags":["hard","set","discriminant"],"status":"published"},{"id":"math-set-representation-022","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":5,"stem":"Let $A = {x | x^{2} − 5x + 6 = 0}$ and $B = {x |\\text{ mx }− 1 = 0}$. If $B ⊆ A$, then the set of possible values of m is","stemZh":"$设 A = {x | x^{2} − 5x + 6 = 0}，B = {x |$ mx $− 1 = 0}。若 B ⊆$ A，则 m 的取值集合为","options":[{"id":"A","text":"{1/2, 1/3}"},{"id":"B","text":"{0, 1/2, 1/3}"},{"id":"C","text":"{1/2}"},{"id":"D","text":"{0, 1/2}"}],"answer":"B","explanation":"$A = {2, 3}$. For $m = 0$ the equation mx $= 1$ has no solution, so $B = ∅$ and $∅ ⊆ A$ holds. For $m ≠ 0, B = {1/m}$, so $1/m = 2$ gives $m = 1/2$ and $1/m = 3$ gives $m = 1/3$. Hence $m ∈ {0, 1/2, 1/3}$; the empty-set case is the one most often missed.","explanationZh":"$A = {2, 3}。当 m = 0$ 时方程 mx $= 1$ 无解，B $= ∅，满足 ∅ ⊆$ A；当 $m ≠ 0 时 B = {1/m}，由 1/m = 2 得 m = 1/2，由 1/m = 3 得 m = 1/3。故 m ∈ {0, 1/2, 1/3}，其中空集情形最易漏掉。$","tags":["hard","set","empty-set"],"status":"published"},{"id":"math-set-representation-023","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":5,"stem":"If the set $A = {x | x^{2} + (a + 2)x + 1 = 0, x ∈ ℝ}$ is empty, find the range of a.","stemZh":"若集合 $A = {x | x^{2} + (a + 2)x + 1 = 0, x ∈ ℝ}$ 为空集，求 a 的取值范围。","options":[{"id":"A","text":"$(−∞, −4) ∪ (0, +∞)$"},{"id":"B","text":"$[−4, 0]$"},{"id":"C","text":"$(0, 4)$"},{"id":"D","text":"$(−4, 0)$"}],"answer":"D","explanation":"The set is empty when the quadratic has no real root, so $Δ < 0: (a + 2)^{2} − 4 < 0$, i.e. $(a + 2)^{2} < 4$, giving $−2 < a + 2 < 2$ and therefore $−4 < a < 0$.","explanationZh":"集合为空说明二次方程无实根，即 $Δ < 0：(a + 2)^{2} − 4 < 0，亦即 (a + 2)^{2} < 4，得 −2 < a + 2 < 2，故 −4 < a < 0。$","tags":["hard","set","discriminant"],"status":"published"},{"id":"math-set-representation-024","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":4,"stem":"For the set $A = {a, a^{2} − a}$ to be well defined, what must a satisfy?","stemZh":"要使集合 $A = {a, a^{2} − a}$ 有意义，a 应满足什么条件？","options":[{"id":"A","text":"$a ≠ 0$"},{"id":"B","text":"$a ≠ 2$"},{"id":"C","text":"$a ≠ 0$ and $a ≠ 2$"},{"id":"D","text":"a can be any real number"}],"answer":"C","explanation":"Elements of a set must be distinct, so $a ≠ a^{2} − a$, i.e. $a^{2} − 2a ≠ 0$, i.e. $a(a − 2) ≠ 0$, giving $a ≠ 0$ and $a ≠ 2$. Both exclusions are needed.","explanationZh":"集合中的元素必须互异，故 $a ≠ a^{2} −$ a，即 $a^{2} − 2a ≠ 0，亦即 a(a − 2) ≠ 0，得 a ≠ 0 且 a ≠ 2，两个条件都要排除。$","tags":["hard","set","distinctness"],"status":"published"},{"id":"math-set-representation-025","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":5,"stem":"Let $A = {x | x^{2} − 3x + 2 ≤ 0}$ and $B = {x | x > a}$. If $A ⊆ B$, find the range of a.","stemZh":"$设 A = {x | x^{2} − 3x + 2 ≤ 0}，B = {x | x > a}。若 A ⊆$ B，求 a 的取值范围。","options":[{"id":"A","text":"$a < 1$"},{"id":"B","text":"$a ≤ 1$"},{"id":"C","text":"$a < 2$"},{"id":"D","text":"$a ≤ 2$"}],"answer":"A","explanation":"$A = [1, 2]$. For $[1, 2] ⊆ (a, +∞)$ every element of A must exceed a, in particular the smallest one, 1, so $a < 1$. Equality $a = 1$ fails because $1 ∈ A$ but $1 ∉ (1, +∞)$.","explanationZh":"$A = [1, 2]。要使 [1, 2] ⊆ (a, +∞)，A$ 中每个元素都要大于 a，特别是最小的元素 $1，故 a < 1。取 a = 1$ 不行，因为 $1 ∈ A 但 1 ∉ (1, +∞)。$","tags":["hard","set","interval"],"status":"published"},{"id":"math-set-representation-026","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":4,"stem":"List the set ${x ∈ ℤ | 0 ≤ 2x − 1 < 5}$.","stemZh":"用列举法表示集合 ${x ∈ ℤ | 0 ≤ 2x − 1 < 5}。$","options":[{"id":"A","text":"{0, 1, 2}"},{"id":"B","text":"{1, 2, 3}"},{"id":"C","text":"{0, 1, 2, 3}"},{"id":"D","text":"{1, 2}"}],"answer":"D","explanation":"$0 ≤ 2x − 1$ gives $x ≥ 1/2$ and $2x − 1 < 5$ gives $x < 3$, so $1/2 ≤ x < 3$. The integers in that range are 1 and 2.","explanationZh":"$由 0 ≤ 2x − 1 得 x ≥ 1/2，由 2x − 1 < 5 得 x < 3，故 1/2 ≤ x < 3，其中的整数为 1 和 2。$","tags":["hard","set","listing"],"status":"published"},{"id":"math-set-representation-027","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":4,"stem":"Let the universal set be $U = ℝ$ and $A = {x | x ≥ 2}$. Write the complement ∁ᵤA as an interval.","stemZh":"设全集 $U = ℝ，A = {x | x ≥ 2}，用区间表示补集$ ∁ᵤA。","options":[{"id":"A","text":"$(−∞, 2]$"},{"id":"B","text":"$(−∞, 2)$"},{"id":"C","text":"$(2, +∞)$"},{"id":"D","text":"$[2, +∞)$"}],"answer":"B","explanation":"$A = [2, +∞)$ contains 2, so its complement contains everything strictly less than 2, namely $(−∞, 2)$. The point 2 belongs to A, so it must be excluded from the complement.","explanationZh":"$A = [2, +∞) 含 2，故其补集为严格小于 2$ 的全体实数，即 $(−∞, 2)。点 2$ 属于 A，必须从补集中排除。","tags":["hard","set","complement"],"status":"published"},{"id":"math-set-representation-028","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":5,"stem":"Let $A = {x | x = k/2 + 1/4, k ∈ ℤ}$ and $B = {x | x = k/4 + 1/2, k ∈ ℤ}$. Which relation holds?","stemZh":"$设 A = {x | x = k/2 + 1/4, k ∈ ℤ}，B = {x | x = k/4 + 1/2, k ∈ ℤ}，二者的关系是？$","options":[{"id":"A","text":"$A = B$"},{"id":"B","text":"$B ⊆ A$"},{"id":"C","text":"$A ⊂ B$"},{"id":"D","text":"cannot be determined"}],"answer":"C","explanation":"$A = {(2k + 1)/4 : k ∈ ℤ}$, the quarters whose numerator is odd. $B = {(k + 2)/4 : k ∈ ℤ}$, all multiples of 1/4. So every element of A lies in B, but B also contains e.g. $1/2 (take k = 0)$, which is not in A. Hence A is a proper subset of B.","explanationZh":"$A = {(2k + 1)/4 : k ∈ ℤ}，即分子为奇数的四分之一；B = {(k + 2)/4 : k ∈ ℤ}，即一切 1/4$ 的整数倍。故 A 中元素都在 B 中，而 B 还含 $1/2（取 k = 0），它不在 A$ 中，所以 A 是 B 的真子集。","tags":["hard","set","set-builder"],"status":"published"},{"id":"math-set-representation-029","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":5,"stem":"Let $A = {1, 2, 3}$. How many sets B satisfy $A ∪ B =$ A?","stemZh":"$设 A = {1, 2, 3}，满足 A ∪ B = A$ 的集合 B 有多少个？","options":[{"id":"A","text":"8"},{"id":"B","text":"7"},{"id":"C","text":"4"},{"id":"D","text":"3"}],"answer":"A","explanation":"$A ∪ B = A$ holds exactly when $B ⊆$ A. A has 3 elements, so it has $2^{3} = 8 subsets$, and each of them works as B.","explanationZh":"$A ∪ B = A$ 当且仅当 $B ⊆$ A。A 有 3 个元素，共有 $2^{3} = 8$ 个子集，每一个都可作为 B。","tags":["hard","set","counting"],"status":"published"},{"id":"math-set-representation-030","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":4,"stem":"Which set-builder notation describes the points of the first quadrant, excluding the axes?","stemZh":"下列哪个描述法表示第一象限内（不含坐标轴）的点集？","options":[{"id":"A","text":"${(x, y) | x ≥ 0, y ≥ 0}$"},{"id":"B","text":"${(x, y) | x > 0, y < 0}$"},{"id":"C","text":"${(x, y) |\\text{ xy }> 0}$"},{"id":"D","text":"${(x, y) | x > 0, y > 0}$"}],"answer":"D","explanation":"Excluding the axes means both coordinates are strictly positive, so $x > 0$ and $y > 0$. Option C describes quadrants I and III together, since xy $> 0$ also holds when both coordinates are negative.","explanationZh":"不含坐标轴意味着两个坐标都严格为正，即 $x > 0 且 y > 0。C$ 描述的是第一和第三象限，因为两坐标同负时也有 xy $> 0。$","tags":["hard","set","coordinates"],"status":"published"},{"id":"math-set-relations-001","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":1,"stem":"Let $A = {2, 4, 6}$ and $B = {2, 4, 6, 8}$. Which statement is true?","stemZh":"已知 $A = {2, 4, 6}，B = {2, 4, 6, 8}，下列关系正确的是？$","options":[{"id":"A","text":"$A ∈ B$"},{"id":"B","text":"$B ⊆ A$"},{"id":"C","text":"$A = B$"},{"id":"D","text":"$A ⊆ B$"}],"answer":"D","explanation":"Every element of $A (2, 4, 6)$ also belongs to B, so $A ⊆ B (in\\text{ fact }A\\text{ is }a\\text{ proper subset of }B). A$ itself is a set, not one of the elements listed inside B, so $A ∈ B$ is false.","explanationZh":"A 中的元素 2、4、6 都属于 B，所以 $A ⊆$ B（且为真子集）。A 是一个集合，不是 B 中的元素，因此 $A ∈ B$ 错误。","tags":["easy","subset","element-relation"],"status":"published"},{"id":"math-set-relations-002","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":1,"stem":"Let $U = {1, 2, 3, 4, 5, 6}$ and $A = {1, 2, 3}$. Find ∁ᵤA.","stemZh":"全集 $U = {1, 2, 3, 4, 5, 6}，A = {1, 2, 3}，求$ ∁ᵤA。","options":[{"id":"A","text":"{4, 5, 6}"},{"id":"B","text":"{1, 2, 3}"},{"id":"C","text":"$∅$"},{"id":"D","text":"{1, 2, 3, 4, 5, 6}"}],"answer":"A","explanation":"The complement of A in U consists of the elements of U that are not in A, namely 4, 5 and 6.","explanationZh":"补集由 U 中不属于 A 的元素组成，即 4、5、6。","tags":["easy","complement","universal-set"],"status":"published"},{"id":"math-set-relations-003","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":2,"stem":"Let $A = {x | x^{2} = 4}$. How many subsets does A have?","stemZh":"已知 $A = {x | x^{2} = 4}，则 A$ 的子集个数为？","options":[{"id":"A","text":"2"},{"id":"B","text":"3"},{"id":"C","text":"4"},{"id":"D","text":"8"}],"answer":"C","explanation":"$x^{2} = 4$ gives $x = −2$ or $x = 2$, so $A = {−2, 2}$ has 2 elements and therefore $2^{2} = 4$ subsets.","explanationZh":"$由 x^{2} = 4 得 x = −2 或 x = 2，故 A = {−2, 2} 有 2$ 个元素，子集个数为 $2^{2} = 4。$","tags":["easy","subsets","counting"],"status":"published"},{"id":"math-set-relations-004","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":1,"stem":"Let $A = {1, 2}$ and $B = {2, 3}$. Find $A ∩$ B.","stemZh":"已知 $A = {1, 2}，B = {2, 3}，求 A ∩$ B。","options":[{"id":"A","text":"{1, 3}"},{"id":"B","text":"{2}"},{"id":"C","text":"{1, 2, 3}"},{"id":"D","text":"$∅$"}],"answer":"B","explanation":"The intersection keeps only the elements common to both sets; 2 is the only one.","explanationZh":"交集只保留两集合的公共元素，只有 2。","tags":["easy","intersection"],"status":"published"},{"id":"math-set-relations-005","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":2,"stem":"Let $A = {1, 2}$ and $B = {2, 3}$. Find $A ∪$ B.","stemZh":"已知 $A = {1, 2}，B = {2, 3}，求 A ∪$ B。","options":[{"id":"A","text":"{2}"},{"id":"B","text":"{1, 3}"},{"id":"C","text":"{1, 2, 3}"},{"id":"D","text":"{1, 2}"}],"answer":"C","explanation":"The union collects every element belonging to at least one of the sets: 1, 2 and 3.","explanationZh":"并集包含属于至少一个集合的所有元素：1、2、3。","tags":["easy","union"],"status":"published"},{"id":"math-set-relations-006","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":1,"stem":"Which relation symbol makes the statement {1, 2} ___ {1, 2} true?","stemZh":"用哪个关系符号可使 {1, 2} ___ {1, 2} 成立？","options":[{"id":"A","text":"$⊆$"},{"id":"B","text":"$⊂$"},{"id":"C","text":"$∈$"},{"id":"D","text":"⊄"}],"answer":"A","explanation":"Every set is a subset of itself, so $⊆$ holds. The proper-subset symbol $⊂$ fails when the two sets are equal, and $∈$ fails because the elements of {1, 2} are the numbers 1 and 2, not the set {1, 2} itself.","explanationZh":"任何集合都是它自己的子集，故 $⊆$ 成立；两集合相等时真子集符号 $⊂ 不成立；{1, 2}$ 的元素是 $1 和 2，而不是集合 {1, 2}$ 本身，故 $∈$ 错误。","tags":["easy","subset","notation"],"status":"published"},{"id":"math-set-relations-007","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":2,"stem":"How many proper subsets does ${a,\\text{ b, }c}$ have?","stemZh":"集合 ${a, b, c}$ 的真子集个数为？","options":[{"id":"A","text":"6"},{"id":"B","text":"8"},{"id":"C","text":"9"},{"id":"D","text":"7"}],"answer":"D","explanation":"A 3-element set has $2^{3} = 8$ subsets. A proper subset excludes the set itself, leaving $8 − 1 = 7$.","explanationZh":"含 3 个元素的集合有 $2^{3} = 8$ 个子集，真子集要去掉集合本身，故为 $8 − 1 = 7。$","tags":["easy","subsets","counting"],"status":"published"},{"id":"math-set-relations-008","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":1,"stem":"Let $A = {x ∈ ℝ | x > 3}$ and $B = {x ∈ ℝ | x > 1}$. Find $A ∩$ B.","stemZh":"已知 $A = {x ∈ ℝ | x > 3}，B = {x ∈ ℝ | x > 1}，求 A ∩$ B。","options":[{"id":"A","text":"${x | x > 1}$"},{"id":"B","text":"${x | x > 3}$"},{"id":"C","text":"$ℝ$"},{"id":"D","text":"$∅$"}],"answer":"B","explanation":"Since $x > 3$ implies $x > 1$, we have $A ⊂ B$, so $A ∩ B = A = {x | x > 3}$.","explanationZh":"$由 x > 3$ 可推出 $x > 1，即 A ⊂$ B，所以 $A ∩ B = A = {x | x > 3}。$","tags":["easy","intersection","intervals"],"status":"published"},{"id":"math-set-relations-009","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":2,"stem":"Take $ℝ$ as the universal set and let $A = {x | x ≤ 1}$. Find ∁ᵤA.","stemZh":"$以 ℝ$ 为全集，A $= {x | x ≤ 1}，求$ ∁ᵤA。","options":[{"id":"A","text":"${x | x ≥ 1}$"},{"id":"B","text":"${x | x < 1}$"},{"id":"C","text":"${x | x > 1}$"},{"id":"D","text":"${x | x > 0}$"}],"answer":"C","explanation":"Removing the part $x ≤ 1$ from $ℝ$ leaves $x > 1$. The endpoint 1 belongs to A, so it must be excluded from the complement.","explanationZh":"$从 ℝ$ 中去掉 $x ≤ 1$ 的部分，剩下 $x > 1；端点 1$ 属于 A，必须从补集中排除。","tags":["easy","complement","intervals"],"status":"published"},{"id":"math-set-relations-010","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":1,"stem":"Let $A = {1, 2, 3}$ and $B = {3}$. Find A − B.","stemZh":"已知 $A = {1, 2, 3}，B = {3}，求 A −$ B。","options":[{"id":"A","text":"{1, 2}"},{"id":"B","text":"{3}"},{"id":"C","text":"{1, 2, 3}"},{"id":"D","text":"$∅$"}],"answer":"A","explanation":"The difference A − B keeps the elements of A that do not belong to B, namely 1 and 2.","explanationZh":"差集 A − B 保留属于 A 但不属于 B 的元素，即 1 和 2。","tags":["easy","difference"],"status":"published"},{"id":"math-set-relations-011","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"Let $A = {x | x^{2} − x − 2 = 0}$ and $B = {−1, 2}$. Which statement is true?","stemZh":"已知 $A = {x | x^{2} − x − 2 = 0}，B = {−1, 2}，下列关系正确的是？$","options":[{"id":"A","text":"$A ⊂ B$"},{"id":"B","text":"$A = B$"},{"id":"C","text":"$B ⊂ A$"},{"id":"D","text":"$A ∩ B = ∅$"}],"answer":"B","explanation":"$x^{2} − x − 2 = (x − 2)(x + 1) = 0$ gives $x = 2$ or $x = −1$, so $A = {−1, 2}$, which is exactly B.","explanationZh":"$由 x^{2} − x − 2 = (x − 2)(x + 1) = 0 得 x = 2 或 x = −1，故 A = {−1, 2} =$ B。","tags":["medium","equality","subset"],"status":"published"},{"id":"math-set-relations-012","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"A finite set has exactly 16 subsets. How many elements does it have?","stemZh":"某有限集合恰有 16 个子集，则该集合有几个元素？","options":[{"id":"A","text":"8"},{"id":"B","text":"5"},{"id":"C","text":"3"},{"id":"D","text":"4"}],"answer":"D","explanation":"A set with n elements has $2^{n}$ subsets. Solving $2^{n} = 16$ gives $n = 4. (16\\text{ is the number of subsets, not twice the number of }elements.)$","explanationZh":"含 n 个元素的集合有 $2^{n}$ 个子集，由 $2^{n} = 16 得 n = 4（16$ 是子集个数，不是元素个数的两倍）。","tags":["medium","subsets","counting"],"status":"published"},{"id":"math-set-relations-013","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"Let $U = {1, 2, 3, 4, 5}, A = {1, 3}$ and $B = {3, 5}$. Find $∁ᵤ(A ∪ B)$.","stemZh":"全集 $U = {1, 2, 3, 4, 5}，A = {1, 3}，B = {3, 5}，求 ∁ᵤ(A ∪ B)。$","options":[{"id":"A","text":"{2, 4}"},{"id":"B","text":"{2}"},{"id":"C","text":"{4}"},{"id":"D","text":"{1, 5}"}],"answer":"A","explanation":"$A ∪ B = {1, 3, 5}$, so the complement inside $U = {1, 2, 3, 4, 5}$ is {2, 4}.","explanationZh":"$A ∪ B = {1, 3, 5}，故其在 U = {1, 2, 3, 4, 5}$ 中的补集为 {2, 4}。","tags":["medium","complement","union"],"status":"published"},{"id":"math-set-relations-014","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"Let $A = {x | −1 < x < 3}$ and $B = {x | 0 ≤ x ≤ 5}$. Find $A ∩$ B.","stemZh":"已知 $A = {x | −1 < x < 3}，B = {x | 0 ≤ x ≤ 5}，求 A ∩$ B。","options":[{"id":"A","text":"$(0, 3)$"},{"id":"B","text":"$(−1, 5]$"},{"id":"C","text":"$[0, 3)$"},{"id":"D","text":"$[0, 5]$"}],"answer":"C","explanation":"Intersect $(−1, 3)$ with $[0, 5]$: the lower endpoint 0 is included because 0 lies in both sets, while 3 is excluded because A is open there.","explanationZh":"$(−1, 3) 与 [0, 5]$ 取交集：下端点 0 同时属于两个集合故取闭，上端点 3 在 A 中是开的故取开，即 $[0, 3)。$","tags":["medium","intersection","intervals"],"status":"published"},{"id":"math-set-relations-015","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"Which of the following statements is true?","stemZh":"下列命题中正确的一个是？","options":[{"id":"A","text":"$∅ ∈ {0}$"},{"id":"B","text":"$∅ ⊆ {0}$"},{"id":"C","text":"${0} ⊆ ∅$"},{"id":"D","text":"$0 ⊆ {0}$"}],"answer":"B","explanation":"The empty set is a subset of every set, so $∅ ⊆ {0}$. It is not an element of {0}: the only element of {0} is the number 0, and 0 is a number, not a set, so it cannot be a subset of anything.","explanationZh":"空集是任何集合的子集，故 $∅ ⊆ {0} 正确；{0}$ 的唯一元素是数字 0，空集不是它的元素；0 是数不是集合，不能作为子集。","tags":["medium","empty-set","element-relation"],"status":"published"},{"id":"math-set-relations-016","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"In a class of 40 students, 25 study Mathematics, 20 study Physics and 10 study both. How many students study neither?","stemZh":"某班 40 人，25 人学数学，20 人学物理，10 人两科都学，两科都不学的有多少人？","options":[{"id":"A","text":"10"},{"id":"B","text":"0"},{"id":"C","text":"15"},{"id":"D","text":"5"}],"answer":"D","explanation":"By the inclusion–exclusion principle $|M ∪ P| = 25 + 20 − 10 = 35$, so $40 − 35 = 5$ students study neither subject.","explanationZh":"由容斥原理 $|M ∪ P| = 25 + 20 − 10 = 35，故两科都不学的有 40 − 35 = 5$ 人。","tags":["medium","venn","counting"],"status":"published"},{"id":"math-set-relations-017","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"For any sets A and B inside a universal set $U, ∁ᵤ(A ∩ B) equals$:","stemZh":"对全集 U 中的任意集合 $A、B，∁ᵤ(A ∩ B)$ 等于？","options":[{"id":"A","text":"∁ᵤA $∪$ ∁ᵤB"},{"id":"B","text":"∁ᵤA $∩$ ∁ᵤB"},{"id":"C","text":"$A ∪ B$"},{"id":"D","text":"$A ∩ B$"}],"answer":"A","explanation":"De Morgan's law: the complement of an intersection is the union of the complements (and the complement of a union is the intersection of the complements).","explanationZh":"德·摩根定律：交集的补集等于补集的并集（并集的补集等于补集的交集）。","tags":["medium","complement","de-morgan"],"status":"published"},{"id":"math-set-relations-018","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"For sets A and B with universal set U, the difference A − B equals:","stemZh":"在全集 U 中，差集 A − B 等于？","options":[{"id":"A","text":"$A ∪$ ∁ᵤB"},{"id":"B","text":"∁ᵤA $∩ B$"},{"id":"C","text":"$A ∩$ ∁ᵤB"},{"id":"D","text":"A − ∁ᵤB"}],"answer":"C","explanation":"A − B keeps the elements of A that are not in B, which is exactly $A ∩$ ∁ᵤB. $(Note A −\\text{ ∁ᵤB }= A ∩\\text{ B, the opposite }operation.)$","explanationZh":"A − B 保留属于 A 但不属于 B 的元素，即 $A ∩$ ∁ᵤB（而 A − ∁ᵤB $= A ∩$ B，正好相反）。","tags":["medium","difference","complement"],"status":"published"},{"id":"math-set-relations-019","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"If $|A| = 3, |B| = 4$ and $|A ∩ B| = 1$, find $|A ∪ B|$.","stemZh":"已知 $|A| = 3，|B| = 4，|A ∩ B| = 1，求 |A ∪ B|。$","options":[{"id":"A","text":"7"},{"id":"B","text":"6"},{"id":"C","text":"5"},{"id":"D","text":"12"}],"answer":"B","explanation":"$|A ∪ B| = |A| + |B| − |A ∩ B| = 3 + 4 − 1 = 6$. Adding without subtracting would double-count the shared element.","explanationZh":"$|A ∪ B| = |A| + |B| − |A ∩ B| = 3 + 4 − 1 = 6；直接相加会把公共元素重复计算一次。$","tags":["medium","union","counting"],"status":"published"},{"id":"math-set-relations-020","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"If $A ∪ B = A$, which statement must be true?","stemZh":"$若 A ∪ B =$ A，则一定成立的是？","options":[{"id":"A","text":"$A ⊆ B$"},{"id":"B","text":"$B = ∅$"},{"id":"C","text":"$A ∩ B = ∅$"},{"id":"D","text":"$B ⊆ A$"}],"answer":"D","explanation":"$A ∪ B = A$ means adding B brings in nothing new, i.e. every element of B already lies in A, so $B ⊆$ A. $B = ∅$ is only one possible case, not a necessary consequence.","explanationZh":"$A ∪ B = A$ 说明并上 B 后没有新增元素，即 B 的元素都在 A 中，故 $B ⊆$ A；B $= ∅$ 只是其中一种可能，并非必然。","tags":["medium","union","subset"],"status":"published"},{"id":"math-set-relations-021","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"Let $M = {x | x^{2} − 5x + 6 = 0}$. How many non-empty subsets does M have?","stemZh":"已知 $M = {x | x^{2} − 5x + 6 = 0}，则 M$ 的非空子集个数为？","options":[{"id":"A","text":"4"},{"id":"B","text":"2"},{"id":"C","text":"3"},{"id":"D","text":"1"}],"answer":"C","explanation":"$x^{2} − 5x + 6 = (x − 2)(x − 3) = 0$, so $M = {2, 3}$. It has $2^{2} = 4 subsets$, and removing the empty set leaves 3 non-empty $ones: {2}, {3}, {2, 3}$.","explanationZh":"$由 (x − 2)(x − 3) = 0 得 M = {2, 3}，共有 2^{2} = 4$ 个子集，去掉空集后非空子集为 ${2}、{3}、{2, 3} 共 3$ 个。","tags":["hard","subsets","counting"],"status":"published"},{"id":"math-set-relations-022","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"Let $A = {x |\\text{ ax }+ 1 = 0}$. If $A ⊆ {1}$, then a equals:","stemZh":"已知 $A = {x |$ ax $+ 1 = 0}，若 A ⊆ {1}，则 a$ 等于？","options":[{"id":"A","text":"0 or −1"},{"id":"B","text":"−1"},{"id":"C","text":"0"},{"id":"D","text":"1"}],"answer":"A","explanation":"Two cases. If $a = 0$ the equation becomes $1 = 0$, so $A = ∅$ and $∅ ⊆ {1}$ holds. If $a ≠ 0$ then $A = {−1/a}$, and $−1/a = 1$ gives $a = −1$. Hence $a = 0$ or $a = −1$; forgetting the empty-set case loses $a = 0$.","explanationZh":"分两种情况：当 $a = 0$ 时方程化为 $1 = 0，A = ∅，而 ∅ ⊆ {1}$ 成立；当 $a ≠ 0 时 A = {−1/a}，由 −1/a = 1 得 a = −1。故 a = 0 或 a = −1；漏掉空集情形就会丢掉 a = 0。$","tags":["hard","empty-set","parameter"],"status":"published"},{"id":"math-set-relations-023","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"Let $A = {x | x < a}$ and $B = {x | x^{2} − 3x + 2 ≤ 0}$. If $A ∩ B = ∅, then$:","stemZh":"已知 $A = {x | x < a}，B = {x | x^{2} − 3x + 2 ≤ 0}，若 A ∩ B = ∅，则？$","options":[{"id":"A","text":"$a < 1$"},{"id":"B","text":"$a ≤ 1$"},{"id":"C","text":"$a ≥ 2$"},{"id":"D","text":"$a ≤ 2$"}],"answer":"B","explanation":"$B = [1, 2]$ and $A = (−∞, a)$. They are disjoint exactly when $a ≤ 1$: at $a = 1$ we get $A = (−∞, 1)$, which contains neither 1 nor 2, so the endpoint is included in the range of a.","explanationZh":"$B = [1, 2]，A = (−∞, a)。两者无公共元素当且仅当 a ≤ 1：当 a = 1 时 A = (−∞, 1)$ 不含 1 也不含 2，因此端点可以取到。","tags":["hard","intervals","parameter"],"status":"published"},{"id":"math-set-relations-024","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"Which of the following identities is correct for all sets A, B and C?","stemZh":"下列集合恒等式中对任意集合 A、B、C 都成立的是？","options":[{"id":"A","text":"$A ∪ (B ∩ C) = (A ∪ B) ∩ C$"},{"id":"B","text":"$A − (B − C) = (A − B) − C$"},{"id":"C","text":"$(A − B) ∪ B = A$"},{"id":"D","text":"$A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)$"}],"answer":"D","explanation":"Intersection distributes over union. The others fail: the correct distributive law is $A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C); A − (B − C) = (A − B) ∪ (A ∩ C) ≠ (A − B) − C$; and $(A − B) ∪ B = A ∪ B$, not A.","explanationZh":"交集对并集可分配。其余均错：正确形式是 $A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)；A − (B − C) = (A − B) ∪ (A ∩ C) ≠ (A − B) − C；(A − B) ∪ B = A ∪ B$ 而非 A。","tags":["hard","identities","de-morgan"],"status":"published"},{"id":"math-set-relations-025","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"How many sets X satisfy ${1} ⊆ X ⊆ {1, 2, 3, 4}$?","stemZh":"满足 ${1} ⊆ X ⊆ {1, 2, 3, 4}$ 的集合 X 有多少个？","options":[{"id":"A","text":"8"},{"id":"B","text":"4"},{"id":"C","text":"16"},{"id":"D","text":"7"}],"answer":"A","explanation":"X must contain 1, and each of 2, 3, 4 may be in or out independently, giving $2^{3} = 8$ possibilities.","explanationZh":"X 必须含有 1，而 2、3、4 各可选或不选，故共有 $2^{3} = 8$ 个。","tags":["hard","subsets","counting"],"status":"published"},{"id":"math-set-relations-026","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"Let $A = {x | x^{2} − 4 ≤ 0}$ and $B = {x | x^{2} − 2x − 3 > 0}$. Find $A ∩$ B.","stemZh":"已知 $A = {x | x^{2} − 4 ≤ 0}，B = {x | x^{2} − 2x − 3 > 0}，求 A ∩$ B。","options":[{"id":"A","text":"$[−2, −1]$"},{"id":"B","text":"$[−2, −1)$"},{"id":"C","text":"$(−2, −1)$"},{"id":"D","text":"$∅$"}],"answer":"B","explanation":"$A = [−2, 2]$ and $B = (−∞, −1) ∪ (3, +∞)$. Their intersection is $[−2, −1): −2$ is included because it satisfies both $x^{2} ≤ 4$ and $x < −1$, while −1 is excluded because $x^{2} − 2x − 3 = 0$ there and the inequality is strict.","explanationZh":"$A = [−2, 2]，B = (−∞, −1) ∪ (3, +∞)，交集为 [−2, −1)：−2$ 同时满足 $x^{2} ≤ 4 与 x < −1 故取闭；−1 处 x^{2} − 2x − 3 = 0，因不等式是严格的故取开。$","tags":["hard","intersection","intervals"],"status":"published"},{"id":"math-set-relations-027","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":5,"stem":"Let $U = {1, 2, 3, 4, 5, 6}$. Given $A ∩$ ∁ᵤB $= {1, 3}$, ∁ᵤA $∩ B = {5}$ and $A ∩ B = {2}$, find $∁ᵤ(A ∪ B)$.","stemZh":"全集 $U = {1, 2, 3, 4, 5, 6}，已知 A ∩$ ∁ᵤB $= {1, 3}，∁ᵤA ∩ B = {5}，A ∩ B = {2}，求 ∁ᵤ(A ∪ B)。$","options":[{"id":"A","text":"{4}"},{"id":"B","text":"{6}"},{"id":"C","text":"{4, 6}"},{"id":"D","text":"{1, 4, 6}"}],"answer":"C","explanation":"The three regions give $A ∪ B = {1, 3} ∪ {2} ∪ {5} = {1, 2, 3, 5}$, so the complement in $U = {1, 2, 3, 4, 5, 6}$ is {4, 6}.","explanationZh":"三部分相加得 $A ∪ B = {1, 3} ∪ {2} ∪ {5} = {1, 2, 3, 5}，故在 U$ 中的补集为 {4, 6}。","tags":["hard","venn","complement"],"status":"published"},{"id":"math-set-relations-028","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"Let $A = {x | x^{2} − 3x − 4 < 0}$ and $B = {x | x^{2} − 2x − 8 ≤ 0}$. Which statement is true?","stemZh":"已知 $A = {x | x^{2} − 3x − 4 < 0}，B = {x | x^{2} − 2x − 8 ≤ 0}，下列关系正确的是？$","options":[{"id":"A","text":"$B ⊆ A$"},{"id":"B","text":"$A ∩ B = ∅$"},{"id":"C","text":"$A ∪ B = ℝ$"},{"id":"D","text":"$A ⊆ B$"}],"answer":"D","explanation":"$A = (−1, 4) (from (x − 4)(x + 1) < 0)$ and $B = [−2, 4] (from (x − 4)(x + 2) ≤ 0)$. Since $(−1, 4)$ lies inside $[−2, 4]$, we get $A ⊂ B$, hence $A ⊆$ B.","explanationZh":"$由 (x − 4)(x + 1) < 0 得 A = (−1, 4)；由 (x − 4)(x + 2) ≤ 0 得 B = [−2, 4]。(−1, 4)$ 包含于 $[−2, 4]，故 A ⊂$ B，即 $A ⊆$ B。","tags":["hard","subset","intervals"],"status":"published"},{"id":"math-set-relations-029","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"Let $A = {x | x = 2k + 1, k ∈ ℤ}$ and $B = {x | x = 4k + 1, k ∈ ℤ}$. Which statement is true?","stemZh":"已知 $A = {x | x = 2k + 1, k ∈ ℤ}，B = {x | x = 4k + 1, k ∈ ℤ}，下列关系正确的是？$","options":[{"id":"A","text":"$A ⊂ B$"},{"id":"B","text":"$B ⊂ A$"},{"id":"C","text":"$A = B$"},{"id":"D","text":"$A ∩ B = ∅$"}],"answer":"B","explanation":"A is the set of odd integers. Every number $4k + 1 = 2(2k) + 1$ is odd, so $B ⊆$ A. But $3 ∈ A$ is not of the form 4k + 1, so the inclusion is $proper: B ⊂$ A.","explanationZh":"$A 是全体奇数。4k + 1 = 2(2k) + 1$ 恒为奇数，故 $B ⊆$ A；但 $3 ∈ A 且 3$ 不能写成 $4k + 1，故为真包含：B ⊂$ A。","tags":["hard","subset","number-sets"],"status":"published"},{"id":"math-set-relations-030","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":5,"stem":"The solution set $A = {x | x^{2} +\\text{ px }+ q = 0}$ consists of exactly one $element, A = {1}$. Find p and q.","stemZh":"$若 A = {x | x^{2} +$ px $+ q = 0}$ 恰有一个元素且 $A = {1}，求$ p、q。","options":[{"id":"A","text":"$p = −2, q = 1$"},{"id":"B","text":"$p = 2, q = 1$"},{"id":"C","text":"$p = −2, q = −1$"},{"id":"D","text":"$p = 1, q = −2$"}],"answer":"A","explanation":"A single-element solution set means the quadratic has a double root at $x = 1$, so $x^{2} +$ px $+ q = (x − 1)^{2} = x^{2} − 2x + 1$. Hence $p = −2$ and $q = 1$.","explanationZh":"解集只有一个元素说明方程有二重根 $x = 1，即 x^{2} +$ px $+ q = (x − 1)^{2} = x^{2} − 2x + 1，故 p = −2，q = 1。$","tags":["hard","parameter","quadratic"],"status":"published"},{"id":"math-quadratic-inequality-001","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":1,"stem":"Solve $x^{2} − x − 6 ≥ 0$.","stemZh":"解不等式 $x^{2} − x − 6 ≥ 0。$","options":[{"id":"A","text":"$(−∞, −2) ∪ (3, +∞)$"},{"id":"B","text":"$[−2, 3]$"},{"id":"C","text":"$(−2, 3)$"},{"id":"D","text":"$(−∞, −2] ∪ [3, +∞)$"}],"answer":"D","explanation":"$x^{2} − x − 6 = (x − 3)(x + 2) ≥ 0$. The parabola opens upward, so the expression is non-negative outside the $roots: x ≤ −2$ or $x ≥ 3$. The roots themselves are included because the inequality is not strict.","explanationZh":"$x^{2} − x − 6 = (x − 3)(x + 2) ≥ 0，开口向上，两根之外非负，即 x ≤ −2 或 x ≥ 3；由于不等式带等号，两根要取到。$","tags":["easy","quadratic","solution-set"],"status":"published"},{"id":"math-quadratic-inequality-002","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":1,"stem":"Solve $x^{2} − 4 < 0$.","stemZh":"解不等式 $x^{2} − 4 < 0。$","options":[{"id":"A","text":"$x < 2$"},{"id":"B","text":"$−2 < x < 2$"},{"id":"C","text":"$x < −2$ or $x > 2$"},{"id":"D","text":"$x ≠ 2$"}],"answer":"B","explanation":"$(x − 2)(x + 2) < 0$ means the two factors have opposite signs, which happens strictly between the $roots: −2 < x < 2$.","explanationZh":"$(x − 2)(x + 2) < 0 表示两因式异号，解集在两根之间：−2 < x < 2。$","tags":["easy","quadratic","factorisation"],"status":"published"},{"id":"math-quadratic-inequality-003","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":1,"stem":"Solve $x^{2} − 5x + 6 ≤ 0$.","stemZh":"解不等式 $x^{2} − 5x + 6 ≤ 0。$","options":[{"id":"A","text":"$[2, 3]$"},{"id":"B","text":"$(2, 3)$"},{"id":"C","text":"$(−∞, 2] ∪ [3, +∞)$"},{"id":"D","text":"$(−∞, 2) ∪ (3, +∞)$"}],"answer":"A","explanation":"$x^{2} − 5x + 6 = (x − 2)(x − 3) ≤ 0$, so x lies between the roots $inclusive: 2 ≤ x ≤ 3$.","explanationZh":"$x^{2} − 5x + 6 = (x − 2)(x − 3) ≤ 0，解集为两根之间（含端点）：2 ≤ x ≤ 3。$","tags":["easy","quadratic","solution-set"],"status":"published"},{"id":"math-quadratic-inequality-004","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":2,"stem":"Solve $(x − 1)(x + 3) > 0$.","stemZh":"解不等式 $(x − 1)(x + 3) > 0。$","options":[{"id":"A","text":"$(−3, 1)$"},{"id":"B","text":"$[−3, 1]$"},{"id":"C","text":"$(−∞, −3) ∪ (1, +∞)$"},{"id":"D","text":"$(−∞, 1) ∪ (3, +∞)$"}],"answer":"C","explanation":"The roots are −3 and 1 and the product is positive outside $them: x < −3$ or $x > 1$. Both endpoints are excluded because the inequality is strict.","explanationZh":"两根为 $−3 与 1，乘积在两根之外为正：x < −3 或 x > 1；不等式严格，端点不取。$","tags":["easy","quadratic","sign-chart"],"status":"published"},{"id":"math-quadratic-inequality-005","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":2,"stem":"Solve $x^{2} + 4x + 4 ≤ 0$.","stemZh":"解不等式 $x^{2} + 4x + 4 ≤ 0。$","options":[{"id":"A","text":"$ℝ$"},{"id":"B","text":"$[−2, 2]$"},{"id":"C","text":"$∅$"},{"id":"D","text":"{−2}"}],"answer":"D","explanation":"$x^{2} + 4x + 4 = (x + 2)^{2}$, which is never negative. It equals 0 only at $x = −2$, so the solution set is the single-element set {−2}.","explanationZh":"$x^{2} + 4x + 4 = (x + 2)^{2} ≥ 0$ 恒成立，仅在 $x = −2$ 时等于 $0，故解集为单元素集 {−2}。$","tags":["easy","quadratic","perfect-square"],"status":"published"},{"id":"math-quadratic-inequality-006","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":1,"stem":"Solve $x^{2} + 5x + 6 < 0$.","stemZh":"解不等式 $x^{2} + 5x + 6 < 0。$","options":[{"id":"A","text":"$(−3, −2)$"},{"id":"B","text":"$(2, 3)$"},{"id":"C","text":"$(−∞, −3) ∪ (−2, +∞)$"},{"id":"D","text":"$[−3, −2]$"}],"answer":"A","explanation":"$x^{2} + 5x + 6 = (x + 2)(x + 3) < 0$ has roots −3 and −2, and the solution lies strictly between $them: −3 < x < −2$.","explanationZh":"$x^{2} + 5x + 6 = (x + 2)(x + 3) < 0，两根为 −3 与 −2，解集在两根之间：−3 < x < −2。$","tags":["easy","quadratic","factorisation"],"status":"published"},{"id":"math-quadratic-inequality-007","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":2,"stem":"Solve $x^{2} + 2x + 1 > 0$.","stemZh":"解不等式 $x^{2} + 2x + 1 > 0。$","options":[{"id":"A","text":"$ℝ$"},{"id":"B","text":"$(−∞, −1)$"},{"id":"C","text":"$(−∞, −1) ∪ (−1, +∞)$"},{"id":"D","text":"$∅$"}],"answer":"C","explanation":"$x^{2} + 2x + 1 = (x + 1)^{2}$ is positive for every x except $x = −1$, where it is 0. So the solution set is $x ≠ −1$, i.e. $(−∞, −1) ∪ (−1, +∞)$.","explanationZh":"$x^{2} + 2x + 1 = (x + 1)^{2} 除 x = −1$ 时等于 0 外恒为正，故解集为 $x ≠ −1，即 (−∞, −1) ∪ (−1, +∞)。$","tags":["easy","quadratic","perfect-square"],"status":"published"},{"id":"math-quadratic-inequality-008","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":1,"stem":"Solve $x(x − 1) < 0$.","stemZh":"解不等式 $x(x − 1) < 0。$","options":[{"id":"A","text":"$x < 0$"},{"id":"B","text":"$0 < x < 1$"},{"id":"C","text":"$x > 1$"},{"id":"D","text":"$x < 0$ or $x > 1$"}],"answer":"B","explanation":"The roots are 0 and 1 and the product is negative only between $them: 0 < x < 1$.","explanationZh":"两根为 $0 与 1，乘积仅在两根之间为负：0 < x < 1。$","tags":["easy","quadratic","sign-chart"],"status":"published"},{"id":"math-quadratic-inequality-009","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":2,"stem":"Solve $x^{2} + 3x ≥ 0$.","stemZh":"解不等式 $x^{2} + 3x ≥ 0。$","options":[{"id":"A","text":"$[−3, 0]$"},{"id":"B","text":"$(−∞, −3) ∪ (0, +∞)$"},{"id":"C","text":"$[0, +∞)$"},{"id":"D","text":"$(−∞, −3] ∪ [0, +∞)$"}],"answer":"D","explanation":"$x(x + 3) ≥ 0$ is non-negative outside the roots −3 and 0, and both roots satisfy the equality, so $x ≤ −3$ or $x ≥ 0$.","explanationZh":"$x(x + 3) ≥ 0$ 在两根 −3 与 0 之外非负，且两交点处等号成立，故 $x ≤ −3 或 x ≥ 0。$","tags":["easy","quadratic","solution-set"],"status":"published"},{"id":"math-quadratic-inequality-010","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":2,"stem":"Solve $x^{2} + x + 1 > 0$.","stemZh":"解不等式 $x^{2} + x + 1 > 0。$","options":[{"id":"A","text":"$ℝ$"},{"id":"B","text":"$∅$"},{"id":"C","text":"$(−∞, −1) ∪ (1, +∞)$"},{"id":"D","text":"$(0, +∞)$"}],"answer":"A","explanation":"The discriminant is $Δ = 1 − 4 = −3 < 0$ and the leading coefficient is positive, so the quadratic is positive for every real x.","explanationZh":"判别式 $Δ = 1 − 4 = −3 < 0，且二次项系数为正，故对一切实数 x$ 该式恒大于 0。","tags":["easy","quadratic","discriminant"],"status":"published"},{"id":"math-quadratic-inequality-011","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $2x^{2} − 5x − 3 ≤ 0$.","stemZh":"解不等式 $2x^{2} − 5x − 3 ≤ 0。$","options":[{"id":"A","text":"$(−∞, −1/2] ∪ [3, +∞)$"},{"id":"B","text":"$[−1/2, 3]$"},{"id":"C","text":"$[−3, 1/2]$"},{"id":"D","text":"$(−1/2, 3)$"}],"answer":"B","explanation":"$2x^{2} − 5x − 3 = (2x + 1)(x − 3)$, with roots −1/2 and 3. Since the leading coefficient is positive, the expression is $≤ 0$ between the $roots: [−1/2, 3]$.","explanationZh":"$2x^{2} − 5x − 3 = (2x + 1)(x − 3)，两根为 −1/2 与 3；二次项系数为正，故两根之间 ≤ 0，解集为 [−1/2, 3]。$","tags":["medium","quadratic","factorisation"],"status":"published"},{"id":"math-quadratic-inequality-012","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $−x^{2} + 2x + 3 ≥ 0$.","stemZh":"解不等式 $−x^{2} + 2x + 3 ≥ 0。$","options":[{"id":"A","text":"$[−1, 3]$"},{"id":"B","text":"$(−∞, −1] ∪ [3, +∞)$"},{"id":"C","text":"$(−1, 3)$"},{"id":"D","text":"$[−3, 1]$"}],"answer":"A","explanation":"Multiplying by −1 reverses the $sign: x^{2} − 2x − 3 ≤ 0$, i.e. $(x − 3)(x + 1) ≤ 0$, giving $−1 ≤ x ≤ 3$.","explanationZh":"两边同乘 $−1 不等号反向：x^{2} − 2x − 3 ≤ 0，即 (x − 3)(x + 1) ≤ 0，得 −1 ≤ x ≤ 3。$","tags":["medium","quadratic","sign-flip"],"status":"published"},{"id":"math-quadratic-inequality-013","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $x^{2} − 4x + 3 < 0$.","stemZh":"解不等式 $x^{2} − 4x + 3 < 0。$","options":[{"id":"A","text":"$(−∞, 1) ∪ (3, +∞)$"},{"id":"B","text":"$[1, 3]$"},{"id":"C","text":"$(−3, −1)$"},{"id":"D","text":"$(1, 3)$"}],"answer":"D","explanation":"$(x − 1)(x − 3) < 0$ gives $1 < x < 3$; both endpoints are excluded.","explanationZh":"$(x − 1)(x − 3) < 0，解得 1 < x < 3，端点不取。$","tags":["medium","quadratic","sign-chart"],"status":"published"},{"id":"math-quadratic-inequality-014","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(2 − x)(x + 1) ≥ 0$.","stemZh":"解不等式 $(2 − x)(x + 1) ≥ 0。$","options":[{"id":"A","text":"$(−∞, −1] ∪ [2, +∞)$"},{"id":"B","text":"$(−∞, −1) ∪ (2, +∞)$"},{"id":"C","text":"$[−1, 2]$"},{"id":"D","text":"$(−1, 2)$"}],"answer":"C","explanation":"Rewrite as $−(x − 2)(x + 1) ≥ 0$, i.e. $(x − 2)(x + 1) ≤ 0$, which holds between the $roots: −1 ≤ x ≤ 2$. The sign of the leading coefficient reverses the usual 'outside the roots' answer.","explanationZh":"化为 $−(x − 2)(x + 1) ≥ 0，即 (x − 2)(x + 1) ≤ 0，解在两根之间：−1 ≤ x ≤ 2。注意首项系数为负时结论不是“两根之外”。$","tags":["medium","quadratic","sign-flip"],"status":"published"},{"id":"math-quadratic-inequality-015","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $x^{2} − 6x + 9 > 0$.","stemZh":"解不等式 $x^{2} − 6x + 9 > 0。$","options":[{"id":"A","text":"$x ≠ 3$"},{"id":"B","text":"$(−∞, 3) ∪ (3, +∞)$"},{"id":"C","text":"$ℝ$"},{"id":"D","text":"$∅$"}],"answer":"B","explanation":"$x^{2} − 6x + 9 = (x − 3)^{2} > 0$ holds for every x except $x = 3$, so the solution set is $(−∞, 3) ∪ (3, +∞)$.","explanationZh":"$x^{2} − 6x + 9 = (x − 3)^{2} > 0$ 对除 $x = 3$ 外的一切实数成立，故解集为 $(−∞, 3) ∪ (3, +∞)。$","tags":["medium","quadratic","perfect-square"],"status":"published"},{"id":"math-quadratic-inequality-016","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $3x^{2} − 12 ≤ 0$.","stemZh":"解不等式 $3x^{2} − 12 ≤ 0。$","options":[{"id":"A","text":"$(−∞, −2] ∪ [2, +∞)$"},{"id":"B","text":"$(−2, 2)$"},{"id":"C","text":"$(−∞, −4] ∪ [4, +∞)$"},{"id":"D","text":"$[−2, 2]$"}],"answer":"D","explanation":"$3x^{2} ≤ 12$ gives $x^{2} ≤ 4$, so $|x| ≤ 2$, i.e. $−2 ≤ x ≤ 2$.","explanationZh":"$3x^{2} ≤ 12 即 x^{2} ≤ 4，故 |x| ≤ 2，解得 −2 ≤ x ≤ 2。$","tags":["medium","quadratic","absolute-value"],"status":"published"},{"id":"math-quadratic-inequality-017","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $x^{2} + 2x − 8 > 0$.","stemZh":"解不等式 $x^{2} + 2x − 8 > 0。$","options":[{"id":"A","text":"$(−∞, −4) ∪ (2, +∞)$"},{"id":"B","text":"$(−4, 2)$"},{"id":"C","text":"$(−∞, −2) ∪ (4, +∞)$"},{"id":"D","text":"$(2, +∞)$"}],"answer":"A","explanation":"$x^{2} + 2x − 8 = (x + 4)(x − 2) > 0$, which is positive outside the roots −4 and $2: x < −4$ or $x > 2$.","explanationZh":"$x^{2} + 2x − 8 = (x + 4)(x − 2) > 0，在两根 −4 与 2$ 之外为正：x $< −4 或 x > 2。$","tags":["medium","quadratic","factorisation"],"status":"published"},{"id":"math-quadratic-inequality-018","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $x^{2} − x − 12 ≤ 0$.","stemZh":"解不等式 $x^{2} − x − 12 ≤ 0。$","options":[{"id":"A","text":"$(−∞, −3] ∪ [4, +∞)$"},{"id":"B","text":"$(−3, 4)$"},{"id":"C","text":"$[−3, 4]$"},{"id":"D","text":"$[−4, 3]$"}],"answer":"C","explanation":"$(x − 4)(x + 3) ≤ 0$ gives $−3 ≤ x ≤ 4$, with both roots included.","explanationZh":"$(x − 4)(x + 3) ≤ 0，解得 −3 ≤ x ≤ 4，两根均取到。$","tags":["medium","quadratic","solution-set"],"status":"published"},{"id":"math-quadratic-inequality-019","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $4x^{2} − 4x + 1 ≥ 0$.","stemZh":"解不等式 $4x^{2} − 4x + 1 ≥ 0。$","options":[{"id":"A","text":"$ℝ$"},{"id":"B","text":"$[1/2, +∞)$"},{"id":"C","text":"$(−∞, 1/2]$"},{"id":"D","text":"{1/2}"}],"answer":"A","explanation":"$4x^{2} − 4x + 1 = (2x − 1)^{2} ≥ 0$ for every real x, so the solution set is all of $ℝ$. The value $x = 1/2$ makes it zero, which still satisfies $≥ 0$.","explanationZh":"$4x^{2} − 4x + 1 = (2x − 1)^{2} ≥ 0$ 对一切实数成立，故解集为 $ℝ；x = 1/2$ 时等于 $0，也满足 ≥ 0。$","tags":["medium","quadratic","perfect-square"],"status":"published"},{"id":"math-quadratic-inequality-020","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $−2x^{2} + 3x + 2 < 0$.","stemZh":"解不等式 $−2x^{2} + 3x + 2 < 0。$","options":[{"id":"A","text":"$(−1/2, 2)$"},{"id":"B","text":"$[−1/2, 2]$"},{"id":"C","text":"$(−∞, −2) ∪ (1/2, +∞)$"},{"id":"D","text":"$(−∞, −1/2) ∪ (2, +∞)$"}],"answer":"D","explanation":"Multiply by $−1: 2x^{2} − 3x − 2 > 0$, i.e. $(2x + 1)(x − 2) > 0$, which holds outside the $roots: x < −1/2$ or $x > 2$.","explanationZh":"同乘 $−1 得 2x^{2} − 3x − 2 > 0，即 (2x + 1)(x − 2) > 0，解在两根之外：x < −1/2 或 x > 2。$","tags":["medium","quadratic","sign-flip"],"status":"published"},{"id":"math-quadratic-inequality-021","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":4,"stem":"For which values of a does $x^{2} − 2x + a > 0$ hold for every real x?","stemZh":"$当 a 取何值时，x^{2} − 2x + a > 0$ 对一切实数 x 恒成立？","options":[{"id":"A","text":"$a ≥ 1$"},{"id":"B","text":"$a > 0$"},{"id":"C","text":"$a > 1$"},{"id":"D","text":"$a < 1$"}],"answer":"C","explanation":"With leading coefficient $1 > 0$ the quadratic is always positive exactly when $Δ < 0: (−2)^{2} − 4a < 0$, so $4 − 4a < 0$ and $a > 1$. At $a = 1$ the expression is $(x − 1)^{2}$, which vanishes at $x = 1$, so the endpoint is excluded.","explanationZh":"二次项系数 $1 > 0，恒正的条件是 Δ < 0：(−2)^{2} − 4a < 0，即 4 − 4a < 0，得 a > 1。a = 1$ 时原式为 $(x − 1)^{2}，在 x = 1$ 处等于 0，故端点不取。","tags":["hard","quadratic","parameter"],"status":"published"},{"id":"math-quadratic-inequality-022","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":4,"stem":"If $x^{2} +$ bx $+ 1 > 0$ for every real x, find the range of b.","stemZh":"$若 x^{2} +$ bx $+ 1 > 0$ 对一切实数 x 恒成立，求 b 的取值范围。","options":[{"id":"A","text":"$b ≤ 2$"},{"id":"B","text":"$−2 0$"}],"answer":"B","explanation":"Always positive requires $Δ < 0: b^{2} − 4 < 0$, i.e. $b^{2} < 4$, giving $−2 < b < 2$. Equality $b = \\pm2$ would give a double root where the expression is 0.","explanationZh":"恒成立需 $Δ < 0：b^{2} − 4 < 0，即 b^{2} < 4，得 −2 < b < 2；b = \\pm2$ 时有二重根，原式可取到 0，不满足严格大于 0。","tags":["hard","quadratic","parameter"],"status":"published"},{"id":"math-quadratic-inequality-023","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":4,"stem":"How many integers x satisfy $x^{2} − 3x − 4 < 0$?","stemZh":"满足 $x^{2} − 3x − 4 < 0$ 的整数 x 有多少个？","options":[{"id":"A","text":"5"},{"id":"B","text":"6"},{"id":"C","text":"3"},{"id":"D","text":"4"}],"answer":"D","explanation":"$(x − 4)(x + 1) < 0$ gives $−1 < x < 4$. The integers strictly between −1 and 4 are 0, 1, 2, 3 — four of them.","explanationZh":"$(x − 4)(x + 1) < 0 得 −1 < x < 4，其间整数为 0、1、2、3，共 4$ 个。","tags":["hard","quadratic","integer-solutions"],"status":"published"},{"id":"math-quadratic-inequality-024","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":4,"stem":"Let $a < 0$ and suppose $ax^{2} +$ bx $+ c = 0$ has two roots $r_{1} < r_{2}$. The solution set of $ax^{2} +$ bx $+ c > 0 is$:","stemZh":"$设 a < 0，且 ax^{2} +$ bx $+ c = 0$ 有两根 $r_{1} < r_{2}，则 ax^{2} +$ bx $+ c > 0$ 的解集为？","options":[{"id":"A","text":"$(r_{1}, r_{2})$"},{"id":"B","text":"$(−∞, r_{1}) ∪ (r_{2}, +∞)$"},{"id":"C","text":"$(−∞, r_{1})$"},{"id":"D","text":"$∅$"}],"answer":"A","explanation":"With $a < 0$ the parabola opens downward, so the expression is positive between the two roots and negative outside them — the opposite of the $a > 0$ case.","explanationZh":"$a < 0$ 时抛物线开口向下，故两根之间为正、两根之外为负，与 $a > 0$ 的情形正好相反。","tags":["hard","quadratic","sign-chart"],"status":"published"},{"id":"math-quadratic-inequality-025","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":4,"stem":"If the solution set of $(x − 1)(x − a) < 0$ is $(1, 3)$, find a.","stemZh":"$若 (x − 1)(x − a) < 0$ 的解集为 $(1, 3)，求$ a。","options":[{"id":"A","text":"−3"},{"id":"B","text":"1"},{"id":"C","text":"3"},{"id":"D","text":"−1"}],"answer":"C","explanation":"The solution of $(x − 1)(x − a) < 0$ is the open interval between 1 and a. Matching $(1, 3)$ gives $a = 3. (If a = 1\\text{ the set would be }empty.)$","explanationZh":"$(x − 1)(x − a) < 0$ 的解集是 1 与 a 之间的开区间，与 $(1, 3)$ 对比得 $a = 3（若 a = 1$ 则解集为空）。","tags":["hard","quadratic","parameter"],"status":"published"},{"id":"math-quadratic-inequality-026","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":5,"stem":"Find the set of x satisfying both $x^{2} − 2x − 3 > 0$ and $x^{2} − 9 < 0$.","stemZh":"求同时满足 $x^{2} − 2x − 3 > 0 与 x^{2} − 9 < 0 的 x$ 的集合。","options":[{"id":"A","text":"$(−3, 3)$"},{"id":"B","text":"$(−3, −1)$"},{"id":"C","text":"$(−3, −1]$"},{"id":"D","text":"$(−1, 3)$"}],"answer":"B","explanation":"$x^{2} − 2x − 3 > 0$ gives $x < −1$ or $x > 3; x^{2} − 9 < 0$ gives $−3 < x < 3. Intersecting: (−3, −1)$. Both endpoints are open since −3 and −1 are excluded by the strict inequalities.","explanationZh":"$由 x^{2} − 2x − 3 > 0 得 x < −1 或 x > 3；由 x^{2} − 9 < 0 得 −3 < x < 3。取交集为 (−3, −1)，两端点因不等式严格均为开。$","tags":["hard","quadratic","system"],"status":"published"},{"id":"math-quadratic-inequality-027","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":5,"stem":"For which values of m does $x^{2} −$ mx $+ 4 ≥ 0$ hold for every real x?","stemZh":"$当 m 取何值时，x^{2} −$ mx $+ 4 ≥ 0$ 对一切实数 x 恒成立？","options":[{"id":"A","text":"$−4 < m < 4$"},{"id":"B","text":"$m ≤ 4$"},{"id":"C","text":"$m ≥ −4$"},{"id":"D","text":"$−4 ≤ m ≤ 4$"}],"answer":"D","explanation":"Because the inequality is non-strict, we only need the quadratic to never go below $zero: Δ ≤ 0$. Here $Δ = m^{2} − 16 ≤ 0$, so $−4 ≤ m ≤ 4. (Strict\\text{ positivity would require strict }inequalities.)$","explanationZh":"由于不等式不严格，只需 $Δ ≤ 0：Δ = m^{2} − 16 ≤ 0，得 −4 ≤ m ≤ 4（若为严格大于 0，则端点不取）。$","tags":["hard","quadratic","parameter"],"status":"published"},{"id":"math-quadratic-inequality-028","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":5,"stem":"The solution set of $x^{2} +$ ax $+ 4 < 0$ is $(1, 4)$. Find a.","stemZh":"$若 x^{2} +$ ax $+ 4 < 0$ 的解集为 $(1, 4)，求$ a。","options":[{"id":"A","text":"−5"},{"id":"B","text":"−4"},{"id":"C","text":"5"},{"id":"D","text":"4"}],"answer":"A","explanation":"The endpoints 1 and 4 are the roots, and their product is 4, matching the constant term. Their sum is $5 = −a$, so $a = −5. Check: x^{2} − 5x + 4 = (x − 1)(x − 4) < 0$ gives $1 < x < 4$.","explanationZh":"端点 1 与 4 即两根，其积为 4 与常数项一致；其和为 $5 =$ −a，故 $a = −5。验证：x^{2} − 5x + 4 = (x − 1)(x − 4) < 0 得 1 < x < 4。$","tags":["hard","quadratic","parameter"],"status":"published"},{"id":"math-quadratic-inequality-029","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":5,"stem":"If x satisfies $x^{2} − 3x + 2 < 0$, what can be said about the value of $x^{2} − 5x + 6$?","stemZh":"若 x 满足 $x^{2} − 3x + 2 < 0，则 x^{2} − 5x + 6$ 的值？","options":[{"id":"A","text":"It is negative"},{"id":"B","text":"It is zero"},{"id":"C","text":"It is positive"},{"id":"D","text":"Its sign cannot be determined"}],"answer":"C","explanation":"$x^{2} − 3x + 2 < 0$ gives $1 < x < 2$. Then $x^{2} − 5x + 6 = (x − 2)(x − 3)$, where both factors are negative on $(1, 2)$, so the product is positive.","explanationZh":"$由 x^{2} − 3x + 2 < 0 得 1 < x < 2；此时 x^{2} − 5x + 6 = (x − 2)(x − 3)$ 的两个因式均为负，故乘积为正。","tags":["hard","quadratic","reasoning"],"status":"published"},{"id":"math-quadratic-inequality-030","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":5,"stem":"The solution set of $ax^{2} +$ bx $+ c < 0$ is $(−∞, 1) ∪ (3, +∞)$. What is the sign of $ax^{2} +$ bx + c at $x = 0$?","stemZh":"$若 ax^{2} +$ bx $+ c < 0$ 的解集为 $(−∞, 1) ∪ (3, +∞)，则该式在 x = 0$ 处的符号为？","options":[{"id":"A","text":"Positive"},{"id":"B","text":"Negative"},{"id":"C","text":"Zero"},{"id":"D","text":"It depends on the value of a"}],"answer":"B","explanation":"Being negative outside the roots 1 and 3 means $a < 0$ and $ax^{2} +$ bx $+ c = a(x − 1)(x − 3)$. At $x = 0$ this is $a(−1)(−3) = 3a < 0$, so the value is negative.","explanationZh":"在两根 1、3 之外为负，说明 $a < 0 且 ax^{2} +$ bx $+ c = a(x − 1)(x − 3)；x = 0$ 时其值为 $a(−1)(−3) = 3a < 0，故为负。$","tags":["hard","quadratic","parameter"],"status":"published"},{"id":"math-rational-inequality-001","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":1,"stem":"Solve $(x − 1)/(x + 2) > 0$.","stemZh":"解不等式 $(x − 1)/(x + 2) > 0。$","options":[{"id":"A","text":"$(−2, 1)$"},{"id":"B","text":"$(−∞, 1) ∪ (2, +∞)$"},{"id":"C","text":"$(−∞, −2] ∪ [1, +∞)$"},{"id":"D","text":"$(−∞, −2) ∪ (1, +∞)$"}],"answer":"D","explanation":"The critical points are $−2 (denominator zero$, always excluded) and $1 (numerator zero). A$ quotient is positive when numerator and denominator share a sign, which happens outside the two $points: x < −2$ or $x > 1$.","explanationZh":"临界点为 $−2（分母为零，必排除）与 1（分子为零）。分式为正即分子分母同号，解在两点之外：x < −2 或 x > 1。$","tags":["easy","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-002","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":1,"stem":"Solve $1/x > 0$.","stemZh":"解不等式 $1/x > 0。$","options":[{"id":"A","text":"$(−∞, 0)$"},{"id":"B","text":"$(0, +∞)$"},{"id":"C","text":"$(−∞, 0) ∪ (0, +∞)$"},{"id":"D","text":"$ℝ$"}],"answer":"B","explanation":"The numerator 1 is positive, so the quotient is positive exactly when $x > 0$. The value $x = 0$ is not allowed.","explanationZh":"分子 1 为正，故分式为正当且仅当 $x > 0；x = 0$ 无意义。","tags":["easy","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-003","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":2,"stem":"Solve $(x + 1)/(x − 3) ≤ 0$.","stemZh":"解不等式 $(x + 1)/(x − 3) ≤ 0。$","options":[{"id":"A","text":"$[−1, 3)$"},{"id":"B","text":"$[−1, 3]$"},{"id":"C","text":"$(−1, 3]$"},{"id":"D","text":"$(−∞, −1] ∪ [3, +∞)$"}],"answer":"A","explanation":"The quotient is non-positive between the critical points −1 and 3. The endpoint −1 is included (numerator zero) but 3 is not, since the denominator vanishes there.","explanationZh":"分式在临界点 −1 与 3 之间非正；−1 处分子为零可取，3 处分母为零必须排除，故为 $[−1, 3)。$","tags":["easy","rational","endpoints"],"status":"published"},{"id":"math-rational-inequality-004","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":1,"stem":"Solve $(x − 2)/(x + 1) < 0$.","stemZh":"解不等式 $(x − 2)/(x + 1) < 0。$","options":[{"id":"A","text":"$(−∞, −1) ∪ (2, +∞)$"},{"id":"B","text":"$(−1, 2]$"},{"id":"C","text":"$(−1, 2)$"},{"id":"D","text":"$[−1, 2)$"}],"answer":"C","explanation":"Numerator and denominator must have opposite signs, which happens strictly between −1 and 2; both endpoints are excluded (the denominator zero at −1, and the inequality is strict at $2)$.","explanationZh":"分子分母异号时位于 −1 与 2 之间；−1 处分母为零、2 处不等式严格，两端点都不取，故为 $(−1, 2)。$","tags":["easy","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-005","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":2,"stem":"Solve $2/(x − 1) ≥ 0$.","stemZh":"解不等式 $2/(x − 1) ≥ 0。$","options":[{"id":"A","text":"$[1, +∞)$"},{"id":"B","text":"$(−∞, 1)$"},{"id":"C","text":"$(−∞, 1]$"},{"id":"D","text":"$(1, +∞)$"}],"answer":"D","explanation":"The numerator 2 is always positive and never zero, so the quotient is positive exactly when $x − 1 > 0$, i.e. $x > 1$. The value $x = 1$ is excluded by the denominator.","explanationZh":"分子 2 恒正且不为零，故分式非负当且仅当 $x − 1 > 0，即 x > 1；x = 1$ 时分母为零须排除。","tags":["easy","rational","endpoints"],"status":"published"},{"id":"math-rational-inequality-006","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":2,"stem":"Solve $(x − 3)/(x − 1) ≥ 0$.","stemZh":"解不等式 $(x − 3)/(x − 1) ≥ 0。$","options":[{"id":"A","text":"$(−∞, 1) ∪ [3, +∞)$"},{"id":"B","text":"$(−∞, 1] ∪ [3, +∞)$"},{"id":"C","text":"$(1, 3]$"},{"id":"D","text":"$[1, 3)$"}],"answer":"A","explanation":"Same signs outside the critical points 1 and 3. The point $x = 3$ is included (numerator zero) but $x = 1$ must be excluded because the denominator is zero there.","explanationZh":"同号区间在临界点 1、3 之外；x $= 3$ 处分子为零可取，x $= 1$ 处分母为零须排除，故为 $(−∞, 1) ∪ [3, +∞)。$","tags":["easy","rational","endpoints"],"status":"published"},{"id":"math-rational-inequality-007","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":1,"stem":"Solve $(x + 2)/(x − 2) ≤ 0$.","stemZh":"解不等式 $(x + 2)/(x − 2) ≤ 0。$","options":[{"id":"A","text":"$(−2, 2)$"},{"id":"B","text":"$[−2, 2)$"},{"id":"C","text":"$[−2, 2]$"},{"id":"D","text":"$(−∞, −2] ∪ [2, +∞)$"}],"answer":"B","explanation":"Between −2 and 2 the quotient is non-positive; −2 is allowed (numerator zero) while 2 is forbidden (denominator zero).","explanationZh":"在 −2 与 2 之间分式非正；−2 处分子为零可取，2 处分母为零须排除，故为 $[−2, 2)。$","tags":["easy","rational","endpoints"],"status":"published"},{"id":"math-rational-inequality-008","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":1,"stem":"Solve $1/(x + 3) < 0$.","stemZh":"解不等式 $1/(x + 3) < 0。$","options":[{"id":"A","text":"$(−3, +∞)$"},{"id":"B","text":"$(−∞, 3)$"},{"id":"C","text":"$(−∞, −3)$"},{"id":"D","text":"$(3, +∞)$"}],"answer":"C","explanation":"The numerator is positive, so the quotient is negative exactly when the denominator is $negative: x + 3 < 0$, i.e. $x < −3$.","explanationZh":"分子为正，分式为负当且仅当分母为负：x $+ 3 < 0，即 x < −3。$","tags":["easy","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-009","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":2,"stem":"Solve $x/(x − 4) > 0$.","stemZh":"解不等式 $x/(x − 4) > 0。$","options":[{"id":"A","text":"$(0, 4)$"},{"id":"B","text":"$(−∞, 0] ∪ [4, +∞)$"},{"id":"C","text":"$(0, +∞)$"},{"id":"D","text":"$(−∞, 0) ∪ (4, +∞)$"}],"answer":"D","explanation":"The critical points are 0 and 4. The quotient is positive outside $them: x < 0$ or $x > 4$; neither endpoint is included.","explanationZh":"临界点为 $0 与 4，同号区间在两点之外：x < 0 或 x > 4，两端点均不取。$","tags":["easy","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-010","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":2,"stem":"Solve $(x − 1)/(2 − x) ≥ 0$.","stemZh":"解不等式 $(x − 1)/(2 − x) ≥ 0。$","options":[{"id":"A","text":"$(1, 2]$"},{"id":"B","text":"$[1, 2]$"},{"id":"C","text":"$[1, 2)$"},{"id":"D","text":"$(−∞, 1] ∪ [2, +∞)$"}],"answer":"A","explanation":"Critical points are 1 and 2. Testing the middle: at $x = 3/2$ the value is $(1/2)/(1/2) = 1 > 0$, so the quotient is non-negative on $(1, 2]$. The point $x = 1$ is included (numerator zero) and $x = 2$ excluded (denominator zero).","explanationZh":"临界点为 $1 与 2。取 x = 3/2 检验：(1/2)/(1/2) = 1 > 0，故解集为 (1, 2]；x = 1$ 处分子为零可取，x $= 2$ 处分母为零须排除。","tags":["easy","rational","endpoints"],"status":"published"},{"id":"math-rational-inequality-011","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(x − 1)/(x + 1) ≤ 1$.","stemZh":"解不等式 $(x − 1)/(x + 1) ≤ 1。$","options":[{"id":"A","text":"$(−∞, −1)$"},{"id":"B","text":"$(−1, +∞)$"},{"id":"C","text":"$[1, +∞)$"},{"id":"D","text":"$(−∞, −1) ∪ [1, +∞)$"}],"answer":"B","explanation":"Move everything to one $side: (x − 1)/(x + 1) − 1 ≤ 0 → −2/(x + 1) ≤ 0 → 2/(x + 1) ≥ 0$. Since $2 > 0$ this needs $x + 1 > 0$, i.e. $x > −1$. Never multiply across by a denominator whose sign is unknown.","explanationZh":"$移项通分：(x − 1)/(x + 1) − 1 ≤ 0 → −2/(x + 1) ≤ 0 → 2/(x + 1) ≥ 0。因 2 > 0，故需 x + 1 > 0，即 x > −1。切勿直接乘以符号未知的分母。$","tags":["medium","rational","rearrangement"],"status":"published"},{"id":"math-rational-inequality-012","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(2x − 1)/(x + 3) > 0$.","stemZh":"解不等式 $(2x − 1)/(x + 3) > 0。$","options":[{"id":"A","text":"$(−3, 1/2)$"},{"id":"B","text":"$(−∞, 1/2) ∪ (3, +∞)$"},{"id":"C","text":"$(−∞, −3) ∪ (1/2, +∞)$"},{"id":"D","text":"$(1/2, +∞)$"}],"answer":"C","explanation":"The numerator vanishes at $x = 1/2$ and the denominator at $x = −3$. Same signs occur outside these $points: x < −3$ or $x > 1/2$.","explanationZh":"分子零点 $x = 1/2，分母零点 x = −3；同号区间在两点之外：x < −3 或 x > 1/2。$","tags":["medium","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-013","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(x + 1)/(x − 2) ≥ 2$.","stemZh":"解不等式 $(x + 1)/(x − 2) ≥ 2。$","options":[{"id":"A","text":"$(−∞, 2) ∪ [5, +∞)$"},{"id":"B","text":"$[2, 5]$"},{"id":"C","text":"$(2, 5)$"},{"id":"D","text":"$(2, 5]$"}],"answer":"D","explanation":"$(x + 1)/(x − 2) − 2 ≥ 0 → (5 − x)/(x − 2) ≥ 0 → (x − 5)/(x − 2) ≤ 0$, which holds on $(2, 5]: x = 2$ is excluded by the $denominator, x = 5$ is included since the numerator vanishes.","explanationZh":"$移项：(x + 1)/(x − 2) − 2 ≥ 0 → (5 − x)/(x − 2) ≥ 0 → (x − 5)/(x − 2) ≤ 0，解集为 (2, 5]；x = 2$ 分母为零须排除，x $= 5$ 分子为零可取。","tags":["medium","rational","rearrangement"],"status":"published"},{"id":"math-rational-inequality-014","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(x − 1)/(x^{2} + 1) > 0$.","stemZh":"解不等式 $(x − 1)/(x^{2} + 1) > 0。$","options":[{"id":"A","text":"$(1, +∞)$"},{"id":"B","text":"$(−∞, 1)$"},{"id":"C","text":"$(−1, 1)$"},{"id":"D","text":"$(−∞, −1) ∪ (1, +∞)$"}],"answer":"A","explanation":"$x^{2} + 1 > 0$ for every real x, so the sign of the fraction is the sign of x − 1: the solution is $x > 1$. Here the denominator can safely be ignored because it is always positive.","explanationZh":"$x^{2} + 1$ 对一切实数恒正，故分式的符号由 x − 1 决定，解为 $x > 1；此处分母恒正，可以放心“忽略”。$","tags":["medium","rational","always-positive-denominator"],"status":"published"},{"id":"math-rational-inequality-015","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(x + 2)/(x − 1) < 2$.","stemZh":"解不等式 $(x + 2)/(x − 1) < 2。$","options":[{"id":"A","text":"$(−∞, 1) ∪ (4, +∞)$"},{"id":"B","text":"$(1, 4)$"},{"id":"C","text":"$(−∞, 1) ∪ [4, +∞)$"},{"id":"D","text":"$(1, 4]$"}],"answer":"A","explanation":"$(x + 2)/(x − 1) − 2 < 0 → (4 − x)/(x − 1) < 0 → (x − 4)/(x − 1) > 0$, so $x < 1$ or $x > 4$, with both endpoints open.","explanationZh":"移项得 $(x + 2)/(x − 1) − 2 < 0 → (4 − x)/(x − 1) < 0 → (x − 4)/(x − 1) > 0，故 x < 1 或 x > 4，两端点均开。$","tags":["medium","rational","rearrangement"],"status":"published"},{"id":"math-rational-inequality-016","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(3x − 1)/(2 − x) ≤ 0$.","stemZh":"解不等式 $(3x − 1)/(2 − x) ≤ 0。$","options":[{"id":"A","text":"$[1/3, 2]$"},{"id":"B","text":"$(−∞, 1/3] ∪ [2, +∞)$"},{"id":"C","text":"$(−∞, 1/3] ∪ (2, +∞)$"},{"id":"D","text":"$(1/3, 2)$"}],"answer":"C","explanation":"Rewrite as $(3x − 1)/(x − 2) ≥ 0 (multiplying\\text{ by }−1)$. Critical points 1/3 and 2; same signs outside them, so $x ≤ 1/3$ or $x > 2$, with $x = 2$ excluded by the denominator.","explanationZh":"同乘 −1 化为 $(3x − 1)/(x − 2) ≥ 0，临界点为 1/3 与 2；同号在两点之外，故 x ≤ 1/3 或 x > 2，其中 x = 2$ 分母为零须排除。","tags":["medium","rational","sign-flip"],"status":"published"},{"id":"math-rational-inequality-017","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(x^{2} − 9)/(x + 1) ≤ 0$.","stemZh":"解不等式 $(x^{2} − 9)/(x + 1) ≤ 0。$","options":[{"id":"A","text":"$[−3, −1) ∪ (3, +∞)$"},{"id":"B","text":"$(−∞, −3] ∪ (−1, 3]$"},{"id":"C","text":"$(−∞, −3] ∪ [−1, 3]$"},{"id":"D","text":"$[−3, −1] ∪ [3, +∞)$"}],"answer":"B","explanation":"Critical points are $−3, −1 (denominator)$ and 3. A sign chart gives: negative on $(−∞, −3)$, zero at −3, positive on $(−3, −1)$, undefined at −1, negative on $(−1, 3)$, zero at 3, positive on $(3, +∞)$. Hence $(−∞, −3] ∪ (−1, 3]$.","explanationZh":"临界点为 $−3、−1（分母）与 3。符号表：(−∞, −3) 为负，−3 处为零，(−3, −1) 为正，−1 处无意义，(−1, 3) 为负，3 处为零，(3, +∞)$ 为正，故解集为 $(−∞, −3] ∪ (−1, 3]。$","tags":["medium","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-018","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $3/(x − 2) ≤ 1$.","stemZh":"解不等式 $3/(x − 2) ≤ 1。$","options":[{"id":"A","text":"$(2, 5]$"},{"id":"B","text":"$(−∞, 2] ∪ [5, +∞)$"},{"id":"C","text":"$[2, 5]$"},{"id":"D","text":"$(−∞, 2) ∪ [5, +∞)$"}],"answer":"D","explanation":"$3/(x − 2) − 1 ≤ 0 → (5 − x)/(x − 2) ≤ 0 → (x − 5)/(x − 2) ≥ 0$, so $x < 2$ or $x ≥ 5$. Note $x = 2$ is excluded and $x = 5$ is included.","explanationZh":"$移项：3/(x − 2) − 1 ≤ 0 → (5 − x)/(x − 2) ≤ 0 → (x − 5)/(x − 2) ≥ 0，故 x < 2 或 x ≥ 5；其中 x = 2$ 排除，x $= 5$ 可取。","tags":["medium","rational","rearrangement"],"status":"published"},{"id":"math-rational-inequality-019","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $1/(x − 1) < 1/(x + 1)$.","stemZh":"解不等式 $1/(x − 1) < 1/(x + 1)。$","options":[{"id":"A","text":"$(1, +∞)$"},{"id":"B","text":"$(−1, 1)$"},{"id":"C","text":"$(−∞, −1)$"},{"id":"D","text":"$(−∞, −1) ∪ (1, +∞)$"}],"answer":"B","explanation":"Bring to one $side: 1/(x − 1) − 1/(x + 1) < 0 → 2/((x − 1)(x + 1)) < 0 → (x − 1)(x + 1) < 0$, giving $−1 < x < 1$.","explanationZh":"$移项通分：1/(x − 1) − 1/(x + 1) < 0 → 2/((x − 1)(x + 1)) < 0 → (x − 1)(x + 1) < 0，得 −1 < x < 1。$","tags":["medium","rational","rearrangement"],"status":"published"},{"id":"math-rational-inequality-020","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(x − 1)/(x + 2) ≥ 1$.","stemZh":"解不等式 $(x − 1)/(x + 2) ≥ 1。$","options":[{"id":"A","text":"$(−2, +∞)$"},{"id":"B","text":"$(−∞, −2]$"},{"id":"C","text":"$∅$"},{"id":"D","text":"$(−∞, −2)$"}],"answer":"D","explanation":"$(x − 1)/(x + 2) − 1 ≥ 0 → −3/(x + 2) ≥ 0 → x + 2 < 0$, so $x < −2$. The endpoint −2 is excluded because the denominator is zero there.","explanationZh":"$移项：(x − 1)/(x + 2) − 1 ≥ 0 → −3/(x + 2) ≥ 0 → x + 2 < 0，故 x < −2；−2$ 处分母为零须排除。","tags":["medium","rational","rearrangement"],"status":"published"},{"id":"math-rational-inequality-021","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":4,"stem":"If the solution set of $(x − a)/(x − 2) < 0$ is $(a, 2)$, what condition must a satisfy?","stemZh":"$若 (x − a)/(x − 2) < 0$ 的解集为 $(a, 2)，则 a$ 应满足？","options":[{"id":"A","text":"$a > 2$"},{"id":"B","text":"$a < 2$"},{"id":"C","text":"$a ≤ 2$"},{"id":"D","text":"$a = 2$"}],"answer":"B","explanation":"The solution always lies between the two critical points a and 2, so we need $a < 2$ for the interval to be $(a, 2)$. If $a > 2$ the interval would be $(2, a)$, and if $a = 2$ the fraction equals 1, giving no solution.","explanationZh":"解集总在两临界点之间，要写成 $(a, 2)$ 必须 $a < 2；若 a > 2$ 则为 $(2, a)，若 a = 2$ 则分式恒为 $1（x ≠ 2），无解。$","tags":["hard","rational","parameter"],"status":"published"},{"id":"math-rational-inequality-022","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":4,"stem":"Solve $(x^{2} − 4)/(x^{2} − 1) ≤ 0$.","stemZh":"解不等式 $(x^{2} − 4)/(x^{2} − 1) ≤ 0。$","options":[{"id":"A","text":"$[−2, −1] ∪ [1, 2]$"},{"id":"B","text":"$(−2, −1) ∪ (1, 2)$"},{"id":"C","text":"$[−2, −1) ∪ (1, 2]$"},{"id":"D","text":"$(−∞, −2] ∪ [2, +∞)$"}],"answer":"C","explanation":"Critical points: −2, −1, 1, 2, where $\\pm1$ come from the denominator and are always excluded. The quotient is $≤ 0$ on $[−2, −1)$ and on $(1, 2]$; the zeros −2 and 2 are included, −1 and 1 are not.","explanationZh":"临界点为 $−2、−1、1、2，其中 \\pm1$ 来自分母必须排除；分式 $≤ 0$ 的区间为 $[−2, −1) 与 (1, 2]（零点 −2、2 可取，−1、1$ 不可取）。","tags":["hard","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-023","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":4,"stem":"Solve $(x − 1)/(x^{2} − x − 6) ≥ 0$.","stemZh":"解不等式 $(x − 1)/(x^{2} − x − 6) ≥ 0。$","options":[{"id":"A","text":"$(−2, 1] ∪ (3, +∞)$"},{"id":"B","text":"$[−2, 1] ∪ [3, +∞)$"},{"id":"C","text":"$(−∞, −2) ∪ (1, 3)$"},{"id":"D","text":"$(−2, 3)$"}],"answer":"A","explanation":"The denominator factors as $(x − 3)(x + 2)$, so the critical points are −2 and $3 (excluded)$ and $1 (included$, numerator zero). A sign chart gives non-negative on $(−2, 1]$ and on $(3, +∞)$.","explanationZh":"分母分解为 $(x − 3)(x + 2)，临界点为 −2、3（须排除）与 1（可取）。符号表显示非负区间为 (−2, 1] 与 (3, +∞)。$","tags":["hard","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-024","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":4,"stem":"How many integers x satisfy $(x − 1)/(x + 3) ≤ 0$?","stemZh":"满足 $(x − 1)/(x + 3) ≤ 0$ 的整数 x 有多少个？","options":[{"id":"A","text":"3"},{"id":"B","text":"4"},{"id":"C","text":"5"},{"id":"D","text":"2"}],"answer":"B","explanation":"The solution set is $(−3, 1]: −3$ is excluded (denominator zero) and 1 is included. The integers are −2, −1, 0, 1, which is 4 of them.","explanationZh":"解集为 $(−3, 1]：−3 处分母为零须排除，1$ 处分子为零可取。整数为 −2、−1、0、1，共 4 个。","tags":["hard","rational","integer-solutions"],"status":"published"},{"id":"math-rational-inequality-025","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":5,"stem":"Solve $(x + 1)/(x − 2) > (x + 1)/(x − 3)$.","stemZh":"解不等式 $(x + 1)/(x − 2) > (x + 1)/(x − 3)。$","options":[{"id":"A","text":"$(−1, 2) ∪ (3, +∞)$"},{"id":"B","text":"$(−∞, −1) ∪ (2, 3]$"},{"id":"C","text":"$(−∞, −1) ∪ (2, 3)$"},{"id":"D","text":"$(−1, 3)$"}],"answer":"C","explanation":"$Subtract: (x + 1)[1/(x − 2) − 1/(x − 3)] > 0 → (x + 1)·(−1)/((x − 2)(x − 3)) > 0 → (x + 1)/((x − 2)(x − 3)) < 0$. With critical points −1, 2, 3 the expression is negative on $(−∞, −1)$ and on $(2, 3)$.","explanationZh":"$作差：(x + 1)[1/(x − 2) − 1/(x − 3)] > 0 → −(x + 1)/((x − 2)(x − 3)) > 0 → (x + 1)/((x − 2)(x − 3)) < 0。临界点 −1、2、3，负值区间为 (−∞, −1) 与 (2, 3)。$","tags":["hard","rational","comparison"],"status":"published"},{"id":"math-rational-inequality-026","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":4,"stem":"Solve $(x^{2} − 3x + 2)/(x − 3) ≤ 0$.","stemZh":"解不等式 $(x^{2} − 3x + 2)/(x − 3) ≤ 0。$","options":[{"id":"A","text":"$(−∞, 1] ∪ [2, 3)$"},{"id":"B","text":"$[1, 2] ∪ [3, +∞)$"},{"id":"C","text":"$(−∞, 1) ∪ (2, 3]$"},{"id":"D","text":"$(1, 2)$"}],"answer":"A","explanation":"The numerator factors as $(x − 1)(x − 2)$. Critical points $1, 2 (included)$ and $3 (excluded)$. The quotient is $≤ 0$ on $(−∞, 1]$ and on $[2, 3)$.","explanationZh":"分子分解为 $(x − 1)(x − 2)，临界点为 1、2（可取）与 3（须排除）。分式 ≤ 0$ 的区间为 $(−∞, 1] 与 [2, 3)。$","tags":["hard","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-027","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":5,"stem":"If the solution set of $(x − a)/(x + 1) ≥ 0$ is $(−∞, −1) ∪ [a, +∞)$, what condition must a satisfy?","stemZh":"$若 (x − a)/(x + 1) ≥ 0$ 的解集为 $(−∞, −1) ∪ [a, +∞)，则 a$ 应满足？","options":[{"id":"A","text":"$a < −1$"},{"id":"B","text":"$a > −1$"},{"id":"C","text":"$a ≥ −1$"},{"id":"D","text":"$a = −1$"}],"answer":"B","explanation":"The solution lies outside the two critical points a and −1. For it to read $(−∞, −1) ∪ [a, +∞)$ the smaller point must be −1, so $a > −1$. If $a < −1$ the set would be $(−∞, a] ∪ (−1, +∞)$.","explanationZh":"解集在两临界点 a 与 −1 之外。要写成 $(−∞, −1) ∪ [a, +∞)，较小者必须是 −1，故 a > −1；若 a < −1，则解集为 (−∞, a] ∪ (−1, +∞)。$","tags":["hard","rational","parameter"],"status":"published"},{"id":"math-rational-inequality-028","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":5,"stem":"Solve $(x − 2)/(x^{2} − 4) ≥ 0$.","stemZh":"解不等式 $(x − 2)/(x^{2} − 4) ≥ 0。$","options":[{"id":"A","text":"$(−2, 2) ∪ (2, +∞)$"},{"id":"B","text":"$(−2, +∞)$"},{"id":"C","text":"$(−2, 2)$"},{"id":"D","text":"$(2, +∞)$"}],"answer":"A","explanation":"The domain excludes both $x = −2$ and $x = 2$. On that domain $(x − 2)/(x^{2} − 4)$ simplifies to $1/(x + 2)$, which is $≥ 0$ when $x + 2 > 0$. So $x > −2$ with $x ≠ 2$, i.e. $(−2, 2) ∪ (2, +∞)$. Cancelling the factor $(x − 2)$ without recording $x ≠ 2$ is the classic error.","explanationZh":"定义域要求 $x ≠ −2 且 x ≠ 2。在定义域内原式化为 1/(x + 2)，其 ≥ 0 当 x + 2 > 0，即 x > −2 且 x ≠ 2，故为 (−2, 2) ∪ (2, +∞)。约去 (x − 2)$ 却忘记 $x ≠ 2$ 是典型错误。","tags":["hard","rational","domain"],"status":"published"},{"id":"math-rational-inequality-029","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":5,"stem":"Solve $(x + 2)/(x − 1) ≥ (x + 3)/(x − 2)$.","stemZh":"解不等式 $(x + 2)/(x − 1) ≥ (x + 3)/(x − 2)。$","options":[{"id":"A","text":"$[−1/2, 1) ∪ (2, +∞)$"},{"id":"B","text":"$(−∞, −1/2) ∪ (1, 2]$"},{"id":"C","text":"$(−∞, −1/2] ∪ (1, 2)$"},{"id":"D","text":"$(−1/2, 1) ∪ (2, +∞)$"}],"answer":"C","explanation":"Subtract and $combine: [(x + 2)(x − 2) − (x + 3)(x − 1)]/((x − 1)(x − 2)) ≥ 0 → (−2x − 1)/((x − 1)(x − 2)) ≥ 0 → (2x + 1)/((x − 1)(x − 2)) ≤ 0$. Critical points $−1/2 (included), 1$ and $2 (excluded)$ give $(−∞, −1/2] ∪ (1, 2)$.","explanationZh":"$作差通分：[(x + 2)(x − 2) − (x + 3)(x − 1)]/((x − 1)(x − 2)) ≥ 0 → (−2x − 1)/((x − 1)(x − 2)) ≥ 0 → (2x + 1)/((x − 1)(x − 2)) ≤ 0。临界点 −1/2（可取）、1 与 2（须排除），得 (−∞, −1/2] ∪ (1, 2)。$","tags":["hard","rational","comparison"],"status":"published"},{"id":"math-rational-inequality-030","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":5,"stem":"How many integers x satisfy both $(x − 1)/(x + 2) ≥ 0$ and $x^{2} ≤ 9$?","stemZh":"同时满足 $(x − 1)/(x + 2) ≥ 0 与 x^{2} ≤ 9$ 的整数 x 有多少个？","options":[{"id":"A","text":"3"},{"id":"B","text":"2"},{"id":"C","text":"5"},{"id":"D","text":"4"}],"answer":"D","explanation":"The first inequality gives $(−∞, −2) ∪ [1, +∞)$; the second gives $[−3, 3]$. Their intersection is $[−3, −2) ∪ [1, 3]$, whose integers are −3, 1, 2, 3 — four values.","explanationZh":"第一个不等式解集为 $(−∞, −2) ∪ [1, +∞)，第二个为 [−3, 3]；交集为 [−3, −2) ∪ [1, 3]，其中整数为 −3、1、2、3，共 4$ 个。","tags":["hard","rational","integer-solutions"],"status":"published"},{"id":"math-absolute-inequality-001","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":1,"stem":"Solve $|x| < 3$.","stemZh":"解不等式 $|x| < 3。$","options":[{"id":"A","text":"$(−∞, −3) ∪ (3, +∞)$"},{"id":"B","text":"$[−3, 3]$"},{"id":"C","text":"$(−3, 3)$"},{"id":"D","text":"$(−∞, 3)$"}],"answer":"C","explanation":"$|x| < 3$ means the distance from x to 0 is less than 3, i.e. $−3 < x < 3$.","explanationZh":"$|x| < 3$ 表示 x 到 0 的距离小于 $3，即 −3 < x < 3。$","tags":["easy","absolute-value","basic"],"status":"published"},{"id":"math-absolute-inequality-002","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":1,"stem":"Solve $|x| ≥ 2$.","stemZh":"解不等式 $|x| ≥ 2。$","options":[{"id":"A","text":"$(−∞, −2] ∪ [2, +∞)$"},{"id":"B","text":"$(−2, 2)$"},{"id":"C","text":"$[−2, 2]$"},{"id":"D","text":"$(−∞, −2) ∪ (2, +∞)$"}],"answer":"A","explanation":"$|x| ≥ 2$ means $x ≤ −2$ or $x ≥ 2$. The endpoints are included because the inequality is not strict.","explanationZh":"$|x| ≥ 2 即 x ≤ −2 或 x ≥ 2；不等式带等号，端点取到。$","tags":["easy","absolute-value","basic"],"status":"published"},{"id":"math-absolute-inequality-003","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":2,"stem":"Solve $|x − 1| ≤ 2$.","stemZh":"解不等式 $|x − 1| ≤ 2。$","options":[{"id":"A","text":"$(−1, 3)$"},{"id":"B","text":"$(−∞, −1] ∪ [3, +∞)$"},{"id":"C","text":"$[−3, 1]$"},{"id":"D","text":"$[−1, 3]$"}],"answer":"D","explanation":"$−2 ≤ x − 1 ≤ 2$, so $−1 ≤ x ≤ 3. Equivalently, x$ is within distance 2 of the point 1.","explanationZh":"$由 −2 ≤ x − 1 ≤ 2 得 −1 ≤ x ≤ 3，即 x 与 1$ 的距离不超过 2。","tags":["easy","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-004","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":1,"stem":"Solve $|x| > −1$.","stemZh":"解不等式 $|x| > −1。$","options":[{"id":"A","text":"$∅$"},{"id":"B","text":"$ℝ$"},{"id":"C","text":"$(0, +∞)$"},{"id":"D","text":"$(−1, 1)$"}],"answer":"B","explanation":"$|x| ≥ 0$ for every real x, so $|x| > −1$ holds for all real numbers. An absolute value compared with a negative bound never gives the empty set.","explanationZh":"对任意实数 $|x| ≥ 0，故 |x| > −1$ 恒成立，解集为 $ℝ；绝对值与负数比较时解集不会是空集。$","tags":["easy","absolute-value","property"],"status":"published"},{"id":"math-absolute-inequality-005","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":2,"stem":"Solve $|x| ≤ 0$.","stemZh":"解不等式 $|x| ≤ 0。$","options":[{"id":"A","text":"{0}"},{"id":"B","text":"$∅$"},{"id":"C","text":"$ℝ$"},{"id":"D","text":"$(−∞, 0]$"}],"answer":"A","explanation":"$|x|$ is never negative, so $|x| ≤ 0$ forces $|x| = 0$, i.e. $x = 0$. The solution set is {0}, not the empty set.","explanationZh":"$|x|$ 不可能为负，故 $|x| ≤ 0$ 只能是 $|x| = 0，即 x = 0，解集为 {0}$ 而非空集。","tags":["easy","absolute-value","property"],"status":"published"},{"id":"math-absolute-inequality-006","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":1,"stem":"Solve $|2x| < 4$.","stemZh":"解不等式 $|2x| < 4。$","options":[{"id":"A","text":"$(−4, 4)$"},{"id":"B","text":"$(−∞, −2) ∪ (2, +∞)$"},{"id":"C","text":"$(−2, 2)$"},{"id":"D","text":"$[−2, 2]$"}],"answer":"C","explanation":"$|2x| = 2|x| < 4$ gives $|x| < 2$, so $−2 < x < 2$. Remember to divide the bound by the coefficient of x.","explanationZh":"$|2x| = 2|x| < 4 即 |x| < 2，故 −2 < x < 2；注意右端也要除以 x$ 的系数。","tags":["easy","absolute-value","basic"],"status":"published"},{"id":"math-absolute-inequality-007","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":2,"stem":"Solve $|x + 1| > 3$.","stemZh":"解不等式 $|x + 1| > 3。$","options":[{"id":"A","text":"$(−4, 2)$"},{"id":"B","text":"$(−∞, −4) ∪ (2, +∞)$"},{"id":"C","text":"$(−∞, −2) ∪ (4, +∞)$"},{"id":"D","text":"$[−4, 2]$"}],"answer":"B","explanation":"$|x + 1| > 3$ means $x + 1 > 3$ or $x + 1 < −3$, giving $x > 2$ or $x < −4$. Distances larger than 3 lie outside the interval $[−4, 2]$.","explanationZh":"$|x + 1| > 3 即 x + 1 > 3 或 x + 1 < −3，得 x > 2 或 x < −4；距离大于 3$ 的点在区间 $[−4, 2]$ 之外。","tags":["easy","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-008","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":1,"stem":"Solve $|x − 2| ≥ 1$.","stemZh":"解不等式 $|x − 2| ≥ 1。$","options":[{"id":"A","text":"$[1, 3]$"},{"id":"B","text":"$(1, 3)$"},{"id":"C","text":"$(−∞, 1) ∪ (3, +∞)$"},{"id":"D","text":"$(−∞, 1] ∪ [3, +∞)$"}],"answer":"D","explanation":"$x − 2 ≥ 1$ or $x − 2 ≤ −1$, i.e. $x ≥ 3$ or $x ≤ 1$. Endpoints included because of the non-strict sign.","explanationZh":"$x − 2 ≥ 1 或 x − 2 ≤ −1，即 x ≥ 3 或 x ≤ 1；因不等式不严格，端点取到。$","tags":["easy","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-009","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":2,"stem":"Solve $|x + 2| ≤ 3$.","stemZh":"解不等式 $|x + 2| ≤ 3。$","options":[{"id":"A","text":"$[−5, 1]$"},{"id":"B","text":"$(−5, 1)$"},{"id":"C","text":"$[−1, 5]$"},{"id":"D","text":"$(−∞, −5] ∪ [1, +∞)$"}],"answer":"A","explanation":"$−3 ≤ x + 2 ≤ 3$, so $−5 ≤ x ≤ 1$.","explanationZh":"$由 −3 ≤ x + 2 ≤ 3 得 −5 ≤ x ≤ 1。$","tags":["easy","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-010","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":2,"stem":"Solve $|x| > 0$.","stemZh":"解不等式 $|x| > 0。$","options":[{"id":"A","text":"$ℝ$"},{"id":"B","text":"$[0, +∞)$"},{"id":"C","text":"$(−∞, 0) ∪ (0, +∞)$"},{"id":"D","text":"$(0, +∞)$"}],"answer":"C","explanation":"$|x| > 0$ fails only at $x = 0$, so the solution set is all non-zero $reals: (−∞, 0) ∪ (0, +∞)$.","explanationZh":"$|x| > 0$ 仅在 $x = 0 处不成立，故解集为一切非零实数：(−∞, 0) ∪ (0, +∞)。$","tags":["easy","absolute-value","property"],"status":"published"},{"id":"math-absolute-inequality-011","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Solve $|2x − 1| ≤ 3$.","stemZh":"解不等式 $|2x − 1| ≤ 3。$","options":[{"id":"A","text":"$[−2, 1]$"},{"id":"B","text":"$[−1, 2]$"},{"id":"C","text":"$(−1, 2)$"},{"id":"D","text":"$(−∞, −1] ∪ [2, +∞)$"}],"answer":"B","explanation":"$−3 ≤ 2x − 1 ≤ 3 → −2 ≤ 2x ≤ 4 → −1 ≤ x ≤ 2$.","explanationZh":"$由 −3 ≤ 2x − 1 ≤ 3 得 −2 ≤ 2x ≤ 4，即 −1 ≤ x ≤ 2。$","tags":["medium","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-012","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Solve $|3 − x| < 2$.","stemZh":"解不等式 $|3 − x| < 2。$","options":[{"id":"A","text":"$(−5, −1)$"},{"id":"B","text":"$(−1, 5)$"},{"id":"C","text":"$[1, 5]$"},{"id":"D","text":"$(1, 5)$"}],"answer":"D","explanation":"$|3 − x| = |x − 3| < 2$ gives $−2 < x − 3 < 2$, so $1 < x < 5$.","explanationZh":"$|3 − x| = |x − 3| < 2，即 −2 < x − 3 < 2，得 1 < x < 5。$","tags":["medium","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-013","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Solve $|x − 1| ≥ 2$.","stemZh":"解不等式 $|x − 1| ≥ 2。$","options":[{"id":"A","text":"$[−1, 3]$"},{"id":"B","text":"$(−∞, −1) ∪ (3, +∞)$"},{"id":"C","text":"$(−1, 3)$"},{"id":"D","text":"$(−∞, −1] ∪ [3, +∞)$"}],"answer":"D","explanation":"$x − 1 ≥ 2$ or $x − 1 ≤ −2$, giving $x ≥ 3$ or $x ≤ −1$, with both endpoints included.","explanationZh":"$x − 1 ≥ 2 或 x − 1 ≤ −2，得 x ≥ 3 或 x ≤ −1，端点均取到。$","tags":["medium","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-014","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Solve $|2x + 4| > 6$.","stemZh":"解不等式 $|2x + 4| > 6。$","options":[{"id":"A","text":"$(−∞, −5) ∪ (1, +∞)$"},{"id":"B","text":"$(−5, 1)$"},{"id":"C","text":"$(−∞, 1) ∪ (5, +∞)$"},{"id":"D","text":"$[−5, 1]$"}],"answer":"A","explanation":"$2x + 4 > 6$ gives $x > 1; 2x + 4 < −6$ gives $x < −5$. So the solution is $(−∞, −5) ∪ (1, +∞)$.","explanationZh":"$2x + 4 > 6 得 x > 1；2x + 4 < −6 得 x < −5，故解集为 (−∞, −5) ∪ (1, +∞)。$","tags":["medium","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-015","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Solve $|x − 2| + 1 ≤ 4$.","stemZh":"解不等式 $|x − 2| + 1 ≤ 4。$","options":[{"id":"A","text":"$(−1, 5)$"},{"id":"B","text":"$[−5, 1]$"},{"id":"C","text":"$[−1, 5]$"},{"id":"D","text":"$(−∞, −1] ∪ [5, +∞)$"}],"answer":"C","explanation":"Isolate the absolute value $first: |x − 2| ≤ 3$, so $−3 ≤ x − 2 ≤ 3$ and $−1 ≤ x ≤ 5$.","explanationZh":"$先移项：|x − 2| ≤ 3，即 −3 ≤ x − 2 ≤ 3，得 −1 ≤ x ≤ 5。$","tags":["medium","absolute-value","rearrangement"],"status":"published"},{"id":"math-absolute-inequality-016","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Find the set of x satisfying both $|x| ≤ 3$ and $|x| > 1$.","stemZh":"求同时满足 $|x| ≤ 3 与 |x| > 1 的 x$ 的集合。","options":[{"id":"A","text":"$[−3, 3]$"},{"id":"B","text":"$[−3, −1) ∪ (1, 3]$"},{"id":"C","text":"$(−3, −1] ∪ [1, 3)$"},{"id":"D","text":"$(−∞, −3] ∪ [3, +∞)$"}],"answer":"B","explanation":"$|x| ≤ 3$ gives $[−3, 3]$ and $|x| > 1$ gives $(−∞, −1) ∪ (1, +∞)$. Their intersection is $[−3, −1) ∪ (1, 3]$.","explanationZh":"$|x| ≤ 3$ 的解集为 $[−3, 3]，|x| > 1$ 的解集为 $(−∞, −1) ∪ (1, +∞)，交集为 [−3, −1) ∪ (1, 3]。$","tags":["medium","absolute-value","system"],"status":"published"},{"id":"math-absolute-inequality-017","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Solve $|x − 1| ≤ 0$.","stemZh":"解不等式 $|x − 1| ≤ 0。$","options":[{"id":"A","text":"{1}"},{"id":"B","text":"$∅$"},{"id":"C","text":"$ℝ$"},{"id":"D","text":"$[0, 2]$"}],"answer":"A","explanation":"An absolute value is never negative, so $|x − 1| ≤ 0$ forces $|x − 1| = 0$, i.e. $x = 1$.","explanationZh":"绝对值非负，故 $|x − 1| ≤ 0$ 只能是 $|x − 1| = 0，即 x = 1。$","tags":["medium","absolute-value","property"],"status":"published"},{"id":"math-absolute-inequality-018","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Solve $|1 − 2x| ≥ 3$.","stemZh":"解不等式 $|1 − 2x| ≥ 3。$","options":[{"id":"A","text":"$[−1, 2]$"},{"id":"B","text":"$(−1, 2)$"},{"id":"C","text":"$(−∞, −1] ∪ [2, +∞)$"},{"id":"D","text":"$(−∞, −2] ∪ [1, +∞)$"}],"answer":"C","explanation":"$|1 − 2x| = |2x − 1| ≥ 3$ gives $2x − 1 ≥ 3 (x ≥ 2)$ or $2x − 1 ≤ −3 (x ≤ −1)$, so $(−∞, −1] ∪ [2, +∞)$.","explanationZh":"$|1 − 2x| = |2x − 1| ≥ 3，即 2x − 1 ≥ 3（x ≥ 2）或 2x − 1 ≤ −3（x ≤ −1），故解集为 (−∞, −1] ∪ [2, +∞)。$","tags":["medium","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-019","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"How many integers x satisfy $|x| ≤ 2$?","stemZh":"满足 $|x| ≤ 2$ 的整数 x 有多少个？","options":[{"id":"A","text":"2"},{"id":"B","text":"3"},{"id":"C","text":"4"},{"id":"D","text":"5"}],"answer":"D","explanation":"$|x| ≤ 2$ gives $−2 ≤ x ≤ 2$, whose integers are −2, −1, 0, 1, 2 — five values.","explanationZh":"$|x| ≤ 2 即 −2 ≤ x ≤ 2，整数为 −2、−1、0、1、2，共 5$ 个。","tags":["medium","absolute-value","integer-solutions"],"status":"published"},{"id":"math-absolute-inequality-020","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"For which real x does $|x| =$ −x hold?","stemZh":"等式 $|x| =$ −x 对哪些实数 x 成立？","options":[{"id":"A","text":"$[0, +∞)$"},{"id":"B","text":"$(−∞, 0]$"},{"id":"C","text":"{0}"},{"id":"D","text":"$∅$"}],"answer":"B","explanation":"Since $|x| ≥ 0$, the right side must satisfy −x $≥ 0$, i.e. $x ≤ 0$. For $x ≤ 0$ we indeed have $|x| = −x$, so the solution is $(−∞, 0]$.","explanationZh":"$由 |x| ≥ 0$ 知需 −x $≥ 0，即 x ≤ 0；当 x ≤ 0$ 时确有 $|x| =$ −x，故解集为 $(−∞, 0]。$","tags":["medium","absolute-value","property"],"status":"published"},{"id":"math-absolute-inequality-021","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":4,"stem":"Solve $|x − 1| + |x − 2| ≤ 1$.","stemZh":"解不等式 $|x − 1| + |x − 2| ≤ 1。$","options":[{"id":"A","text":"$(1, 2)$"},{"id":"B","text":"$(−∞, 1] ∪ [2, +∞)$"},{"id":"C","text":"$[1, 2]$"},{"id":"D","text":"$∅$"}],"answer":"C","explanation":"$|x − 1| + |x − 2|$ is the sum of the distances from x to 1 and to 2. Its minimum is the distance between the two points, 1, attained precisely on the segment $[1, 2]$. Outside it the sum exceeds 1.","explanationZh":"$|x − 1| + |x − 2|$ 表示 x 到 1 与 2 的距离之和，其最小值等于两点间距离 $1，且恰在 [1, 2]$ 上取得；区间外距离和大于 1。","tags":["hard","absolute-value","distance-reasoning"],"status":"published"},{"id":"math-absolute-inequality-022","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":5,"stem":"Solve $|x − 3| + |x + 1| ≥ 4$.","stemZh":"解不等式 $|x − 3| + |x + 1| ≥ 4。$","options":[{"id":"A","text":"$[−1, 3]$"},{"id":"B","text":"$(−∞, −1] ∪ [3, +∞)$"},{"id":"C","text":"$∅$"},{"id":"D","text":"$ℝ$"}],"answer":"D","explanation":"The sum of the distances from x to 3 and to −1 is at least the distance between the two points, which is 4, with equality on $[−1, 3]$. So the inequality holds for every real x.","explanationZh":"x 到 3 与 −1 的距离之和不小于两点间距离 $4（在 [−1, 3]$ 上取等号），故对一切实数成立，解集为 $ℝ。$","tags":["hard","absolute-value","distance-reasoning"],"status":"published"},{"id":"math-absolute-inequality-023","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":5,"stem":"Solve $|x − 1| < 2x$.","stemZh":"解不等式 $|x − 1| < 2x。$","options":[{"id":"A","text":"$(0, +∞)$"},{"id":"B","text":"$(1/3, +∞)$"},{"id":"C","text":"$(1/2, +∞)$"},{"id":"D","text":"$(−1, +∞)$"}],"answer":"B","explanation":"Since $|x − 1| ≥ 0$ we need $2x > 0$. Then $|x − 1| < 2x$ is equivalent to $−2x < x − 1 < 2x$, i.e. $3x > 1$ and $x > −1$. Together with $x > 0$ this gives $x > 1/3$.","explanationZh":"$因 |x − 1| ≥ 0，须 2x > 0；此时 |x − 1| < 2x$ 等价于 $−2x < x − 1 < 2x，即 3x > 1 且 x > −1。结合 x > 0 得 x > 1/3。$","tags":["hard","absolute-value","variable-bound"],"status":"published"},{"id":"math-absolute-inequality-024","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":4,"stem":"If $|x − 2| = 2 − x, then$:","stemZh":"$若 |x − 2| = 2 −$ x，则？","options":[{"id":"A","text":"$x ≤ 2$"},{"id":"B","text":"$x ≥ 2$"},{"id":"C","text":"$x < 2$"},{"id":"D","text":"$x = 2$"}],"answer":"A","explanation":"Since $2 − x = −(x − 2)$, the equation says $|x − 2| = −(x − 2)$, which holds exactly when $x − 2 ≤ 0$, i.e. $x ≤ 2 (the\\text{ case }x = 2\\text{ gives }0 = 0)$.","explanationZh":"$因 2 − x = −(x − 2)，等式即 |x − 2| = −(x − 2)，当且仅当 x − 2 ≤ 0，即 x ≤ 2（x = 2 时 0 = 0$ 也成立）。","tags":["hard","absolute-value","property"],"status":"published"},{"id":"math-absolute-inequality-025","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":4,"stem":"How many integers x satisfy $|x − 1| < 2$ and $|x| ≥ 1$?","stemZh":"同时满足 $|x − 1| < 2 与 |x| ≥ 1$ 的整数 x 有多少个？","options":[{"id":"A","text":"3"},{"id":"B","text":"2"},{"id":"C","text":"4"},{"id":"D","text":"1"}],"answer":"B","explanation":"$|x − 1| < 2$ gives $−1 < x < 3; |x| ≥ 1$ gives $x ≤ −1$ or $x ≥ 1$. The intersection is $[1, 3)$, whose integers are 1 and 2 — two values.","explanationZh":"$|x − 1| < 2 得 −1 < x < 3；|x| ≥ 1 得 x ≤ −1 或 x ≥ 1。交集为 [1, 3)，整数为 1 和 2，共 2$ 个。","tags":["hard","absolute-value","integer-solutions"],"status":"published"},{"id":"math-absolute-inequality-026","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":5,"stem":"Solve $|x − 1| ≤ 2x − 1$.","stemZh":"解不等式 $|x − 1| ≤ 2x − 1。$","options":[{"id":"A","text":"$[2/3, +∞)$"},{"id":"B","text":"$(2/3, +∞)$"},{"id":"C","text":"$[1/2, +∞)$"},{"id":"D","text":"$(−∞, 1/3] ∪ [2, +∞)$"}],"answer":"A","explanation":"First the right side must be $non-negative: 2x − 1 ≥ 0$, so $x ≥ 1/2$. Then $−(2x − 1) ≤ x − 1 ≤ 2x − 1$ gives $x ≥ 0$ and $3x ≥ 2$, i.e. $x ≥ 2/3. Combining: x ≥ 2/3$.","explanationZh":"$首先右端须非负：2x − 1 ≥ 0，即 x ≥ 1/2；再由 −(2x − 1) ≤ x − 1 ≤ 2x − 1 得 x ≥ 0 且 3x ≥ 2，即 x ≥ 2/3。综合得 x ≥ 2/3。$","tags":["hard","absolute-value","variable-bound"],"status":"published"},{"id":"math-absolute-inequality-027","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":5,"stem":"Solve $|x| + |x − 2| < 3$.","stemZh":"解不等式 $|x| + |x − 2| < 3。$","options":[{"id":"A","text":"$(−3/2, 3/2)$"},{"id":"B","text":"$(−1, 3)$"},{"id":"C","text":"$(0, 2)$"},{"id":"D","text":"$(−1/2, 5/2)$"}],"answer":"D","explanation":"On $[0, 2]$ the sum equals $2 < 3$. For $x > 2$ it is $x + (x − 2) = 2x − 2 < 3$, so $x < 5/2$. For $x < 0$ it is −x $+ (2 − x) = 2 − 2x < 3$, so $x > −1/2$. Hence $(−1/2, 5/2)$.","explanationZh":"$在 [0, 2]$ 上距离和为 $2 < 3；当 x > 2 时 x + (x − 2) = 2x − 2 < 3，得 x < 5/2；当 x < 0 时$ −x $+ (2 − x) = 2 − 2x < 3，得 x > −1/2。故解集为 (−1/2, 5/2)。$","tags":["hard","absolute-value","distance-reasoning"],"status":"published"},{"id":"math-absolute-inequality-028","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":4,"stem":"If the solution set of $|x − a| ≤ 2$ is $[−1, 3]$, find a.","stemZh":"$若 |x − a| ≤ 2$ 的解集为 $[−1, 3]，求$ a。","options":[{"id":"A","text":"−1"},{"id":"B","text":"2"},{"id":"C","text":"1"},{"id":"D","text":"3"}],"answer":"C","explanation":"$|x − a| ≤ 2$ has solution $[a − 2, a + 2]$. Matching $[−1, 3]$ gives $a − 2 = −1$ and $a + 2 = 3$, so $a = 1$.","explanationZh":"$|x − a| ≤ 2$ 的解集为 $[a − 2, a + 2]，与 [−1, 3]$ 对比得 $a − 2 = −1、a + 2 = 3，故 a = 1。$","tags":["hard","absolute-value","parameter"],"status":"published"},{"id":"math-absolute-inequality-029","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":5,"stem":"Solve $|2x − 1| ≥$ x.","stemZh":"解不等式 $|2x − 1| ≥$ x。","options":[{"id":"A","text":"$(−∞, 1/3) ∪ (1, +∞)$"},{"id":"B","text":"$(−∞, 1/3] ∪ [1, +∞)$"},{"id":"C","text":"$[1/3, 1]$"},{"id":"D","text":"$(0, 1/3] ∪ [1, +∞)$"}],"answer":"B","explanation":"For $x ≤ 0$ the inequality is automatic since $|2x − 1| ≥ 0 ≥$ x. For $x > 0$ we need $2x − 1 ≥ x$ or $2x − 1 ≤ −x$, i.e. $x ≥ 1$ or $x ≤ 1/3$. Altogether $(−∞, 1/3] ∪ [1, +∞)$.","explanationZh":"$当 x ≤ 0 时，|2x − 1| ≥ 0 ≥ x$ 自动成立；当 $x > 0$ 时需 $2x − 1 ≥ x 或 2x − 1 ≤$ −x，即 $x ≥ 1 或 x ≤ 1/3。合并得 (−∞, 1/3] ∪ [1, +∞)。$","tags":["hard","absolute-value","variable-bound"],"status":"published"},{"id":"math-absolute-inequality-030","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":5,"stem":"How many integers x satisfy $|x − 1| + |x + 1| ≤ 4$?","stemZh":"满足 $|x − 1| + |x + 1| ≤ 4$ 的整数 x 有多少个？","options":[{"id":"A","text":"5"},{"id":"B","text":"4"},{"id":"C","text":"6"},{"id":"D","text":"3"}],"answer":"A","explanation":"On $[−1, 1]$ the sum is $2 ≤ 4$. For $x > 1$ it is $2x ≤ 4$, so $x ≤ 2$; for $x < −1$ it is $−2x ≤ 4$, so $x ≥ −2$. Hence the solution set is $[−2, 2]$, containing the five integers −2, −1, 0, 1, 2.","explanationZh":"$在 [−1, 1]$ 上距离和为 $2 ≤ 4；x > 1$ 时为 $2x ≤ 4，得 x ≤ 2；x < −1$ 时为 $−2x ≤ 4，得 x ≥ −2。故解集为 [−2, 2]，含整数 −2、−1、0、1、2 共 5$ 个。","tags":["hard","absolute-value","integer-solutions"],"status":"published"},{"id":"math-domain-001","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":1,"stem":"What is the domain of $f(x) = \\sqrt[3]{}x + 1$?","stemZh":"函数 $f(x) = \\sqrt[3]{}x + 1$ 的定义域是什么？","options":[{"id":"A","text":"$(−∞, +∞)$"},{"id":"B","text":"$[1, +∞)$"},{"id":"C","text":"$[0, +∞)$"},{"id":"D","text":"$(−∞, 0]$"}],"answer":"A","explanation":"A cube root is an odd root, so its radicand may be any real number $(negative$, zero or positive). Adding 1 places no further restriction, so the domain is all real $numbers, (−∞, +∞)$.","explanationZh":"立方根是奇次根式，被开方数可以为任意实数（负、零、正均可），再加 1 也不产生新的限制，所以定义域为全体实数 $(−∞, +∞)。$","tags":["easy","domain","radical"],"status":"published"},{"id":"math-domain-002","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":1,"stem":"What is the domain of $f(x) = 1/(x − 2)$?","stemZh":"函数 $f(x) = 1/(x − 2)$ 的定义域是什么？","options":[{"id":"A","text":"$x > 2$"},{"id":"B","text":"$x ≠ 2$"},{"id":"C","text":"$x ≥ 2$"},{"id":"D","text":"$x ≤ 2$"}],"answer":"B","explanation":"The only restriction for a fraction is that the denominator is not $zero: x − 2 ≠ 0$, so $x ≠ 2$. Values below and above 2 are both allowed.","explanationZh":"分式唯一的限制是分母不为零：x $− 2 ≠ 0，即 x ≠ 2。小于 2$ 和大于 2 的数都可以取。","tags":["easy","domain","rational"],"status":"published"},{"id":"math-domain-003","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":1,"stem":"What is the domain of $f(x) = \\sqrt{x − 4}$?","stemZh":"函数 $f(x) = \\sqrt{x − 4}$ 的定义域是什么？","options":[{"id":"A","text":"$(4, +∞)$"},{"id":"B","text":"$(−∞, 4]$"},{"id":"C","text":"$[4, +∞)$"},{"id":"D","text":"$(−∞, +∞)$"}],"answer":"C","explanation":"For an even root the radicand must satisfy $x − 4 ≥ 0$, so $x ≥ 4$. The endpoint $x = 4$ is included because $\\sqrt{0} = 0$ is defined, giving $[4, +∞)$.","explanationZh":"偶次根式要求被开方数非负：x $− 4 ≥ 0，即 x ≥ 4。端点 x = 4$ 要取到，因为 $\\sqrt{0} = 0$ 有意义，所以定义域为 $[4, +∞)。$","tags":["easy","domain","radical"],"status":"published"},{"id":"math-domain-004","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":1,"stem":"What is the domain of $f(x) = x^{2} − 2x − 3$?","stemZh":"函数 $f(x) = x^{2} − 2x − 3$ 的定义域是什么？","options":[{"id":"A","text":"$[0, +∞)$"},{"id":"B","text":"$[−4, +∞)$"},{"id":"C","text":"$(−∞, −1] ∪ [3, +∞)$"},{"id":"D","text":"$(−∞, +∞)$"}],"answer":"D","explanation":"A polynomial is defined for every real number, so the domain is $(−∞, +∞)$. Option B is the range (the vertex value is $−4)$ and option C is the solution set of $x^{2} − 2x − 3 ≥ 0 —$ neither is the domain.","explanationZh":"多项式对一切实数都有意义，定义域为 $(−∞, +∞)。选项 B$ 是值域（顶点处取到 −4），选项 C 是不等式 $x^{2} − 2x − 3 ≥ 0$ 的解集，二者都不是定义域。","tags":["easy","domain","polynomial"],"status":"published"},{"id":"math-domain-005","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":1,"stem":"In $y = 3x − 1$, which statement about the variables is correct?","stemZh":"$在 y = 3x − 1$ 中，下列关于变量的说法哪个正确？","options":[{"id":"A","text":"x is independent; y is dependent"},{"id":"B","text":"y is independent; x is dependent"},{"id":"C","text":"both are independent"},{"id":"D","text":"both are dependent"}],"answer":"A","explanation":"In $y = 3x − 1$ the value of x is chosen freely, so x is the independent variable; the value of y is then determined by the rule, so y is the dependent variable.","explanationZh":"$在 y = 3x − 1$ 中，x 的值可以自由选取，因此 x 是自变量；y 的值由对应关系确定，因此 y 是因变量。","tags":["easy","function-concept","variables"],"status":"published"},{"id":"math-domain-006","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":2,"stem":"Let $A = {1, 2, 3}$ and $B = {2, 4, 6}$. Which of the following relations is a function from A to B?","stemZh":"$设 A = {1, 2, 3}，B = {2, 4, 6}。下列哪个关系是从 A 到 B$ 的函数？","options":[{"id":"A","text":"${(1, 2), (1, 4), (2, 6), (3, 2)}$"},{"id":"B","text":"${(1, 2), (2, 4), (3, 6)}$"},{"id":"C","text":"${(1, 2), (2, 4)}$"},{"id":"D","text":"${(1, 4), (2, 4), (3, 6), (3, 2)}$"}],"answer":"B","explanation":"A function must assign exactly one element of B to every element of A. In A the element 1 has two images; in C the element 3 has no image; in D the element 3 has two images. Only B gives each of 1, 2, 3 exactly one image.","explanationZh":"函数要求 A 中每一个元素在 B 中都有唯一确定的像。A 中 1 有两个像；C 中 3 没有像；D 中 3 有两个像。只有 B 使 1、2、3 各有唯一像。","tags":["easy","function-concept","mapping"],"status":"published"},{"id":"math-domain-007","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":2,"stem":"What is the domain of $f(x) = \\log_{_{2}}(x − 1)$?","stemZh":"函数 $f(x) = \\log_{_{2}}(x − 1)$ 的定义域是什么？","options":[{"id":"A","text":"$[1, +∞)$"},{"id":"B","text":"$(−∞, 1)$"},{"id":"C","text":"$(1, +∞)$"},{"id":"D","text":"$(−∞, +∞)$"}],"answer":"C","explanation":"The argument of a logarithm must be strictly $positive: x − 1 > 0$, so $x > 1$. The endpoint $x = 1$ is excluded because $\\log_{_{2}}0$ is undefined.","explanationZh":"对数的真数必须严格大于 $0：x − 1 > 0，即 x > 1。端点 x = 1$ 不能取，因为 $\\log_{_{2}}0$ 无意义。","tags":["easy","domain","logarithm"],"status":"published"},{"id":"math-domain-008","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":2,"stem":"What is the domain of $f(x) = \\sqrt{4 − x}$?","stemZh":"函数 $f(x) = \\sqrt{4 − x}$ 的定义域是什么？","options":[{"id":"A","text":"$[4, +∞)$"},{"id":"B","text":"$(−∞, 4)$"},{"id":"C","text":"$(4, +∞)$"},{"id":"D","text":"$(−∞, 4]$"}],"answer":"D","explanation":"The radicand must be $non-negative: 4 − x ≥ 0$, so $x ≤ 4$. The endpoint $x = 4$ is included, giving $(−∞, 4]$.","explanationZh":"$被开方数必须非负：4 − x ≥ 0，即 x ≤ 4。端点 x = 4$ 可以取到，所以定义域为 $(−∞, 4]。$","tags":["easy","domain","radical"],"status":"published"},{"id":"math-domain-009","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":2,"stem":"What is the domain of $f(x) = \\log_{3}(2 − x)$?","stemZh":"函数 $f(x) = \\log_{3}(2 − x)$ 的定义域是什么？","options":[{"id":"A","text":"$(−∞, 2)$"},{"id":"B","text":"$(2, +∞)$"},{"id":"C","text":"$(−∞, 2]$"},{"id":"D","text":"$(−∞, +∞)$"}],"answer":"A","explanation":"The argument must satisfy $2 − x > 0$, so $x < 2$. Dividing by −1 reverses nothing here because the inequality is $strict: x < 2$, i.e. $(−∞, 2)$. The endpoint is not included.","explanationZh":"真数需满足 $2 − x > 0，即 x < 2，也就是 (−∞, 2)。端点不包含在内。$","tags":["easy","domain","logarithm"],"status":"published"},{"id":"math-domain-010","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":2,"stem":"What is the domain of $f(x) = (x − 3)/(x + 2)$?","stemZh":"函数 $f(x) = (x − 3)/(x + 2)$ 的定义域是什么？","options":[{"id":"A","text":"$x ≠ 3$"},{"id":"B","text":"$x ≠ −2$"},{"id":"C","text":"$x ≠ 3$ and $x ≠ −2$"},{"id":"D","text":"$x > −2$"}],"answer":"B","explanation":"Only the denominator $matters: x + 2 ≠ 0$, so $x ≠ −2$. The numerator may be zero (that only makes the value $0)$, so $x = 3$ is allowed.","explanationZh":"只需考虑分母：x $+ 2 ≠ 0，即 x ≠ −2。分子可以为零（此时函数值为 0），所以 x = 3$ 是允许的。","tags":["easy","domain","rational"],"status":"published"},{"id":"math-domain-011","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":3,"stem":"What is the domain of $f(x) = 1/\\sqrt{x − 1}$?","stemZh":"函数 $f(x) = 1/\\sqrt{x − 1}$ 的定义域是什么？","options":[{"id":"A","text":"$[1, +∞)$"},{"id":"B","text":"$[0, +∞)$"},{"id":"C","text":"$(1, +∞)$"},{"id":"D","text":"$(−∞, +∞)$"}],"answer":"C","explanation":"Two conditions apply at once: the radicand needs $x − 1 ≥ 0$, and the denominator must not be zero, so $\\sqrt{x − 1} ≠ 0$, i.e. $x − 1 > 0$. Hence $x > 1$, giving $(1, +∞)$. Forgetting the denominator condition leads to the wrong answer $[1, +∞)$.","explanationZh":"这里有两个限制：被开方数需 $x − 1 ≥ 0，同时分母不能为零，即 \\sqrt{x − 1} ≠ 0，从而 x − 1 > 0。因此 x > 1，定义域为 (1, +∞)。若只想到根式条件会得到错误的 [1, +∞)。$","tags":["medium","domain","radical"],"status":"published"},{"id":"math-domain-012","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":3,"stem":"What is the domain of $f(x) = \\sqrt{x − 1} + \\sqrt{3 − x}$?","stemZh":"函数 $f(x) = \\sqrt{x − 1} + \\sqrt{3 − x}$ 的定义域是什么？","options":[{"id":"A","text":"$(1, 3)$"},{"id":"B","text":"$[1, +∞)$"},{"id":"C","text":"$(−∞, 3]$"},{"id":"D","text":"$[1, 3]$"}],"answer":"D","explanation":"Both radicals must be $defined: x − 1 ≥ 0$ gives $x ≥ 1$, and $3 − x ≥ 0$ gives $x ≤ 3$. The intersection is $1 ≤ x ≤ 3$, i.e. $[1, 3]$. Both endpoints are included since both radicands are then 0 or positive.","explanationZh":"两个根式都要有意义：x $− 1 ≥ 0 得 x ≥ 1，3 − x ≥ 0 得 x ≤ 3。取交集得 1 ≤ x ≤ 3，即 [1, 3]。两端点处被开方数分别取 0$ 或正数，都应包含。","tags":["medium","domain","radical"],"status":"published"},{"id":"math-domain-013","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":3,"stem":"What is the domain of $f(x) = \\log_{_{2}}(x^{2} − 4)$?","stemZh":"函数 $f(x) = \\log_{_{2}}(x^{2} − 4)$ 的定义域是什么？","options":[{"id":"A","text":"$(−∞, −2) ∪ (2, +∞)$"},{"id":"B","text":"$(−∞, −2] ∪ [2, +∞)$"},{"id":"C","text":"$(−2, 2)$"},{"id":"D","text":"$[−2, 2]$"}],"answer":"A","explanation":"The argument must be strictly $positive: x^{2} − 4 > 0$, so $(x − 2)(x + 2) > 0$, giving $x < −2$ or $x > 2$. The endpoints $\\pm2$ make the argument 0, where the logarithm is undefined, so they are excluded.","explanationZh":"真数必须严格大于 $0：x^{2} − 4 > 0，即 (x − 2)(x + 2) > 0，解得 x < −2 或 x > 2。端点 \\pm2$ 使真数为 0，对数无意义，故不含端点。","tags":["medium","domain","logarithm"],"status":"published"},{"id":"math-domain-014","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":3,"stem":"What is the domain of $f(x) = \\sqrt{x^{2} − 4}$?","stemZh":"函数 $f(x) = \\sqrt{x^{2} − 4}$ 的定义域是什么？","options":[{"id":"A","text":"$(−∞, −2) ∪ (2, +∞)$"},{"id":"B","text":"$(−∞, −2] ∪ [2, +∞)$"},{"id":"C","text":"$[−2, 2]$"},{"id":"D","text":"$[2, +∞)$"}],"answer":"B","explanation":"The radicand must be $non-negative: x^{2} − 4 ≥ 0$, so $x ≤ −2$ or $x ≥ 2$. Unlike the logarithm case, the endpoints are included because $\\sqrt{0} = 0$ is defined, giving $(−∞, −2] ∪ [2, +∞)$.","explanationZh":"$被开方数需非负：x^{2} − 4 ≥ 0，解得 x ≤ −2 或 x ≥ 2。与对数情形不同，这里端点要取，因为 \\sqrt{0} = 0$ 有意义，故为 $(−∞, −2] ∪ [2, +∞)。$","tags":["medium","domain","radical"],"status":"published"},{"id":"math-domain-015","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":3,"stem":"What is the domain of $f(x) = 1/(x − 1) + \\sqrt{x + 2}$?","stemZh":"函数 $f(x) = 1/(x − 1) + \\sqrt{x + 2}$ 的定义域是什么？","options":[{"id":"A","text":"$[−2, +∞)$"},{"id":"B","text":"$(−2, 1) ∪ (1, +∞)$"},{"id":"C","text":"$[−2, 1] ∪ (1, +∞)$"},{"id":"D","text":"$[−2, 1) ∪ (1, +∞)$"}],"answer":"D","explanation":"The radical needs $x + 2 ≥ 0$, so $x ≥ −2$; the fraction needs $x − 1 ≠ 0$, so $x ≠ 1$. Intersecting gives $[−2, 1) ∪ (1, +∞)$. Note that $x = −2$ is allowed $(\\sqrt{0} = 0)$ but $x = 1$ is not.","explanationZh":"根式要求 $x + 2 ≥ 0，即 x ≥ −2；分式要求 x − 1 ≠ 0，即 x ≠ 1。取交集得 [−2, 1) ∪ (1, +∞)。注意 x = −2 可以取（\\sqrt{0} = 0），但 x = 1$ 不能取。","tags":["medium","domain","combined"],"status":"published"},{"id":"math-domain-016","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":3,"stem":"What is the domain of $f(x) = \\sqrt{\\log_{2}(x − 1)}$?","stemZh":"函数 $f(x) = \\sqrt{\\log_{2}(x − 1)}$ 的定义域是什么？","options":[{"id":"A","text":"$(1, +∞)$"},{"id":"B","text":"$[1, +∞)$"},{"id":"C","text":"$[2, +∞)$"},{"id":"D","text":"$(2, +∞)$"}],"answer":"C","explanation":"Two nested conditions: the logarithm needs $x − 1 > 0$, and the square root needs $\\log_{_{2}}(x − 1) ≥ 0$. Since the base $2 > 1, \\log_{_{2}}(x − 1) ≥ 0$ means $x − 1 ≥ 1$, i.e. $x ≥ 2$. This automatically satisfies $x > 1$, so the domain is $[2, +∞)$.","explanationZh":"这里有两层限制：对数要求 $x − 1 > 0，根式要求 \\log_{_{2}}(x − 1) ≥ 0。由于底数 2 > 1，\\log_{_{2}}(x − 1) ≥ 0$ 等价于 $x − 1 ≥ 1，即 x ≥ 2。这已蕴含 x > 1，故定义域为 [2, +∞)。$","tags":["medium","domain","logarithm"],"status":"published"},{"id":"math-domain-017","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":3,"stem":"What is the domain of $f(x) = \\log(3 − x)$ with base $(x − 1)$?","stemZh":"函数 $f(x) = 以 (x − 1)$ 为底、真数为 $(3 − x)$ 的对数的定义域是什么？","options":[{"id":"A","text":"$(1, 2) ∪ (2, 3)$"},{"id":"B","text":"$(1, 3)$"},{"id":"C","text":"$[1, 2) ∪ (2, 3]$"},{"id":"D","text":"$(1, 2) ∪ (2, +∞)$"}],"answer":"A","explanation":"A logarithm needs three things: the base is positive $(x − 1 > 0,\\text{ so }x > 1)$, the base is not $1 (x − 1 ≠ 1,\\text{ so }x ≠ 2)$, and the argument is positive $(3 − x > 0,\\text{ so }x < 3). Together: (1, 2) ∪ (2, 3)$.","explanationZh":"对数有三条要求：底数大于 $0（x − 1 > 0，即 x > 1）、底数不等于 1（x − 1 ≠ 1，即 x ≠ 2）、真数大于 0（3 − x > 0，即 x < 3）。综合得 (1, 2) ∪ (2, 3)。$","tags":["medium","domain","logarithm"],"status":"published"},{"id":"math-domain-018","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":3,"stem":"What is the domain of $f(x) = \\sqrt{x − 2}/(x − 3)$?","stemZh":"函数 $f(x) = \\sqrt{x − 2}/(x − 3)$ 的定义域是什么？","options":[{"id":"A","text":"$[2, +∞)$"},{"id":"B","text":"$[2, 3) ∪ (3, +∞)$"},{"id":"C","text":"$(2, 3) ∪ (3, +∞)$"},{"id":"D","text":"$[3, +∞)$"}],"answer":"B","explanation":"The radicand gives $x − 2 ≥ 0$, so $x ≥ 2$; the denominator gives $x ≠ 3. Intersecting: [2, 3) ∪ (3, +∞)$. The endpoint $x = 2$ is kept because $\\sqrt{0} = 0$ is allowed.","explanationZh":"被开方数要求 $x − 2 ≥ 0，即 x ≥ 2；分母要求 x ≠ 3。取交集得 [2, 3) ∪ (3, +∞)。端点 x = 2$ 保留，因为 $\\sqrt{0} = 0$ 有意义。","tags":["medium","domain","combined"],"status":"published"},{"id":"math-domain-019","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":3,"stem":"What is the domain of $f(x) = \\tan$ x?","stemZh":"函数 $f(x) = \\tan x$ 的定义域是什么？","options":[{"id":"A","text":"$\\{x \\mid x \\neq \\pi/2 + k\\pi,\\; k \\in \\mathbb{Z}\\}$"},{"id":"B","text":"${x | x ≠\\text{ kπ, }k ∈ ℤ}$"},{"id":"C","text":"${x | x ≠ π/2}$"},{"id":"D","text":"All real numbers"}],"answer":"A","explanation":"Since $\\tan x = \\frac{\\sin x}{\\cos x}$, we need $\\cos x \\neq 0$. The cosine is zero at $x = \\pi/2 + k\\pi$ for every integer k, so those values — and only those — are excluded.","explanationZh":"因为 $\\tan x = \\frac{\\sin x}{\\cos x}，需要 \\cos x \\neq 0。余弦在 x = \\pi/2 + k\\pi（k \\in \\mathbb{Z}）处为零，因此恰好这些值要去掉。$","tags":["medium","domain","trigonometric"],"status":"published"},{"id":"math-domain-020","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":3,"stem":"A rectangle has perimeter 20 cm. If one side has length x cm, its area is $A(x) = x(10 − x)$. What is the domain of A in this context?","stemZh":"矩形周长为 20 cm。若一边长为 x cm，则面积 $A(x) = x(10 − x)。在该实际背景下 A$ 的定义域是什么？","options":[{"id":"A","text":"$[0, 10]$"},{"id":"B","text":"$(0, 10)$"},{"id":"C","text":"$(0, +∞)$"},{"id":"D","text":"$(−∞, 10)$"}],"answer":"B","explanation":"The formula alone is a polynomial, but in context both side lengths must be $positive: x > 0$ and $10 − x > 0$, so $0 < x < 10$. The endpoints are excluded because a side of length 0 gives no rectangle.","explanationZh":"单看解析式是多项式，但实际背景下两边长都必须为正：x $> 0 且 10 − x > 0，即 0 < x < 10。端点不取，因为边长为 0$ 时不成矩形。","tags":["medium","domain","context"],"status":"published"},{"id":"math-domain-021","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":4,"stem":"What is the domain of $f(x) = \\sqrt{4 − x^{2}} + 1/(x − 1)$?","stemZh":"函数 $f(x) = \\sqrt{4 − x^{2}} + 1/(x − 1)$ 的定义域是什么？","options":[{"id":"A","text":"$[−2, 2]$"},{"id":"B","text":"$[−2, 1) ∪ (1, 2)$"},{"id":"C","text":"$[−2, 1) ∪ (1, 2]$"},{"id":"D","text":"$(−2, 1) ∪ (1, 2]$"}],"answer":"C","explanation":"The radical needs $4 − x^{2} ≥ 0$, i.e. $x^{2} ≤ 4$, so $−2 ≤ x ≤ 2$. The fraction needs $x ≠ 1$. Removing $x = 1$ from $[−2, 2]$ leaves $[−2, 1) ∪ (1, 2]$; the endpoints −2 and 2 are still allowed because $\\sqrt{0} = 0$.","explanationZh":"根式要求 $4 − x^{2} ≥ 0，即 x^{2} ≤ 4，得 −2 ≤ x ≤ 2；分式要求 x ≠ 1。从 [−2, 2]$ 中去掉 $x = 1，得 [−2, 1) ∪ (1, 2]；端点 −2 和 2$ 仍可取，因为 $\\sqrt{0} = 0。$","tags":["hard","domain","combined"],"status":"published"},{"id":"math-domain-022","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":4,"stem":"What is the domain of $f(x) = \\log_{_{2}}[(x − 1)/(x + 2)]$?","stemZh":"函数 $f(x) = \\log_{_{2}}[(x − 1)/(x + 2)]$ 的定义域是什么？","options":[{"id":"A","text":"$(−2, 1)$"},{"id":"B","text":"$(−∞, −2) ∪ [1, +∞)$"},{"id":"C","text":"$(−∞, −2] ∪ (1, +∞)$"},{"id":"D","text":"$(−∞, −2) ∪ (1, +∞)$"}],"answer":"D","explanation":"The whole fraction is the logarithm's argument, so it must be strictly $positive: (x − 1)/(x + 2) > 0$. The critical values are −2 and 1; testing intervals gives $x < −2$ or $x > 1$. The endpoints are excluded $(at x = 1\\text{ the argument is }0,\\text{ at }x = −2\\text{ it is }undefined)$.","explanationZh":"整个分式是对数的真数，必须严格大于 $0：(x − 1)/(x + 2) > 0。临界点为 −2 与 1，分区间讨论得 x < −2 或 x > 1。两端点均不含（x = 1$ 时真数为 $0，x = −2$ 时无意义）。","tags":["hard","domain","logarithm"],"status":"published"},{"id":"math-domain-023","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":4,"stem":"What is the domain of $f(x) = \\sqrt{[(x − 1)/(x + 2)]}$?","stemZh":"函数 $f(x) = \\sqrt{[(x − 1)/(x + 2)]}$ 的定义域是什么？","options":[{"id":"A","text":"$(−∞, −2) ∪ (1, +∞)$"},{"id":"B","text":"$(−∞, −2) ∪ [1, +∞)$"},{"id":"C","text":"$(−∞, −2] ∪ [1, +∞)$"},{"id":"D","text":"$[−2, 1]$"}],"answer":"B","explanation":"Now the fraction is a radicand, so it must be $non-negative: (x − 1)/(x + 2) ≥ 0$, giving $x < −2$ or $x ≥ 1$. Compared with the logarithm $version, x = 1$ is now included (the radicand is $0)$ but $x = −2$ is still excluded (the denominator vanishes).","explanationZh":"$此时分式是被开方数，需非负：(x − 1)/(x + 2) ≥ 0，解得 x < −2 或 x ≥ 1。与对数情形相比，x = 1$ 现在可以取（被开方数为 $0），但 x = −2$ 仍不可取（分母为零）。","tags":["hard","domain","combined"],"status":"published"},{"id":"math-domain-024","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":4,"stem":"What is the domain of $f(x) = 1/[\\log_{_{2}}(x − 1)]$?","stemZh":"函数 $f(x) = 1/[\\log_{_{2}}(x − 1)]$ 的定义域是什么？","options":[{"id":"A","text":"$(1, 2) ∪ (2, +∞)$"},{"id":"B","text":"$(1, +∞)$"},{"id":"C","text":"$(2, +∞)$"},{"id":"D","text":"$[1, 2) ∪ (2, +∞)$"}],"answer":"A","explanation":"The logarithm needs $x − 1 > 0$, so $x > 1$. In addition the denominator must not be $zero: \\log_{_{2}}(x − 1) ≠ 0$, so $x − 1 ≠ 1$, i.e. $x ≠ 2. Combining: (1, 2) ∪ (2, +∞)$.","explanationZh":"对数要求 $x − 1 > 0，即 x > 1；同时分母不能为零：\\log_{_{2}}(x − 1) ≠ 0，即 x − 1 ≠ 1，x ≠ 2。综合得 (1, 2) ∪ (2, +∞)。$","tags":["hard","domain","logarithm"],"status":"published"},{"id":"math-domain-025","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":5,"stem":"What is the domain of $f(x) = \\sqrt{x − 1}/(x − 2) + \\log_3(4 − x)$?","stemZh":"函数 $f(x) = \\sqrt{x − 1}/(x − 2) + \\log_3(4 − x)$ 的定义域是什么？","options":[{"id":"A","text":"$[1, 4)$"},{"id":"B","text":"$(1, 2) ∪ (2, 4)$"},{"id":"C","text":"$[1, 2) ∪ (2, 4)$"},{"id":"D","text":"$[1, 2) ∪ (2, 4]$"}],"answer":"C","explanation":"Three conditions must hold $simultaneously: x − 1 ≥ 0 (so x ≥ 1), x − 2 ≠ 0 (so x ≠ 2)$, and $4 − x > 0 (so x < 4)$. The intersection is $[1, 2) ∪ (2, 4)$. Watch the $endpoints: x = 1$ is included $(\\sqrt{0} = 0)$, while $x = 4$ is not $(\\log_30 undefined)$.","explanationZh":"三个条件需同时满足：x $− 1 ≥ 0（x ≥ 1）、x − 2 ≠ 0（x ≠ 2）、4 − x > 0（x < 4）。取交集得 [1, 2) ∪ (2, 4)。注意端点：x = 1 要取（\\sqrt{0} = 0），x = 4 不取（\\log_30$ 无意义）。","tags":["hard","domain","combined"],"status":"published"},{"id":"math-domain-026","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":4,"stem":"What is the domain of $f(x) = (x − 1)^{0} + \\sqrt{x + 1}$?","stemZh":"函数 $f(x) = (x − 1)^{0} + \\sqrt{x + 1}$ 的定义域是什么？","options":[{"id":"A","text":"$[−1, +∞)$"},{"id":"B","text":"$(−1, 1) ∪ (1, +∞)$"},{"id":"C","text":"$[−1, 1] ∪ (1, +∞)$"},{"id":"D","text":"$[−1, 1) ∪ (1, +∞)$"}],"answer":"D","explanation":"The zero power $(x − 1)^{0}$ requires a non-zero $base: x − 1 ≠ 0$, so $x ≠ 1$. The radical requires $x + 1 ≥ 0$, so $x ≥ −1$. Intersecting gives $[−1, 1) ∪ (1, +∞)$.","explanationZh":"零次幂 $(x − 1)^{0}$ 要求底数不为零：x $− 1 ≠ 0，即 x ≠ 1；根式要求 x + 1 ≥ 0，即 x ≥ −1。取交集得 [−1, 1) ∪ (1, +∞)。$","tags":["hard","domain","combined"],"status":"published"},{"id":"math-domain-027","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":4,"stem":"The graph of $y = f(x)$ consists of the line segment joining $(−2, 1)$ to $(1, 4)$, together with the single point $(3, 2)$. What is the domain of f?","stemZh":"函数 $y = f(x)$ 的图象由连接 $(−2, 1) 与 (1, 4)$ 的线段以及单独一点 $(3, 2)$ 组成。f 的定义域是什么？","options":[{"id":"A","text":"$[−2, 3]$"},{"id":"B","text":"$[−2, 1] ∪ {3}$"},{"id":"C","text":"$[1, 4] ∪ {2}$"},{"id":"D","text":"$[−2, 1) ∪ (1, 3]$"}],"answer":"B","explanation":"The domain is the set of x-values covered by the graph. The segment covers every x from −2 to 1 inclusive, and the isolated point contributes $x = 3 only$, so the domain is $[−2, 1] ∪ {3}$. Option C is the range, not the domain.","explanationZh":"定义域是图象覆盖到的 x 的取值集合。线段覆盖 −2 到 1（含端点）的全部 x，孤立点只贡献 $x = 3，故定义域为 [−2, 1] ∪ {3}。选项 C$ 是值域而非定义域。","tags":["hard","domain","graph"],"status":"published"},{"id":"math-domain-028","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":5,"stem":"The function f is defined on $[−1, 3]$. What is the domain of $g(x) = f(x^{2})$?","stemZh":"函数 f 的定义域为 $[−1, 3]。则 g(x) = f(x^{2})$ 的定义域是什么？","options":[{"id":"A","text":"$[−\\sqrt{3}, \\sqrt{3}]$"},{"id":"B","text":"$[−1, \\sqrt{3}]$"},{"id":"C","text":"$[0, \\sqrt{3}]$"},{"id":"D","text":"$[−3, 3]$"}],"answer":"A","explanation":"For $g(x) = f(x^{2})$ to be defined, the input $x^{2}$ must lie in the domain of f, i.e. $−1 ≤ x^{2} ≤ 3$. Since $x^{2} ≥ 0 always$, the left inequality is automatic and only $x^{2} ≤ 3 matters$, giving $|x| ≤ \\sqrt{3}$, i.e. $[−\\sqrt{3}, \\sqrt{3}]$.","explanationZh":"要使 $g(x) = f(x^{2})$ 有意义，需 $x^{2}$ 落在 f 的定义域内，即 $−1 ≤ x^{2} ≤ 3。由于 x^{2} ≥ 0$ 恒成立，左边不等式自动满足，只需 $x^{2} ≤ 3，得 |x| ≤ \\sqrt{3}，即 [−\\sqrt{3}, \\sqrt{3}]。$","tags":["hard","domain","composition"],"status":"published"},{"id":"math-domain-029","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":4,"stem":"What is the domain of $f(x) = \\sqrt{4 − |x|} + \\log_{2}x$?","stemZh":"函数 $f(x) = \\sqrt{4 − |x|} + \\log_{2}x$ 的定义域是什么？","options":[{"id":"A","text":"$[−4, 4]$"},{"id":"B","text":"$[0, 4]$"},{"id":"C","text":"$(0, 4]$"},{"id":"D","text":"$(0, 4)$"}],"answer":"C","explanation":"The radical needs $4 − |x| ≥ 0$, so $|x| ≤ 4$, i.e. $−4 ≤ x ≤ 4$. The logarithm needs $x > 0$. Intersecting gives $0 < x ≤ 4$, i.e. $(0, 4]: x = 0$ is excluded by the logarithm and $x = 4$ is allowed by the radical.","explanationZh":"根式要求 $4 − |x| ≥ 0，即 |x| ≤ 4，−4 ≤ x ≤ 4；对数要求 x > 0。取交集得 0 < x ≤ 4，即 (0, 4]：x = 0$ 被对数排除，x $= 4$ 对根式允许。","tags":["hard","domain","combined"],"status":"published"},{"id":"math-domain-030","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"domain","type":"single-choice","difficulty":5,"stem":"What is the domain of $f(x) = \\sqrt{x^{2} − 2x − 3} + \\log_{2}(6 − x − x^{2})$?","stemZh":"函数 $f(x) = \\sqrt{x^{2} − 2x − 3} + \\log_{2}(6 − x − x^{2})$ 的定义域是什么？","options":[{"id":"A","text":"$[−3, −1]$"},{"id":"B","text":"$(−3, −1)$"},{"id":"C","text":"$(−∞, −1] ∪ [3, +∞)$"},{"id":"D","text":"$(−3, −1]$"}],"answer":"D","explanation":"$First: x^{2} − 2x − 3 ≥ 0$ gives $(x − 3)(x + 1) ≥ 0$, so $x ≤ −1$ or $x ≥ 3. Second: 6 − x − x^{2} > 0$ gives $x^{2} + x − 6 < 0$, so $−3 < x < 2$. The intersection of $(−∞, −1] ∪ [3, +∞)$ with $(−3, 2)$ is $(−3, −1]$.","explanationZh":"$其一：x^{2} − 2x − 3 ≥ 0 得 (x − 3)(x + 1) ≥ 0，即 x ≤ −1 或 x ≥ 3。其二：6 − x − x^{2} > 0 得 x^{2} + x − 6 < 0，即 −3 < x < 2。两集合交集为 (−3, −1]。$","tags":["hard","domain","combined"],"status":"published"},{"id":"math-range-001","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":1,"stem":"What is the range of $y = x^{2} + 1$?","stemZh":"函数 $y = x^{2} + 1$ 的值域是什么？","options":[{"id":"A","text":"$[1, +∞)$"},{"id":"B","text":"$(1, +∞)$"},{"id":"C","text":"$[0, +∞)$"},{"id":"D","text":"$(−∞, +∞)$"}],"answer":"A","explanation":"Since $x^{2} ≥ 0$ for every real x, we have $y = x^{2} + 1 ≥ 1$, and the value 1 is attained at $x = 0$. Every value above 1 is also attained, so the range is $[1, +∞)$.","explanationZh":"因为对任意实数 x 都有 $x^{2} ≥ 0，所以 y = x^{2} + 1 ≥ 1，且 x = 0$ 时取到 1。大于 1 的值也都能取到，故值域为 $[1, +∞)。$","tags":["easy","range","quadratic"],"status":"published"},{"id":"math-range-002","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":1,"stem":"What is the range of $y = |x|$?","stemZh":"函数 $y = |x|$ 的值域是什么？","options":[{"id":"A","text":"$(0, +∞)$"},{"id":"B","text":"$[0, +∞)$"},{"id":"C","text":"$(−∞, 0]$"},{"id":"D","text":"$(−∞, +∞)$"}],"answer":"B","explanation":"An absolute value is never negative, and $|0| = 0$, so the smallest value is 0 and it is attained. The range is $[0, +∞)$.","explanationZh":"绝对值恒非负，且 $|0| = 0，所以最小值为 0$ 且能取到，值域为 $[0, +∞)。$","tags":["easy","range","absolute-value"],"status":"published"},{"id":"math-range-003","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":1,"stem":"What is the range of $y = 2^{x}$?","stemZh":"函数 $y = 2^{x}$ 的值域是什么？","options":[{"id":"A","text":"$[0, +∞)$"},{"id":"B","text":"$[1, +∞)$"},{"id":"C","text":"$(0, +∞)$"},{"id":"D","text":"$(−∞, +∞)$"}],"answer":"C","explanation":"An exponential with positive base is always strictly positive and never reaches 0: as $x → −∞$ it approaches 0 but stays above it. Hence the range is $(0, +∞)$.","explanationZh":"底数为正的指数函数恒大于 0 且取不到 $0：当 x → −∞$ 时趋近于 0 但始终大于 $0。故值域为 (0, +∞)。$","tags":["easy","range","exponential"],"status":"published"},{"id":"math-range-004","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":1,"stem":"What is the range of $y = \\log_{_{2}}x$?","stemZh":"函数 $y = \\log_{_{2}}x$ 的值域是什么？","options":[{"id":"A","text":"$(0, +∞)$"},{"id":"B","text":"$[0, +∞)$"},{"id":"C","text":"$[1, +∞)$"},{"id":"D","text":"$(−∞, +∞)$"}],"answer":"D","explanation":"As x runs over $(0, +∞)$, the logarithm takes every real value: it tends to $−∞$ as $x → 0^{+}$ and to $+∞$ as $x → +∞$. So the range is all real $numbers, (−∞, +∞)$.","explanationZh":"当 x 取遍 $(0, +∞)$ 时，对数可取遍一切实数：x $→ 0^{+}$ 时趋于 $−∞，x → +∞$ 时趋于 $+∞。故值域为全体实数 (−∞, +∞)。$","tags":["easy","range","logarithm"],"status":"published"},{"id":"math-range-005","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":1,"stem":"What is the range of $y = \\sin$ x?","stemZh":"函数 $y = \\sin x$ 的值域是什么？","options":[{"id":"A","text":"$[−1, 1]$"},{"id":"B","text":"$(−1, 1)$"},{"id":"C","text":"$[0, 1]$"},{"id":"D","text":"$[−1, 0]$"}],"answer":"A","explanation":"The sine function oscillates between −1 and 1 and attains both extremes $(at x = −π/2 + 2kπ\\text{ and }x = π/2 + 2kπ)$. So the range is $[−1, 1]$.","explanationZh":"正弦函数在 −1 与 1 之间振荡，并且两个最值都能取到（分别在 $x = −π/2 + 2kπ 与 x = π/2 + 2kπ$ 处），故值域为 $[−1, 1]。$","tags":["easy","range","trigonometric"],"status":"published"},{"id":"math-range-006","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":2,"stem":"What is the range of $y = −x^{2}$?","stemZh":"函数 $y = −x^{2}$ 的值域是什么？","options":[{"id":"A","text":"$(−∞, 0)$"},{"id":"B","text":"$(−∞, 0]$"},{"id":"C","text":"$[0, +∞)$"},{"id":"D","text":"$(−∞, +∞)$"}],"answer":"B","explanation":"Since $x^{2} ≥ 0$, we get $−x^{2} ≤ 0$, with equality at $x = 0$. The parabola opens downward with maximum 0, so the range is $(−∞, 0]$.","explanationZh":"因为 $x^{2} ≥ 0，所以 −x^{2} ≤ 0，且 x = 0$ 时取等号。抛物线开口向下、最大值为 $0，故值域为 (−∞, 0]。$","tags":["easy","range","quadratic"],"status":"published"},{"id":"math-range-007","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":2,"stem":"What is the range of $y = \\sqrt{x} + 1$?","stemZh":"函数 $y = \\sqrt{x} + 1$ 的值域是什么？","options":[{"id":"A","text":"$[0, +∞)$"},{"id":"B","text":"$(1, +∞)$"},{"id":"C","text":"$[1, +∞)$"},{"id":"D","text":"$[−1, +∞)$"}],"answer":"C","explanation":"The square root satisfies $\\sqrt{x} ≥ 0$ on its domain $x ≥ 0$, so $y ≥ 1$, with $y = 1$ attained at $x = 0$. The range is $[1, +∞)$. Option A is the domain, not the range.","explanationZh":"在定义域 $x ≥ 0$ 上有 $\\sqrt{x} ≥ 0，故 y ≥ 1，且 x = 0$ 时取到 $y = 1，值域为 [1, +∞)。选项 A$ 是定义域而非值域。","tags":["easy","range","radical"],"status":"published"},{"id":"math-range-008","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":2,"stem":"What is the range of $y = x^{2} − 2x + 3$?","stemZh":"函数 $y = x^{2} − 2x + 3$ 的值域是什么？","options":[{"id":"A","text":"$[0, +∞)$"},{"id":"B","text":"$[3, +∞)$"},{"id":"C","text":"$[1, +∞)$"},{"id":"D","text":"$[2, +∞)$"}],"answer":"D","explanation":"Complete the $square: y = (x − 1)^{2} + 2$. The parabola opens upward with vertex $(1, 2)$, so the minimum value is 2 and the range is $[2, +∞)$.","explanationZh":"配方得 $y = (x − 1)^{2} + 2。抛物线开口向上，顶点为 (1, 2)，最小值为 2，故值域为 [2, +∞)。$","tags":["easy","range","quadratic"],"status":"published"},{"id":"math-range-009","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":2,"stem":"What is the range of $y = −2x^{2} + 4x − 1$?","stemZh":"函数 $y = −2x^{2} + 4x − 1$ 的值域是什么？","options":[{"id":"A","text":"$(−∞, 1]$"},{"id":"B","text":"$(−∞, 1)$"},{"id":"C","text":"$[1, +∞)$"},{"id":"D","text":"$[−1, +∞)$"}],"answer":"A","explanation":"Complete the $square: y = −2(x − 1)^{2} + 1$. Since $a = −2 < 0$ the parabola opens downward with vertex $(1, 1)$, so the maximum is 1 and the range is $(−∞, 1]$.","explanationZh":"配方得 $y = −2(x − 1)^{2} + 1。由于 a = −2 < 0，抛物线开口向下，顶点为 (1, 1)，最大值为 1，故值域为 (−∞, 1]。$","tags":["easy","range","quadratic"],"status":"published"},{"id":"math-range-010","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":2,"stem":"What is the range of $y = 1/x$?","stemZh":"函数 $y = 1/x$ 的值域是什么？","options":[{"id":"A","text":"$[0, +∞)$"},{"id":"B","text":"$(−∞, 0) ∪ (0, +∞)$"},{"id":"C","text":"$(−∞, +∞)$"},{"id":"D","text":"$(−∞, 1) ∪ (1, +∞)$"}],"answer":"B","explanation":"A fraction with non-zero numerator can never be 0, so $y ≠ 0$. Every other real value is attained $(take x = 1/y)$. Hence the range is $(−∞, 0) ∪ (0, +∞)$.","explanationZh":"分子不为零的分式恒不为 $0，故 y ≠ 0；其余实数都能取到（取 x = 1/y$ 即可）。因此值域为 $(−∞, 0) ∪ (0, +∞)。$","tags":["easy","range","rational"],"status":"published"},{"id":"math-range-011","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":3,"stem":"What is the range of $y = x^{2} − 2x − 3$?","stemZh":"函数 $y = x^{2} − 2x − 3$ 的值域是什么？","options":[{"id":"A","text":"$[−3, +∞)$"},{"id":"B","text":"$[0, +∞)$"},{"id":"C","text":"$[−4, +∞)$"},{"id":"D","text":"$(−∞, −4]$"}],"answer":"C","explanation":"Complete the $square: y = (x − 1)^{2} − 4$. The vertex is $(1, −4)$ and the parabola opens upward, so the minimum is −4 and the range is $[−4, +∞)$. Using the constant term −3 instead of the vertex value is a common slip.","explanationZh":"配方得 $y = (x − 1)^{2} − 4。顶点为 (1, −4)，开口向上，最小值是 −4，值域为 [−4, +∞)。误用常数项 −3$ 是常见错误。","tags":["medium","range","quadratic"],"status":"published"},{"id":"math-range-012","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":3,"stem":"What is the range of $y = x^{2} − 4x + 1$ on the domain $[0, 3]$?","stemZh":"函数 $y = x^{2} − 4x + 1$ 在定义域 $[0, 3]$ 上的值域是什么？","options":[{"id":"A","text":"$[−3, −2]$"},{"id":"B","text":"$[−2, 1]$"},{"id":"C","text":"$[−3, +∞)$"},{"id":"D","text":"$[−3, 1]$"}],"answer":"D","explanation":"The vertex is at $x = 2$, which lies in $[0, 3]$, giving the minimum $4 − 8 + 1 = −3$. At the $endpoints, y(0) = 1$ and $y(3) = −2$, so the maximum is 1. The range is $[−3, 1]$.","explanationZh":"顶点在 $x = 2，落在 [0, 3]$ 内，最小值为 $4 − 8 + 1 = −3；端点处 y(0) = 1，y(3) = −2，故最大值为 1。值域为 [−3, 1]。$","tags":["medium","range","quadratic"],"status":"published"},{"id":"math-range-013","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":3,"stem":"What is the range of $y = 2^{x} + 1$?","stemZh":"函数 $y = 2^{x} + 1$ 的值域是什么？","options":[{"id":"A","text":"$(1, +∞)$"},{"id":"B","text":"$(0, +∞)$"},{"id":"C","text":"$[1, +∞)$"},{"id":"D","text":"$(−∞, +∞)$"}],"answer":"A","explanation":"Since $2^{x} > 0$, we have $y > 1$. The horizontal asymptote moves up with the $graph: y = 1$ is approached as $x → −∞$ but never reached, so the range is $(1, +∞)$, not $[1, +∞)$.","explanationZh":"$由 2^{x} > 0 得 y > 1。渐近线随图象一起上移：x → −∞ 时 y$ 趋近 1 但取不到，故值域为 $(1, +∞)，而非 [1, +∞)。$","tags":["medium","range","exponential"],"status":"published"},{"id":"math-range-014","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":3,"stem":"What is the range of $y = −2^{x}$?","stemZh":"函数 $y = −2^{x}$ 的值域是什么？","options":[{"id":"A","text":"$(−∞, 0]$"},{"id":"B","text":"$(−∞, 0)$"},{"id":"C","text":"$(0, +∞)$"},{"id":"D","text":"$(−∞, +∞)$"}],"answer":"B","explanation":"Since $2^{x} > 0$, we have $y = −2^{x} < 0$. As $x → −∞, 2^{x} → 0$, so $y → 0$ from below but never equals 0. The range is $(−∞, 0)$, open at 0.","explanationZh":"$由 2^{x} > 0 得 y = −2^{x} < 0。当 x → −∞ 时 2^{x} → 0，y$ 从下方趋近 0 但取不到 $0。故值域为 (−∞, 0)，0$ 处是开区间。","tags":["medium","range","exponential"],"status":"published"},{"id":"math-range-015","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":3,"stem":"What is the range of $y = |x| − 2$?","stemZh":"函数 $y = |x| − 2$ 的值域是什么？","options":[{"id":"A","text":"$[0, +∞)$"},{"id":"B","text":"$(−2, +∞)$"},{"id":"C","text":"$[−2, +∞)$"},{"id":"D","text":"$[2, +∞)$"}],"answer":"C","explanation":"Since $|x| ≥ 0$, we get $y ≥ −2$, and the value −2 is attained at $x = 0$. The graph is the V-shape $y = |x|$ shifted down by 2, so the range is $[−2, +∞)$.","explanationZh":"$由 |x| ≥ 0 得 y ≥ −2，且 x = 0$ 时取到 $−2。图象是 y = |x|$ 向下平移 2 个单位，故值域为 $[−2, +∞)。$","tags":["medium","range","absolute-value"],"status":"published"},{"id":"math-range-016","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":3,"stem":"What is the range of $y = 1/(x^{2} + 1)$?","stemZh":"函数 $y = 1/(x^{2} + 1)$ 的值域是什么？","options":[{"id":"A","text":"$(0, 1]$"},{"id":"B","text":"$[0, 1]$"},{"id":"C","text":"$(0, +∞)$"},{"id":"D","text":"$[1, +∞)$"}],"answer":"A","explanation":"The denominator satisfies $x^{2} + 1 ≥ 1$, so $0 < y ≤ 1$, with the maximum $y = 1$ attained at $x = 0$. As $|x| → ∞$ the value tends to 0 but never equals it, so 0 is $excluded: (0, 1]$.","explanationZh":"分母满足 $x^{2} + 1 ≥ 1，故 0 < y ≤ 1，最大值 y = 1 在 x = 0$ 处取到。当 $|x| → ∞$ 时函数值趋近 0 但取不到 0，故 0 不含，值域为 $(0, 1]。$","tags":["medium","range","rational"],"status":"published"},{"id":"math-range-017","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":3,"stem":"What is the range of $y = 3 − 2\\sin$ x?","stemZh":"函数 $y = 3 − 2\\sin x$ 的值域是什么？","options":[{"id":"A","text":"$[−1, 1]$"},{"id":"B","text":"$[3, 5]$"},{"id":"C","text":"$[2, 4]$"},{"id":"D","text":"$[1, 5]$"}],"answer":"D","explanation":"Since $\\sin x ∈ [−1, 1]$, we have $−2\\sin x ∈ [−2, 2]$, so $y = 3 − 2\\sin x ∈ [1, 5]$. Both endpoints occur $(at \\sin x = 1$ and $\\sin x = −1 respectively)$.","explanationZh":"$由 \\sin x ∈ [−1, 1] 得 −2\\sin x ∈ [−2, 2]，故 y = 3 − 2\\sin x ∈ [1, 5]。两端点都能取到（分别对应 \\sin x = 1 与 \\sin x = −1）。$","tags":["medium","range","trigonometric"],"status":"published"},{"id":"math-range-018","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":3,"stem":"What is the range of $y = 1/(x^{2} + 2)$?","stemZh":"函数 $y = 1/(x^{2} + 2)$ 的值域是什么？","options":[{"id":"A","text":"$(0, 1/2)$"},{"id":"B","text":"$(0, 1/2]$"},{"id":"C","text":"$[0, 1/2]$"},{"id":"D","text":"$(0, +∞)$"}],"answer":"B","explanation":"The denominator satisfies $x^{2} + 2 ≥ 2$, so $0 < y ≤ 1/2$, with the maximum 1/2 attained at $x = 0$. The value 0 is only a limit as $|x| → ∞$, so the range is $(0, 1/2]$.","explanationZh":"分母满足 $x^{2} + 2 ≥ 2，故 0 < y ≤ 1/2，最大值 1/2 在 x = 0 处取到。0$ 只是 $|x| → ∞$ 时的极限值，取不到，故值域为 $(0, 1/2]。$","tags":["medium","range","rational"],"status":"published"},{"id":"math-range-019","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":3,"stem":"What is the range of $y = 4 − \\sqrt{x + 1}$?","stemZh":"函数 $y = 4 − \\sqrt{x + 1}$ 的值域是什么？","options":[{"id":"A","text":"$[4, +∞)$"},{"id":"B","text":"$(−∞, 4)$"},{"id":"C","text":"$(−∞, 4]$"},{"id":"D","text":"$[0, +∞)$"}],"answer":"C","explanation":"On its domain $x ≥ −1$ we have $\\sqrt{x + 1} ≥ 0$, so $y ≤ 4$, with the maximum 4 attained at $x = −1$. As $x → +∞$ the value tends to $−∞$, so the range is $(−∞, 4]$.","explanationZh":"在定义域 $x ≥ −1 上 \\sqrt{x + 1} ≥ 0，故 y ≤ 4，最大值 4 在 x = −1$ 处取到。x $→ +∞$ 时函数值趋于 $−∞，故值域为 (−∞, 4]。$","tags":["medium","range","radical"],"status":"published"},{"id":"math-range-020","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":3,"stem":"Let $f(x) = x^{2}$ for $−2 ≤ x < 0$, and $f(x) = 2 − x$ for $0 ≤ x ≤ 3$. What is the range of f?","stemZh":"$设 f(x) = x^{2}（−2 ≤ x < 0），f(x) = 2 − x（0 ≤ x ≤ 3）。f$ 的值域是什么？","options":[{"id":"A","text":"$[−1, 4]$"},{"id":"B","text":"$(0, 4]$"},{"id":"C","text":"$[−1, 2]$"},{"id":"D","text":"$[0, 4]$"}],"answer":"A","explanation":"On $[−2, 0)$ the values of $x^{2}$ run over $(0, 4]$. On $[0, 3]$ the expression 2 − x decreases from 2 to −1, giving $[−1, 2]$. The union $(0, 4] ∪ [−1, 2]$ is $[−1, 4]$.","explanationZh":"$在 [−2, 0) 上 x^{2}$ 的取值为 $(0, 4]；在 [0, 3] 上 2 − x 从 2$ 递减到 $−1，取值为 [−1, 2]。取并集 (0, 4] ∪ [−1, 2] = [−1, 4]。$","tags":["medium","range","piecewise"],"status":"published"},{"id":"math-range-021","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":4,"stem":"What is the range of $y = (x − 1)/(x + 1)$?","stemZh":"函数 $y = (x − 1)/(x + 1)$ 的值域是什么？","options":[{"id":"A","text":"$(−∞, +∞)$"},{"id":"B","text":"$(−∞, −1) ∪ (−1, +∞)$"},{"id":"C","text":"$[1, +∞)$"},{"id":"D","text":"$(−∞, 1) ∪ (1, +∞)$"}],"answer":"D","explanation":"Rewrite $y = (x + 1 − 2)/(x + 1) = 1 − 2/(x + 1)$. Since $2/(x + 1)$ is never 0, y can never equal 1, and every other real value is attained. Hence the range is $(−∞, 1) ∪ (1, +∞)$.","explanationZh":"改写为 $y = (x + 1 − 2)/(x + 1) = 1 − 2/(x + 1)。由于 2/(x + 1)$ 恒不为 0，y 永远取不到 $1，而其余实数都能取到。故值域为 (−∞, 1) ∪ (1, +∞)。$","tags":["hard","range","rational"],"status":"published"},{"id":"math-range-022","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":5,"stem":"What is the range of $y = x + 1/x$?","stemZh":"函数 $y = x + 1/x$ 的值域是什么？","options":[{"id":"A","text":"$[2, +∞)$"},{"id":"B","text":"$(−∞, −2] ∪ [2, +∞)$"},{"id":"C","text":"$(−∞, −2) ∪ (2, +∞)$"},{"id":"D","text":"$(−∞, +∞)$"}],"answer":"B","explanation":"For $x > 0$, AM–GM gives $x + 1/x ≥ 2$, with equality at $x = 1$. For $x < 0$, writing $x =$ −t with $t > 0$ gives $y = −(t + 1/t) ≤ −2$, with equality at $x = −1$. The two branches together give $(−∞, −2] ∪ [2, +∞)$.","explanationZh":"$当 x > 0$ 时，由均值不等式 $x + 1/x ≥ 2，x = 1$ 时取等号；当 $x < 0$ 时，令 $x =$ −t（t $> 0），则 y = −(t + 1/t) ≤ −2，x = −1$ 时取等号。两支合并得 $(−∞, −2] ∪ [2, +∞)。$","tags":["hard","range","rational"],"status":"published"},{"id":"math-range-023","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":5,"stem":"What is the range of $y = 2x/(x^{2} + 1)$?","stemZh":"函数 $y = 2x/(x^{2} + 1)$ 的值域是什么？","options":[{"id":"A","text":"$(−1, 1)$"},{"id":"B","text":"$[0, 1]$"},{"id":"C","text":"$[−1, 1]$"},{"id":"D","text":"$(−∞, +∞)$"}],"answer":"C","explanation":"Since $x^{2} + 1 ≥ 2|x| (equivalently (|x| − 1)^{2} ≥ 0)$, we get $|y| = 2|x|/(x^{2} + 1) ≤ 1$. Equality holds at $x = 1 (y = 1)$ and $x = −1 (y = −1)$, and all values in between are attained, so the range is $[−1, 1]$.","explanationZh":"$由 x^{2} + 1 ≥ 2|x|（等价于 (|x| − 1)^{2} ≥ 0）得 |y| = 2|x|/(x^{2} + 1) ≤ 1。x = 1 时 y = 1，x = −1 时 y = −1，均取到，中间值也都能取到，故值域为 [−1, 1]。$","tags":["hard","range","rational"],"status":"published"},{"id":"math-range-024","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":4,"stem":"What is the range of $y = −x^{2} + 2x + 3$ on the domain $[0, 3]$?","stemZh":"函数 $y = −x^{2} + 2x + 3$ 在定义域 $[0, 3]$ 上的值域是什么？","options":[{"id":"A","text":"$[0, 4]$"},{"id":"B","text":"$[3, 4]$"},{"id":"C","text":"$(−∞, 4]$"},{"id":"D","text":"$[0, 3]$"}],"answer":"A","explanation":"The vertex is at $x = 1 (inside [0, 3])$ with value $−1 + 2 + 3 = 4$, the maximum. At the endpoints $y(0) = 3$ and $y(3) = −9 + 6 + 3 = 0$, so the minimum is 0. The range is $[0, 4]$.","explanationZh":"顶点在 $x = 1（位于 [0, 3]$ 内），值为 $−1 + 2 + 3 = 4，是最大值；端点处 y(0) = 3，y(3) = −9 + 6 + 3 = 0，故最小值为 0。值域为 [0, 4]。$","tags":["hard","range","quadratic"],"status":"published"},{"id":"math-range-025","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":4,"stem":"What is the range of $y = \\sqrt{x^{2} − 4x + 5}$?","stemZh":"函数 $y = \\sqrt{x^{2} − 4x + 5}$ 的值域是什么？","options":[{"id":"A","text":"$[0, +∞)$"},{"id":"B","text":"$[\\sqrt{5}, +∞)$"},{"id":"C","text":"$(1, +∞)$"},{"id":"D","text":"$[1, +∞)$"}],"answer":"D","explanation":"Complete the square $inside: x^{2} − 4x + 5 = (x − 2)^{2} + 1 ≥ 1$. So the radicand is always at least 1 and the square root is at least 1, attained at $x = 2$. The range is $[1, +∞)$.","explanationZh":"$根号内配方：x^{2} − 4x + 5 = (x − 2)^{2} + 1 ≥ 1。因此被开方数不小于 1，平方根不小于 1，且在 x = 2$ 处取到 $1。值域为 [1, +∞)。$","tags":["hard","range","radical"],"status":"published"},{"id":"math-range-026","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":4,"stem":"What is the range of $y = 2^{x}$ on the restricted domain $x ≤ 0$?","stemZh":"函数 $y = 2^{x}$ 在限定定义域 $x ≤ 0$ 上的值域是什么？","options":[{"id":"A","text":"$(0, +∞)$"},{"id":"B","text":"$(0, 1]$"},{"id":"C","text":"$[0, 1]$"},{"id":"D","text":"$(−∞, 1]$"}],"answer":"B","explanation":"On $x ≤ 0$ the exponential is increasing, so $0 < 2^{x} ≤ 2^{0} = 1$. The value 1 is attained at $x = 0$, while 0 is only approached as $x → −∞$. The range is $(0, 1]$.","explanationZh":"$在 x ≤ 0$ 上指数函数单调递增，故 $0 < 2^{x} ≤ 2^{0} = 1。x = 0$ 时取到 1，而 0 只是 $x → −∞$ 时的极限，取不到。值域为 $(0, 1]。$","tags":["hard","range","exponential"],"status":"published"},{"id":"math-range-027","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":4,"stem":"What is the range of $y = \\log_{_{2}}x$ on the domain $[1, 8]$?","stemZh":"函数 $y = \\log_{_{2}}x$ 在定义域 $[1, 8]$ 上的值域是什么？","options":[{"id":"A","text":"$[0, 3]$"},{"id":"B","text":"$(0, 3)$"},{"id":"C","text":"$[1, 3]$"},{"id":"D","text":"$[0, +∞)$"}],"answer":"A","explanation":"The base $2 > 1$, so $\\log_{_{2}}x$ is increasing: the minimum is $\\log_{_{2}}1 = 0$ and the maximum is $\\log_{_{2}}8 = 3$. Both endpoints are attained, giving $[0, 3]$.","explanationZh":"底数 $2 > 1，\\log_{_{2}}x$ 单调递增：最小值 $\\log_{_{2}}1 = 0，最大值 \\log_{_{2}}8 = 3。两个端点都能取到，故值域为 [0, 3]。$","tags":["hard","range","logarithm"],"status":"published"},{"id":"math-range-028","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":4,"stem":"What is the range of $y = \\cos x$ on the interval $[π/3, π]$?","stemZh":"函数 $y = \\cos x$ 在区间 $[π/3, π]$ 上的值域是什么？","options":[{"id":"A","text":"$[−1, 1]$"},{"id":"B","text":"$[1/2, 1]$"},{"id":"C","text":"$[−1, −1/2]$"},{"id":"D","text":"$[−1, 1/2]$"}],"answer":"D","explanation":"Cosine is decreasing on $[0, π]$, so on $[π/3, π]$ its maximum is $\\cos(π/3) = 1/2$ and its minimum is $\\cos π = −1$. Both are attained, so the range is $[−1, 1/2]$.","explanationZh":"余弦在 $[0, π]$ 上单调递减，因此在 $[π/3, π]$ 上最大值是 $\\cos(π/3) = 1/2，最小值是 \\cos π = −1，二者都能取到，值域为 [−1, 1/2]。$","tags":["hard","range","trigonometric"],"status":"published"},{"id":"math-range-029","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":5,"stem":"What is the range of $y = |x − 1| + |x + 1|$?","stemZh":"函数 $y = |x − 1| + |x + 1|$ 的值域是什么？","options":[{"id":"A","text":"$[0, +∞)$"},{"id":"B","text":"$[1, +∞)$"},{"id":"C","text":"$[2, +∞)$"},{"id":"D","text":"$(2, +∞)$"}],"answer":"C","explanation":"The expression is the sum of the distances from x to 1 and to −1 on the number line. By the triangle inequality this sum is at least the distance between −1 and 1, which is 2, with equality for every x in $[−1, 1]$. Hence the range is $[2, +∞)$.","explanationZh":"该式表示数轴上 x 到 1 与到 −1 的距离之和。由三角不等式，和不小于 −1 与 1 之间的距离 $2，且当 x ∈ [−1, 1]$ 时取等号。故值域为 $[2, +∞)。$","tags":["hard","range","absolute-value"],"status":"published"},{"id":"math-range-030","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"range","type":"single-choice","difficulty":5,"stem":"What is the range of $y = 4^{x} − 2^{x}$?","stemZh":"函数 $y = 4^{x} − 2^{x}$ 的值域是什么？","options":[{"id":"A","text":"$[−1/2, +∞)$"},{"id":"B","text":"$[−1/4, +∞)$"},{"id":"C","text":"$(−1/4, +∞)$"},{"id":"D","text":"$[0, +∞)$"}],"answer":"B","explanation":"Substitute $t = 2^{x}$, so $t > 0$ and $y = t^{2} − t = (t − 1/2)^{2} − 1/4$. The minimum −1/4 occurs at $t = 1/2$, i.e. $2^{x} = 1/2$, so $x = −1$. As $t → 0^{+}$ or $t → +∞$ the value grows without bound, so the range is $[−1/4, +∞)$.","explanationZh":"$令 t = 2^{x}，则 t > 0 且 y = t^{2} − t = (t − 1/2)^{2} − 1/4。最小值 −1/4 在 t = 1/2 即 2^{x} = 1/2（x = −1）时取得。当 t → 0^{+} 或 t → +∞$ 时函数值无界增大，故值域为 $[−1/4, +∞)。$","tags":["hard","range","exponential"],"status":"published"},{"id":"math-monotonicity-001","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":1,"stem":"Which of the following functions is increasing on $ℝ$?","stemZh":"下列函数中，在 $ℝ$ 上单调递增的是？","options":[{"id":"A","text":"$y = −2x + 1$"},{"id":"B","text":"$y = x^{2}$"},{"id":"C","text":"$y = (1/3)^x$"},{"id":"D","text":"$y = 3x − 1$"}],"answer":"D","explanation":"$y = 3x − 1$ has slope $3 > 0$, so it increases on $ℝ. y = −2x + 1$ has negative $slope, y = x^{2}$ decreases then increases, and $y = (1/3)^x$ decreases because its base lies in $(0, 1)$.","explanationZh":"$y = 3x − 1$ 的斜率为 $3 > 0，故在 ℝ$ 上单调递增；y $= −2x + 1$ 斜率为负；y $= x^{2}$ 先减后增；y $= (1/3)^x$ 底数在 $(0,1)$ 内，单调递减。","tags":["easy","monotonicity","linear"],"status":"published"},{"id":"math-monotonicity-002","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":2,"stem":"On $ℝ$, the function $y =$ −x + 4 is:","stemZh":"函数 $y =$ −x $+ 4 在 ℝ$ 上：","options":[{"id":"A","text":"Increasing"},{"id":"B","text":"Constant"},{"id":"C","text":"Decreasing"},{"id":"D","text":"Increasing on $(−∞, 0]$ and decreasing on $[0, +∞)$"}],"answer":"C","explanation":"The slope is $−1 < 0$, so $y =$ −x + 4 is strictly decreasing on the whole real line.","explanationZh":"斜率为 $−1 < 0，故 y =$ −x + 4 在整个实数集上严格单调递减。","tags":["easy","monotonicity","linear"],"status":"published"},{"id":"math-monotonicity-003","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":1,"stem":"For the exponential function $y = a^x (a > 0, a ≠ 1)$, it is decreasing on $ℝ when$:","stemZh":"指数函数 $y = a^x（a > 0，a ≠ 1）在 ℝ$ 上单调递减的条件是：","options":[{"id":"A","text":"$0 < a < 1$"},{"id":"B","text":"$a > 1$"},{"id":"C","text":"$a < 0$"},{"id":"D","text":"$a = 1$"}],"answer":"A","explanation":"$y = a^x$ increases when $a > 1$ and decreases when $0 < a < 1$; the base must be positive and not equal to 1.","explanationZh":"$a > 1 时 y = a^x 单调递增，0 < a < 1$ 时单调递减；底数必须为正且不等于 1。","tags":["easy","monotonicity","exponential"],"status":"published"},{"id":"math-monotonicity-004","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":2,"stem":"On $ℝ$, the function $y = 2^x is$:","stemZh":"函数 $y = 2^x 在 ℝ$ 上：","options":[{"id":"A","text":"Increasing"},{"id":"B","text":"Decreasing"},{"id":"C","text":"Decreasing on $(−∞, 0]$ and increasing on $[0, +∞)$"},{"id":"D","text":"Neither increasing nor decreasing"}],"answer":"A","explanation":"The base $2 > 1$, so $y = 2^x$ is strictly increasing on all of $ℝ$.","explanationZh":"底数 $2 > 1，故 y = 2^x$ 在整个 $ℝ$ 上严格单调递增。","tags":["easy","monotonicity","exponential"],"status":"published"},{"id":"math-monotonicity-005","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":1,"stem":"On its domain $(0, +∞)$, the function $y = \\log_{_{2}}x is$:","stemZh":"函数 $y = \\log_{_{2}}x$ 在其定义域 $(0, +∞)$ 上：","options":[{"id":"A","text":"Decreasing"},{"id":"B","text":"Increasing"},{"id":"C","text":"Neither increasing nor decreasing"},{"id":"D","text":"Increasing on $(0, 1)$ and decreasing on $(1, +∞)$"}],"answer":"B","explanation":"The base $2 > 1$, so $y = \\log_{_{2}}x$ is strictly increasing on $(0, +∞)$.","explanationZh":"底数 $2 > 1，故 y = \\log_{_{2}}x 在 (0, +∞)$ 上严格单调递增。","tags":["easy","monotonicity","logarithm"],"status":"published"},{"id":"math-monotonicity-006","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":2,"stem":"On its domain $(0, +∞)$, the function $y = \\log_{1/2}x is$:","stemZh":"函数 $y = \\log_{1/2}x$ 在其定义域 $(0, +∞)$ 上：","options":[{"id":"A","text":"Increasing"},{"id":"B","text":"Neither increasing nor decreasing"},{"id":"C","text":"Increasing on $(0, 1)$ and decreasing on $(1, +∞)$"},{"id":"D","text":"Decreasing"}],"answer":"D","explanation":"The base 1/2 lies in $(0, 1)$, so $y = \\log_{1/2}x$ is strictly decreasing on $(0, +∞)$.","explanationZh":"底数 $1/2 ∈ (0,1)，故 y = \\log_{1/2}x 在 (0, +∞)$ 上严格单调递减。","tags":["easy","monotonicity","logarithm"],"status":"published"},{"id":"math-monotonicity-007","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":1,"stem":"The function $y = x^{2}$ is decreasing on:","stemZh":"函数 $y = x^{2}$ 的单调递减区间是：","options":[{"id":"A","text":"$[0, +∞)$"},{"id":"B","text":"$(0, +∞)$"},{"id":"C","text":"All of $ℝ$"},{"id":"D","text":"$(−∞, 0]$"}],"answer":"D","explanation":"The parabola $y = x^{2}$ has vertex axis $x = 0$ and opens upward: it decreases on $(−∞, 0]$ and increases on $[0, +∞)$.","explanationZh":"抛物线 $y = x^{2}$ 开口向上、对称轴为 $x = 0，在 (−∞, 0]$ 上递减，在 $[0, +∞)$ 上递增。","tags":["easy","monotonicity","quadratic"],"status":"published"},{"id":"math-monotonicity-008","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":2,"stem":"The increasing interval of $y = x^{2} − 2x is$:","stemZh":"函数 $y = x^{2} − 2x$ 的单调递增区间是：","options":[{"id":"A","text":"$(−∞, 1]$"},{"id":"B","text":"$(−∞, 0]$"},{"id":"C","text":"$[1, +∞)$"},{"id":"D","text":"$[0, +∞)$"}],"answer":"C","explanation":"$y = (x − 1)^{2} − 1$ has axis $x = 1$ and opens upward, so it decreases on $(−∞, 1]$ and increases on $[1, +∞)$.","explanationZh":"$y = (x − 1)^{2} − 1$ 开口向上、对称轴 $x = 1，故在 (−∞, 1] 上递减、[1, +∞)$ 上递增。","tags":["easy","monotonicity","quadratic"],"status":"published"},{"id":"math-monotonicity-009","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":1,"stem":"Which statement about $y = x^{3}$ on $ℝ$ is correct?","stemZh":"关于 $y = x^{3} 在 ℝ$ 上的单调性，正确的是：","options":[{"id":"A","text":"It is decreasing"},{"id":"B","text":"It is increasing"},{"id":"C","text":"It is constant"},{"id":"D","text":"It decreases first and then increases"}],"answer":"B","explanation":"For $x_{1} < x_{2}$ we always have $x_{1}^{3} < x_{2}^{3}$, so $y = x^{3}$ is strictly increasing on $ℝ$.","explanationZh":"$当 x_{1} < x_{2}$ 时恒有 $x_{1}^{3} < x_{2}^{3}，故 y = x^{3} 在 ℝ$ 上严格单调递增。","tags":["easy","monotonicity","power"],"status":"published"},{"id":"math-monotonicity-010","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":2,"stem":"Which statement about $y = 1/x$ is correct?","stemZh":"关于 $y = 1/x$ 的单调性，正确的是：","options":[{"id":"A","text":"It decreases on $(−∞, 0)$ and decreases on $(0, +∞)$"},{"id":"B","text":"It decreases on $(−∞, 0) ∪ (0, +∞)$"},{"id":"C","text":"It increases on $(−∞, 0)$ and increases on $(0, +∞)$"},{"id":"D","text":"It increases on $ℝ \\ {0}$"}],"answer":"A","explanation":"On each branch 1/x decreases. It is NOT decreasing on the union: take $x_{1} = −1 < x_{2} = 1$, then $f(−1) = −1 < 1 = f(1)$, which contradicts decreasing. Monotonicity is only stated separately on each interval.","explanationZh":"$y = 1/x$ 在每一支上分别递减，但不能说在 $(−∞,0) ∪ (0,+∞)$ 上递减：取 $x_{1} = −1 < x_{2} = 1，有 f(−1) = −1 < 1 = f(1)，与递减矛盾。单调区间必须分开写。$","tags":["easy","monotonicity","rational"],"status":"published"},{"id":"math-monotonicity-011","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":3,"stem":"The increasing interval of $f(x) = −x^{2} + 4x − 1 is$:","stemZh":"函数 $f(x) = −x^{2} + 4x − 1$ 的单调递增区间是：","options":[{"id":"A","text":"$[2, +∞)$"},{"id":"B","text":"$(−∞, 2]$"},{"id":"C","text":"$(−∞, 4]$"},{"id":"D","text":"$[4, +∞)$"}],"answer":"B","explanation":"The parabola opens downward and its axis is $x = 2$, so f increases on $(−∞, 2]$ and decreases on $[2, +∞)$.","explanationZh":"抛物线开口向下，对称轴 $x = 2，故 f 在 (−∞, 2]$ 上递增、在 $[2, +∞)$ 上递减。","tags":["medium","monotonicity","quadratic"],"status":"published"},{"id":"math-monotonicity-012","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":3,"stem":"f is increasing on $ℝ$. If $f(2a − 1) < f(a + 3), then$:","stemZh":"$f 在 ℝ$ 上单调递增，若 $f(2a − 1) < f(a + 3)，则：$","options":[{"id":"A","text":"$a > 4$"},{"id":"B","text":"$a < 2$"},{"id":"C","text":"$a > 2$"},{"id":"D","text":"$a < 4$"}],"answer":"D","explanation":"An increasing function preserves order of the inputs, so $2a − 1 < a + 3$, which gives $a < 4$.","explanationZh":"增函数保持自变量的大小顺序，故 $2a − 1 < a + 3，解得 a < 4。$","tags":["medium","monotonicity","comparison"],"status":"published"},{"id":"math-monotonicity-013","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":3,"stem":"$f(x) = x^{2} − 2ax + 1$ is increasing on $[1, +∞)$. What is the range of a?","stemZh":"$f(x) = x^{2} − 2ax + 1 在 [1, +∞)$ 上单调递增，则 a 的取值范围是：","options":[{"id":"A","text":"$a ≥ 1$"},{"id":"B","text":"$a ≥ 2$"},{"id":"C","text":"$a ≤ 1$"},{"id":"D","text":"$a ≤ 0$"}],"answer":"C","explanation":"$f(x) = (x − a)^{2} + 1 − a^{2}$ increases on $[a, +∞)$. For $[1, +∞)$ to lie inside that interval we need $a ≤ 1$.","explanationZh":"$f(x) = (x − a)^{2} + 1 − a^{2}$ 的递增区间为 $[a, +∞)，要使 [1, +∞)$ 包含于其中，需 $a ≤ 1。$","tags":["medium","monotonicity","parameter"],"status":"published"},{"id":"math-monotonicity-014","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":3,"stem":"The function $y = \\sqrt{x − 1}$ is increasing on:","stemZh":"函数 $y = \\sqrt{x − 1}$ 的单调递增区间是：","options":[{"id":"A","text":"$[1, +∞)$"},{"id":"B","text":"$(1, +∞)$"},{"id":"C","text":"$(−∞, 1]$"},{"id":"D","text":"All of $ℝ$"}],"answer":"A","explanation":"The domain is $x ≥ 1$, and $\\sqrt{t}$ increases with t, so the function increases on its whole domain $[1, +∞)$.","explanationZh":"定义域为 $x ≥ 1，且 \\sqrt{t} 随 t$ 递增，故函数在定义域 $[1, +∞)$ 上递增。","tags":["medium","monotonicity","radical"],"status":"published"},{"id":"math-monotonicity-015","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":3,"stem":"Which statement about $y = −1/x$ is correct?","stemZh":"关于 $y = −1/x$ 的单调性，正确的是：","options":[{"id":"A","text":"It decreases on $(−∞, 0)$ and decreases on $(0, +∞)$"},{"id":"B","text":"It increases on $(−∞, 0) ∪ (0, +∞)$"},{"id":"C","text":"It increases on $(−∞, 0)$ and increases on $(0, +∞)$"},{"id":"D","text":"It decreases on $(−∞, 0)$ and increases on $(0, +∞)$"}],"answer":"C","explanation":"For $x_{1} < x_{2}$ inside one $branch, −1/x_{1} < −1/x_{2}$, so the function increases on each of $(−∞, 0)$ and $(0, +∞) —$ again the two intervals must be stated separately, not as a union.","explanationZh":"在同一支上 $x_{1} < x_{2}$ 时有 $−1/x_{1} < −1/x_{2}，故在 (−∞,0) 与 (0,+∞)$ 上分别递增，仍须分开写而不能写成并集。","tags":["medium","monotonicity","rational"],"status":"published"},{"id":"math-monotonicity-016","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":3,"stem":"The decreasing interval of $f(x) = |x − 1| is$:","stemZh":"函数 $f(x) = |x − 1|$ 的单调递减区间是：","options":[{"id":"A","text":"$[1, +∞)$"},{"id":"B","text":"$(−∞, 1]$"},{"id":"C","text":"$[0, +∞)$"},{"id":"D","text":"$(−∞, 0]$"}],"answer":"B","explanation":"$f(x) = 1 − x$ for $x ≤ 1$ and $f(x) = x − 1$ for $x ≥ 1$, so the graph falls up to $x = 1$ and rises afterwards.","explanationZh":"$x ≤ 1 时 f(x) = 1 −$ x，x $≥ 1 时 f(x) = x − 1，故在 (−∞, 1]$ 上递减，在 $[1, +∞)$ 上递增。","tags":["medium","monotonicity","absolute-value"],"status":"published"},{"id":"math-monotonicity-017","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":3,"stem":"The decreasing interval of $y = 2^{x^{2} − 2x} is$:","stemZh":"函数 $y = 2^(x^{2} − 2x)$ 的单调递减区间是：","options":[{"id":"A","text":"$[1, +∞)$"},{"id":"B","text":"$[0, +∞)$"},{"id":"C","text":"$(−∞, 0]$"},{"id":"D","text":"$(−∞, 1]$"}],"answer":"D","explanation":"Since $y = 2^t$ increases with t, the composite decreases exactly where $t = x^{2} − 2x decreases$, that is on $(−∞, 1]$.","explanationZh":"$因 y = 2^t$ 关于 t 递增，故复合函数的单调性由 $t = x^{2} − 2x$ 决定：t $在 (−∞, 1]$ 上递减，所以原函数在 $(−∞, 1]$ 上递减。","tags":["medium","monotonicity","composite"],"status":"published"},{"id":"math-monotonicity-018","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":3,"stem":"For $f(x) = x^{2} − 2x$ on the interval $[0, 3]$, which statement is true?","stemZh":"对于区间 $[0, 3]$ 上的 $f(x) = x^{2} − 2x，下列说法正确的是：$","options":[{"id":"A","text":"The minimum is 0, attained at $x = 0$"},{"id":"B","text":"The minimum is −1, attained at $x = 1$"},{"id":"C","text":"The minimum is −1, attained at $x = 3$"},{"id":"D","text":"The minimum is 3, attained at $x = 3$"}],"answer":"B","explanation":"$f(x) = (x − 1)^{2} − 1$ decreases on $[0, 1]$ and increases on $[1, 3]$, so the minimum is $f(1) = −1 (and f(3) = 3\\text{ is the }maximum)$.","explanationZh":"$f(x) = (x − 1)^{2} − 1 在 [0,1] 上递减、[1,3]$ 上递增，故最小值为 $f(1) = −1（f(3) = 3$ 为最大值）。","tags":["medium","monotonicity","quadratic"],"status":"published"},{"id":"math-monotonicity-019","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":3,"stem":"f is decreasing on $ℝ$. If $f(1 − a) > f(2a + 3), then$:","stemZh":"$f 在 ℝ$ 上单调递减，若 $f(1 − a) > f(2a + 3)，则：$","options":[{"id":"A","text":"$a < −2/3$"},{"id":"B","text":"$a > 2/3$"},{"id":"C","text":"$a > −2/3$"},{"id":"D","text":"$a < −4/3$"}],"answer":"C","explanation":"A decreasing function reverses the order of the inputs, so $1 − a < 2a + 3$, i.e. $3a > −2$ and $a > −2/3$.","explanationZh":"减函数使自变量大小关系反向，故 $1 − a < 2a + 3，即 3a > −2，得 a > −2/3。$","tags":["medium","monotonicity","comparison"],"status":"published"},{"id":"math-monotonicity-020","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":3,"stem":"A function f satisfies $f(−2) = 5, f(−1) = 3, f(0) = 1, f(1) = 3$ and $f(2) = 5$. Which conclusion is best supported by these values?","stemZh":"已知函数 f 满足 $f(−2) = 5，f(−1) = 3，f(0) = 1，f(1) = 3，f(2) = 5。由这些值最能得到的结论是：$","options":[{"id":"A","text":"f decreases on $[−2, 0]$ and increases on $[0, 2]$"},{"id":"B","text":"f increases on $[−2, 2]$"},{"id":"C","text":"f decreases on $[−2, 2]$"},{"id":"D","text":"f increases on $[−2, 0]$ and decreases on $[0, 2]$"}],"answer":"A","explanation":"The values fall from 5 to 1 as x goes from −2 to 0, then rise from 1 to 5 as x goes from 0 to 2, so f decreases on $[−2, 0]$ and increases on $[0, 2]$.","explanationZh":"x 由 −2 到 0 时函数值由 5 降到 1，x 由 0 到 2 时又由 1 升到 $5，故 f 在 [−2, 0]$ 上递减、在 $[0, 2]$ 上递增。","tags":["medium","monotonicity","graph"],"status":"published"},{"id":"math-monotonicity-021","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":4,"stem":"On $(0, +∞)$, the function $f(x) = x + 1/x is$:","stemZh":"函数 $f(x) = x + 1/x 在 (0, +∞)$ 上：","options":[{"id":"A","text":"Increasing"},{"id":"B","text":"Decreasing on $(0, 1]$ and increasing on $[1, +∞)$"},{"id":"C","text":"Decreasing"},{"id":"D","text":"Increasing on $(0, 1]$ and decreasing on $[1, +∞)$"}],"answer":"B","explanation":"$f′(x) = 1 − 1/x^{2}$ is negative on $(0, 1)$ and positive on $(1, +∞)$, so f decreases on $(0, 1]$, increases on $[1, +∞)$ and has minimum $f(1) = 2$.","explanationZh":"$f′(x) = 1 − 1/x^{2} 在 (0,1)$ 上为负、在 $(1,+∞)$ 上为正，故 $f 在 (0,1] 上递减、[1,+∞)$ 上递增，最小值为 $f(1) = 2。$","tags":["hard","monotonicity","rational"],"status":"published"},{"id":"math-monotonicity-022","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":4,"stem":"The increasing interval of $f(x) = \\log_{_{2}}(x^{2} − 2x − 3) is$:","stemZh":"函数 $f(x) = \\log_{_{2}}(x^{2} − 2x − 3)$ 的单调递增区间是：","options":[{"id":"A","text":"$(−∞, −1)$"},{"id":"B","text":"$(1, +∞)$"},{"id":"C","text":"$(3, +∞)$"},{"id":"D","text":"$(−∞, 1)$"}],"answer":"C","explanation":"$Domain: x^{2} − 2x − 3 > 0$ gives $x < −1$ or $x > 3. Inside, t = (x − 1)^{2} − 4$ increases for $x ≥ 1$ and $\\log_{_{2}}t$ increases in t, so f increases on $(3, +∞)$ and decreases on $(−∞, −1)$.","explanationZh":"$定义域：x^{2} − 2x − 3 > 0，得 x < −1 或 x > 3。内层 t = (x − 1)^{2} − 4 在 x ≥ 1$ 时递增，且 $\\log_{_{2}}t$ 关于 t 递增，故 $f 在 (3, +∞)$ 上递增，在 $(−∞, −1)$ 上递减。","tags":["hard","monotonicity","logarithm"],"status":"published"},{"id":"math-monotonicity-023","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":4,"stem":"The increasing interval of $y = (1/3)^{x^{2} − 2x} is$:","stemZh":"函数 $y = (1/3)^(x^{2} − 2x)$ 的单调递增区间是：","options":[{"id":"A","text":"$[1, +∞)$"},{"id":"B","text":"$[0, +∞)$"},{"id":"C","text":"$(−∞, 0]$"},{"id":"D","text":"$(−∞, 1]$"}],"answer":"D","explanation":"The base $1/3 < 1$, so $y = (1/3)^t$ decreases as t grows. Therefore y increases where $t = x^{2} − 2x decreases$, namely on $(−∞, 1]$.","explanationZh":"底数 $1/3 < 1，y = (1/3)^t$ 关于 t 递减，故原函数在 $t = x^{2} − 2x$ 递减的区间上递增，即 $(−∞, 1]。$","tags":["hard","monotonicity","composite"],"status":"published"},{"id":"math-monotonicity-024","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":5,"stem":"f is decreasing on $ℝ$. For every real number a, which relation always holds?","stemZh":"$f 在 ℝ$ 上单调递减，则对任意实数 a 恒有：","options":[{"id":"A","text":"$f(a^{2} + 1) ≥ f(2a)$"},{"id":"B","text":"$f(a^{2} + 1) > f(2a)$"},{"id":"C","text":"$f(a^{2} + 1) ≤ f(2a)$"},{"id":"D","text":"The order cannot be determined"}],"answer":"C","explanation":"$a^{2} + 1 − 2a = (a − 1)^{2} ≥ 0$, so $a^{2} + 1 ≥ 2a$. Since f decreases, the larger input gives the smaller $value: f(a^{2} + 1) ≤ f(2a)$, with equality only when $a = 1 (so\\text{ strict }>\\text{ never always }holds)$.","explanationZh":"$a^{2} + 1 − 2a = (a − 1)^{2} ≥ 0，故 a^{2} + 1 ≥ 2a。f$ 递减，自变量越大函数值越小，所以 $f(a^{2} + 1) ≤ f(2a)，仅当 a = 1$ 时取等号，因此严格大于不成立。","tags":["hard","monotonicity","comparison"],"status":"published"},{"id":"math-monotonicity-025","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":5,"stem":"$f(x) = x^{2} − 2(a − 1)x + 2$ is decreasing on $(−∞, 3]$. What is the range of a?","stemZh":"$f(x) = x^{2} − 2(a − 1)x + 2 在 (−∞, 3]$ 上单调递减，则 a 的取值范围是：","options":[{"id":"A","text":"$a ≤ 4$"},{"id":"B","text":"$a ≤ 3$"},{"id":"C","text":"$a ≥ 4$"},{"id":"D","text":"$a ≥ −4$"}],"answer":"C","explanation":"The axis is $x = a − 1$ and the parabola opens upward, so f decreases on $(−∞, a − 1]$. For f to decrease on all of $(−∞, 3]$ we need $3 ≤ a − 1$, i.e. $a ≥ 4$.","explanationZh":"对称轴为 $x = a − 1，开口向上，递减区间为 (−∞, a − 1]。要在 (−∞, 3]$ 上递减，需 $3 ≤ a − 1，即 a ≥ 4。$","tags":["hard","monotonicity","parameter"],"status":"published"},{"id":"math-monotonicity-026","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":5,"stem":"For $f(x) = \\log_a(2 − ax)$ to be decreasing on $[0, 1]$, the base a must lie in:","stemZh":"$若 f(x) = \\log_a(2 − ax) 在 [0, 1]$ 上单调递减，则底数 a 的取值范围是：","options":[{"id":"A","text":"$(0, 1)$"},{"id":"B","text":"$(2, +∞)$"},{"id":"C","text":"$(0, 2)$"},{"id":"D","text":"$(1, 2)$"}],"answer":"D","explanation":"The inner function 2 − ax is decreasing in $x (a > 0)$, so the composite decreases only when $\\log_a$ is increasing, i.e. $a > 1$. The domain condition 2 − ax $> 0$ on $[0, 1]$ gives $2 − a > 0$, i.e. $a < 2$. Hence $1 < a < 2$.","explanationZh":"内层 2 − ax（a $> 0）关于 x$ 递减，要使复合函数递减，外层 $\\log_a$ 必须递增，即 $a > 1；又由定义域 2 −$ ax $> 0 在 [0,1]$ 上恒成立得 $2 − a > 0，即 a < 2。综上 1 < a < 2。$","tags":["hard","monotonicity","logarithm"],"status":"published"},{"id":"math-monotonicity-027","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":4,"stem":"The increasing interval of $f(x) = \\sqrt{x^{2} − 2x − 3} is$:","stemZh":"函数 $f(x) = \\sqrt{x^{2} − 2x − 3}$ 的单调递增区间是：","options":[{"id":"A","text":"$(−∞, −1]$"},{"id":"B","text":"$[1, +∞)$"},{"id":"C","text":"$(−∞, 1]$"},{"id":"D","text":"$[3, +∞)$"}],"answer":"D","explanation":"The domain is $x^{2} − 2x − 3 ≥ 0$, i.e. $x ≤ −1$ or $x ≥ 3$. Since $\\sqrt{t}$ increases in t and $t = (x − 1)^{2} − 4$ increases for $x ≥ 1, f$ increases on $[3, +∞)$ and decreases on $(−∞, −1]$.","explanationZh":"定义域 $x^{2} − 2x − 3 ≥ 0，即 x ≤ −1 或 x ≥ 3。因 \\sqrt{t}$ 关于 t 递增且 $t = (x − 1)^{2} − 4 在 x ≥ 1$ 时递增，故 $f 在 [3, +∞)$ 上递增，在 $(−∞, −1]$ 上递减。","tags":["hard","monotonicity","radical"],"status":"published"},{"id":"math-monotonicity-028","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":5,"stem":"Which statement about $f(x) = x|x|$ on $ℝ$ is correct?","stemZh":"关于 $f(x) = x|x| 在 ℝ$ 上的单调性，正确的是：","options":[{"id":"A","text":"f is increasing on $ℝ$"},{"id":"B","text":"f is decreasing on $(−∞, 0]$ and increasing on $[0, +∞)$"},{"id":"C","text":"f is decreasing on $ℝ$"},{"id":"D","text":"f is increasing on $(−∞, 0]$ and decreasing on $[0, +∞)$"}],"answer":"A","explanation":"For $x < 0, f(x) = −x^{2}$ whose derivative −2x is positive; for $x ≥ 0, f(x) = x^{2}$ whose derivative 2x is non-negative. Being continuous at 0, f increases on the whole of $ℝ$.","explanationZh":"$x < 0 时 f(x) = −x^{2}，导数 −2x > 0；x ≥ 0 时 f(x) = x^{2}，导数 2x ≥ 0；且 f 在 0$ 处连续，故 f 在整个 $ℝ$ 上递增。","tags":["hard","monotonicity","absolute-value"],"status":"published"},{"id":"math-monotonicity-029","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":4,"stem":"The increasing interval of $f(x) = \\log_{1/2}(x^{2} − 2x) is$:","stemZh":"函数 $f(x) = \\log_{1/2}(x^{2} − 2x)$ 的单调递增区间是：","options":[{"id":"A","text":"$(−∞, 0)$"},{"id":"B","text":"$(2, +∞)$"},{"id":"C","text":"$(−∞, 1)$"},{"id":"D","text":"$(1, 2)$"}],"answer":"A","explanation":"$Domain: x^{2} − 2x > 0$ gives $x < 0$ or $x > 2$. With base $1/2 < 1$ the logarithm decreases in its argument, so f increases where $t = (x − 1)^{2} − 1 decreases$, i.e. on $(−∞, 1]$; intersected with the domain this gives $(−∞, 0)$.","explanationZh":"$定义域：x^{2} − 2x > 0，得 x < 0 或 x > 2。底数 1/2 < 1，对数关于真数递减，故 f 在 t = (x − 1)^{2} − 1$ 递减处递增，即 $(−∞, 1]，与定义域取交集得 (−∞, 0)。$","tags":["hard","monotonicity","logarithm"],"status":"published"},{"id":"math-monotonicity-030","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"monotonicity","type":"single-choice","difficulty":4,"stem":"If f is increasing on $ℝ$ and g is decreasing on $ℝ$, then $h(x) = f(g(x)) is$:","stemZh":"$若 f 在 ℝ$ 上递增、g $在 ℝ$ 上递减，则 $h(x) = f(g(x))：$","options":[{"id":"A","text":"Increasing on $ℝ$"},{"id":"B","text":"Decreasing on $ℝ$"},{"id":"C","text":"Impossible to determine"},{"id":"D","text":"Increasing on some intervals and decreasing on others"}],"answer":"B","explanation":"Take $x_{1} < x_{2}$. Since $g decreases, g(x_{1}) > g(x_{2})$; since $f increases, f(g(x_{1})) > f(g(x_{2}))$. Hence $h(x_{1}) > h(x_{2})$ and h is decreasing on $ℝ$.","explanationZh":"任取 $x_{1} < x_{2}，由 g$ 递减得 $g(x_{1}) > g(x_{2})，再由 f$ 递增得 $f(g(x_{1})) > f(g(x_{2}))，故 h 在 ℝ$ 上递减。","tags":["hard","monotonicity","composite"],"status":"published"},{"id":"math-parity-001","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":1,"stem":"The function $f(x) = x^{2} is$:","stemZh":"函数 $f(x) = x^{2}$ 是：","options":[{"id":"A","text":"Odd"},{"id":"B","text":"Even"},{"id":"C","text":"Neither odd nor even"},{"id":"D","text":"Both odd and even"}],"answer":"B","explanation":"The domain $ℝ$ is symmetric about 0 and $f(−x) = (−x)^{2} = x^{2} = f(x)$, so f is even.","explanationZh":"定义域 $ℝ$ 关于原点对称，且 $f(−x) = (−x)^{2} = x^{2} = f(x)，故为偶函数。$","tags":["easy","parity","power"],"status":"published"},{"id":"math-parity-002","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":1,"stem":"The function $f(x) = x^{3} is$:","stemZh":"函数 $f(x) = x^{3}$ 是：","options":[{"id":"A","text":"Even"},{"id":"B","text":"Neither odd nor even"},{"id":"C","text":"Odd"},{"id":"D","text":"Both odd and even"}],"answer":"C","explanation":"$f(−x) = (−x)^{3} = −x^{3} = −f(x)$, so f is odd.","explanationZh":"$f(−x) = (−x)^{3} = −x^{3} = −f(x)，故为奇函数。$","tags":["easy","parity","power"],"status":"published"},{"id":"math-parity-003","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":1,"stem":"The function $f(x) = \\sin x is$:","stemZh":"函数 $f(x) = \\sin x$ 是：","options":[{"id":"A","text":"Odd"},{"id":"B","text":"Even"},{"id":"C","text":"Neither odd nor even"},{"id":"D","text":"Both odd and even"}],"answer":"A","explanation":"$\\sin(−x) = −\\sin x$ for every x, so the sine function is odd.","explanationZh":"对任意 $x 有 \\sin(−x) = −\\sin$ x，故为正弦奇函数。","tags":["easy","parity","trigonometric"],"status":"published"},{"id":"math-parity-004","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":2,"stem":"The function $f(x) = \\cos x is$:","stemZh":"函数 $f(x) = \\cos x$ 是：","options":[{"id":"A","text":"Odd"},{"id":"B","text":"Neither odd nor even"},{"id":"C","text":"Both odd and even"},{"id":"D","text":"Even"}],"answer":"D","explanation":"$\\cos(−x) = \\cos x$ for every x, so the cosine function is even.","explanationZh":"对任意 $x 有 \\cos(−x) = \\cos$ x，故余弦函数为偶函数。","tags":["easy","parity","trigonometric"],"status":"published"},{"id":"math-parity-005","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":1,"stem":"The function $f(x) = |x| is$:","stemZh":"函数 $f(x) = |x|$ 是：","options":[{"id":"A","text":"Even"},{"id":"B","text":"Odd"},{"id":"C","text":"Neither odd nor even"},{"id":"D","text":"Even for $x ≥ 0$ and odd for $x < 0$"}],"answer":"A","explanation":"$|−x| = |x|$ for every x, so f is even.","explanationZh":"对任意 $x 有 |−x| = |x|，故为偶函数。$","tags":["easy","parity","absolute-value"],"status":"published"},{"id":"math-parity-006","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":2,"stem":"The function $f(x) = 1/x is$:","stemZh":"函数 $f(x) = 1/x$ 是：","options":[{"id":"A","text":"Even"},{"id":"B","text":"Both odd and even"},{"id":"C","text":"Neither odd nor even"},{"id":"D","text":"Odd"}],"answer":"D","explanation":"The domain $ℝ \\ {0}$ is symmetric about 0 and $f(−x) = 1/(−x) = −1/x = −f(x)$, so f is odd.","explanationZh":"定义域 $ℝ \\ {0}$ 关于原点对称，且 $f(−x) = 1/(−x) = −1/x = −f(x)，故为奇函数。$","tags":["easy","parity","rational"],"status":"published"},{"id":"math-parity-007","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":1,"stem":"The graph of an even function is symmetric about:","stemZh":"偶函数的图象关于什么对称？","options":[{"id":"A","text":"The y-axis"},{"id":"B","text":"The origin"},{"id":"C","text":"The x-axis"},{"id":"D","text":"The line $y = x$"}],"answer":"A","explanation":"Even means $f(−x) = f(x)$, i.e. the points $(x, y)$ and $(−x, y)$ both lie on the graph, which is symmetry about the y-axis.","explanationZh":"偶函数满足 $f(−x) = f(x)，即点 (x, y) 与 (−x, y)$ 都在图象上，故关于 y 轴对称。","tags":["easy","parity","graph"],"status":"published"},{"id":"math-parity-008","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":2,"stem":"The graph of an odd function is symmetric about:","stemZh":"奇函数的图象关于什么对称？","options":[{"id":"A","text":"The y-axis"},{"id":"B","text":"The origin"},{"id":"C","text":"The line $y = x$"},{"id":"D","text":"The x-axis"}],"answer":"B","explanation":"Odd means $f(−x) = −f(x)$, i.e. the points $(x, y)$ and $(−x, −y)$ both lie on the graph, which is symmetry about the origin.","explanationZh":"奇函数满足 $f(−x) = −f(x)，即点 (x, y) 与 (−x, −y)$ 都在图象上，故关于原点对称。","tags":["easy","parity","graph"],"status":"published"},{"id":"math-parity-009","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":1,"stem":"The function $f(x) = x + 1 is$:","stemZh":"函数 $f(x) = x + 1$ 是：","options":[{"id":"A","text":"Odd"},{"id":"B","text":"Even"},{"id":"C","text":"Neither odd nor even"},{"id":"D","text":"Both odd and even"}],"answer":"C","explanation":"$f(−x) =$ −x + 1, which equals neither $f(x) = x + 1$ nor $−f(x) =$ −x − 1, so f is neither odd nor even.","explanationZh":"$f(−x) =$ −x $+ 1，既不等于 f(x) = x + 1，也不等于 −f(x) =$ −x − 1，故非奇非偶。","tags":["easy","parity","linear"],"status":"published"},{"id":"math-parity-010","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":2,"stem":"The function $f(x) = x^{2} + x is$:","stemZh":"函数 $f(x) = x^{2} + x$ 是：","options":[{"id":"A","text":"Odd"},{"id":"B","text":"Even"},{"id":"C","text":"Both odd and even"},{"id":"D","text":"Neither odd nor even"}],"answer":"D","explanation":"$f(−x) = x^{2} − x$, which is neither $f(x) = x^{2} + x$ nor $−f(x) = −x^{2} − x$, so f is neither odd nor even.","explanationZh":"$f(−x) = x^{2} −$ x，既不等于 $f(x) = x^{2} +$ x，也不等于 $−f(x) = −x^{2} −$ x，故非奇非偶。","tags":["easy","parity","quadratic"],"status":"published"},{"id":"math-parity-011","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":3,"stem":"The function $f(x) = x^{2} + |x| is$:","stemZh":"函数 $f(x) = x^{2} + |x|$ 是：","options":[{"id":"A","text":"Odd"},{"id":"B","text":"Even"},{"id":"C","text":"Neither odd nor even"},{"id":"D","text":"Odd on $(−∞, 0)$ and even on $(0, +∞)$"}],"answer":"B","explanation":"Both $x^{2}$ and $|x|$ are even, and $f(−x) = (−x)^{2} + |−x| = x^{2} + |x| = f(x)$, so f is even.","explanationZh":"$x^{2} 与 |x| 都是偶函数，f(−x) = (−x)^{2} + |−x| = x^{2} + |x| = f(x)，故为偶函数。$","tags":["medium","parity","absolute-value"],"status":"published"},{"id":"math-parity-012","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":3,"stem":"The function $f(x) = x/(x^{2} + 1) is$:","stemZh":"函数 $f(x) = x/(x^{2} + 1)$ 是：","options":[{"id":"A","text":"Even"},{"id":"B","text":"Neither odd nor even"},{"id":"C","text":"Odd"},{"id":"D","text":"Both odd and even"}],"answer":"C","explanation":"The denominator is never 0, so the domain is $ℝ; f(−x) = (−x)/((−x)^{2} + 1) = −x/(x^{2} + 1) = −f(x)$, hence f is odd.","explanationZh":"分母恒不为 $0，定义域为 ℝ；f(−x) = (−x)/((−x)^{2} + 1) = −x/(x^{2} + 1) = −f(x)，故为奇函数。$","tags":["medium","parity","rational"],"status":"published"},{"id":"math-parity-013","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":3,"stem":"The function $f(x) = |x − 1| + |x + 1| is$:","stemZh":"函数 $f(x) = |x − 1| + |x + 1|$ 是：","options":[{"id":"A","text":"Even"},{"id":"B","text":"Odd"},{"id":"C","text":"Neither odd nor even"},{"id":"D","text":"Both odd and even"}],"answer":"A","explanation":"$f(−x) = |−x − 1| + |−x + 1| = |x + 1| + |x − 1| = f(x)$, so f is even.","explanationZh":"$f(−x) = |−x − 1| + |−x + 1| = |x + 1| + |x − 1| = f(x)，故为偶函数。$","tags":["medium","parity","absolute-value"],"status":"published"},{"id":"math-parity-014","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":3,"stem":"The function $f(x) = x|x| is$:","stemZh":"函数 $f(x) = x|x|$ 是：","options":[{"id":"A","text":"Even"},{"id":"B","text":"Neither odd nor even"},{"id":"C","text":"Both odd and even"},{"id":"D","text":"Odd"}],"answer":"D","explanation":"$f(−x) = (−x)|−x| = −x|x| = −f(x)$, so f is odd.","explanationZh":"$f(−x) = (−x)|−x| = −x|x| = −f(x)，故为奇函数。$","tags":["medium","parity","absolute-value"],"status":"published"},{"id":"math-parity-015","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":3,"stem":"The function $f(x) = x^{3} + x^{2} is$:","stemZh":"函数 $f(x) = x^{3} + x^{2}$ 是：","options":[{"id":"A","text":"Odd"},{"id":"B","text":"Even"},{"id":"C","text":"Neither odd nor even"},{"id":"D","text":"Both odd and even"}],"answer":"C","explanation":"$f(−x) = −x^{3} + x^{2}$, which equals neither $f(x) = x^{3} + x^{2}$ nor $−f(x) = −x^{3} − x^{2}$, so f is neither odd nor even.","explanationZh":"$f(−x) = −x^{3} + x^{2}，既不等于 f(x) = x^{3} + x^{2}，也不等于 −f(x) = −x^{3} − x^{2}，故非奇非偶。$","tags":["medium","parity","polynomial"],"status":"published"},{"id":"math-parity-016","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":3,"stem":"f is an odd function and $f(2) = −5$. What is $f(−2)$?","stemZh":"f 为奇函数且 $f(2) = −5，则 f(−2)$ 等于：","options":[{"id":"A","text":"−5"},{"id":"B","text":"5"},{"id":"C","text":"0"},{"id":"D","text":"It cannot be determined"}],"answer":"B","explanation":"For an odd function $f(−2) = −f(2) = −(−5) = 5$.","explanationZh":"奇函数满足 $f(−2) = −f(2) = −(−5) = 5。$","tags":["medium","parity","computation"],"status":"published"},{"id":"math-parity-017","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":3,"stem":"For which value of a is $f(x) = x^{2} + (a − 1)x + 3$ an even function?","stemZh":"$当 a 取何值时，f(x) = x^{2} + (a − 1)x + 3$ 为偶函数？","options":[{"id":"A","text":"$a = 0$"},{"id":"B","text":"$a = −1$"},{"id":"C","text":"$a = 1$"},{"id":"D","text":"$a = 2$"}],"answer":"C","explanation":"$f(−x) = x^{2} − (a − 1)x + 3$. For $f(−x) = f(x)$ the coefficient of the odd term must $vanish: a − 1 = 0$, so $a = 1$.","explanationZh":"$f(−x) = x^{2} − (a − 1)x + 3。要使 f(−x) = f(x)，一次项系数必须为 0，即 a − 1 = 0，得 a = 1。$","tags":["medium","parity","parameter"],"status":"published"},{"id":"math-parity-018","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":3,"stem":"Consider $f(x) = x^{2}$ with domain $[−1, 2]$. This function is:","stemZh":"函数 $f(x) = x^{2}，定义域为 [−1, 2]，该函数：$","options":[{"id":"A","text":"Even"},{"id":"B","text":"Odd"},{"id":"C","text":"Both odd and even"},{"id":"D","text":"Neither, because its domain is not symmetric about 0"}],"answer":"D","explanation":"Parity requires the domain to be symmetric about 0 first. Here 2 lies in the domain but −2 does not, so f(−x) is not even defined for every x and the function has no parity.","explanationZh":"判断奇偶性首先要看定义域是否关于原点对称。此处 2 在定义域内而 $−2 不在，f(−x)$ 并非处处有意义，故该函数非奇非偶。","tags":["medium","parity","domain"],"status":"published"},{"id":"math-parity-019","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":3,"stem":"f is an odd function defined on $ℝ$. Which statement must be true?","stemZh":"f 是定义在 $ℝ$ 上的奇函数，下列结论一定成立的是：","options":[{"id":"A","text":"$f(0) = 1$"},{"id":"B","text":"$f(0) = 0$"},{"id":"C","text":"$f(0)$ can be any real number"},{"id":"D","text":"$f(0)$ is not defined"}],"answer":"B","explanation":"From $f(−0) = −f(0)$ we get $f(0) = −f(0)$, hence $2f(0) = 0$ and $f(0) = 0$.","explanationZh":"$由 f(−0) = −f(0) 得 f(0) = −f(0)，即 2f(0) = 0，故 f(0) = 0。$","tags":["medium","parity","computation"],"status":"published"},{"id":"math-parity-020","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":3,"stem":"The function $f(x) = (2^x − 2^{−x})/2 is$:","stemZh":"函数 $f(x) = (2^x − 2^{−x})/2$ 是：","options":[{"id":"A","text":"Odd"},{"id":"B","text":"Even"},{"id":"C","text":"Neither odd nor even"},{"id":"D","text":"Odd only for $x > 0$"}],"answer":"A","explanation":"$f(−x) = (2^{−x} − 2^x)/2 = −(2^x − 2^{−x})/2 = −f(x)$ on the symmetric domain $ℝ$, so f is odd.","explanationZh":"定义域 $ℝ$ 关于原点对称，且 $f(−x) = (2^(−x) − 2^x)/2 = −(2^x − 2^(−x))/2 = −f(x)，故为奇函数。$","tags":["medium","parity","exponential"],"status":"published"},{"id":"math-parity-021","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":4,"stem":"The function $f(x) = lg(x + \\sqrt{x^{2} + 1}) is$:","stemZh":"函数 $f(x) = lg(x + \\sqrt{x^{2} + 1})$ 是：","options":[{"id":"A","text":"Even"},{"id":"B","text":"Odd"},{"id":"C","text":"Neither odd nor even"},{"id":"D","text":"Both odd and even"}],"answer":"B","explanation":"The domain is $ℝ (since \\sqrt{x^{2} + 1} > |x|)$. Using $(\\sqrt{x^{2} + 1} − x)(\\sqrt{x^{2} + 1} + x) = 1, f(−x) = lg(\\sqrt{x^{2} + 1} − x) = lg(1/(\\sqrt{x^{2} + 1} + x)) = −f(x)$, so f is odd.","explanationZh":"$由 \\sqrt{x^{2} + 1} > |x|$ 知定义域为 $ℝ。利用 (\\sqrt{x^{2} + 1} − x)(\\sqrt{x^{2} + 1} + x) = 1，得 f(−x) = lg(\\sqrt{x^{2} + 1} − x) = lg(1/(\\sqrt{x^{2} + 1} + x)) = −f(x)，故为奇函数。$","tags":["hard","parity","logarithm"],"status":"published"},{"id":"math-parity-022","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":5,"stem":"The function $f(x) = (x^{2} + x)/(x + 1) is$:","stemZh":"函数 $f(x) = (x^{2} + x)/(x + 1)$ 是：","options":[{"id":"A","text":"Odd, since $f(x) = x$ after simplification"},{"id":"B","text":"Even"},{"id":"C","text":"Neither, because its domain is not symmetric about 0"},{"id":"D","text":"Both odd and even"}],"answer":"C","explanation":"Although the rule simplifies to x, the domain is $ℝ \\ {−1}: 1$ belongs to the domain but −1 does not. Since the domain is not symmetric about 0, the function is neither odd nor even.","explanationZh":"虽然化简后为 x，但定义域是 $ℝ \\ {−1}：1$ 在定义域内而 −1 不在。定义域不关于原点对称，故该函数非奇非偶。","tags":["hard","parity","domain"],"status":"published"},{"id":"math-parity-023","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":4,"stem":"The function $f(x) = 1/(2^x − 1) + 1/2 is$:","stemZh":"函数 $f(x) = 1/(2^x − 1) + 1/2$ 是：","options":[{"id":"A","text":"Odd"},{"id":"B","text":"Even"},{"id":"C","text":"Neither odd nor even"},{"id":"D","text":"Odd only on $(0, +∞)$"}],"answer":"A","explanation":"The domain is $ℝ \\ {0}$, symmetric about 0. Writing $u = 2^x$ gives $f(x) = (u + 1)/(2(u − 1))$ and $f(−x) = (u + 1)/(2(1 − u)) = −f(x)$, so f is odd.","explanationZh":"定义域 $ℝ \\ {0}$ 关于原点对称。令 $u = 2^x，则 f(x) = (u + 1)/(2(u − 1))，f(−x) = (u + 1)/(2(1 − u)) = −f(x)，故为奇函数。$","tags":["hard","parity","exponential"],"status":"published"},{"id":"math-parity-024","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":4,"stem":"The function $f(x) = |x + 1| − |x − 1| is$:","stemZh":"函数 $f(x) = |x + 1| − |x − 1|$ 是：","options":[{"id":"A","text":"Even"},{"id":"B","text":"Neither odd nor even"},{"id":"C","text":"Both odd and even"},{"id":"D","text":"Odd"}],"answer":"D","explanation":"$f(−x) = |−x + 1| − |−x − 1| = |x − 1| − |x + 1| = −f(x)$, so f is odd.","explanationZh":"$f(−x) = |−x + 1| − |−x − 1| = |x − 1| − |x + 1| = −f(x)，故为奇函数。$","tags":["hard","parity","absolute-value"],"status":"published"},{"id":"math-parity-025","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":5,"stem":"f is an odd function on $ℝ$ and $f(x) = x^{2} − 2x$ for $x > 0$. What is $f(−3)$?","stemZh":"$f 是 ℝ$ 上的奇函数，且当 $x > 0 时 f(x) = x^{2} − 2x，则 f(−3)$ 等于：","options":[{"id":"A","text":"−3"},{"id":"B","text":"3"},{"id":"C","text":"15"},{"id":"D","text":"−15"}],"answer":"A","explanation":"First $f(3) = 3^{2} − 2·3 = 3$. Since f is $odd, f(−3) = −f(3) = −3. (A\\text{ common error is to substitute }x = −3\\text{ directly into }x^{2} − 2x,\\text{ which is only valid for }x > 0.)$","explanationZh":"先求 $f(3) = 3^{2} − 2·3 = 3，由奇函数得 f(−3) = −f(3) = −3。（常见错误是直接把 x = −3$ 代入 $x^{2} − 2x，该式只对 x > 0$ 成立。）","tags":["hard","parity","computation"],"status":"published"},{"id":"math-parity-026","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":5,"stem":"For a function f whose domain is symmetric about 0, let $g(x) = f(x) + f(−x)$ and $h(x) = f(x) − f(−x). Then$:","stemZh":"设 f 的定义域关于原点对称，令 $g(x) = f(x) + f(−x)，h(x) = f(x) − f(−x)，则：$","options":[{"id":"A","text":"g is odd and h is odd"},{"id":"B","text":"g is even and h is odd"},{"id":"C","text":"g is even and h is even"},{"id":"D","text":"g is odd and h is even"}],"answer":"B","explanation":"$g(−x) = f(−x) + f(x) = g(x)$, so g is $even; h(−x) = f(−x) − f(x) = −h(x)$, so h is odd. Every such f splits into an even part and an odd part.","explanationZh":"$g(−x) = f(−x) + f(x) = g(x)，故 g 为偶函数；h(−x) = f(−x) − f(x) = −h(x)，故 h$ 为奇函数。任意这样的 f 都可分解为偶函数部分与奇函数部分之和。","tags":["hard","parity","structure"],"status":"published"},{"id":"math-parity-027","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":4,"stem":"If f and g are both odd functions on $ℝ$, then $h(x) = f(g(x)) is$:","stemZh":"若 f、g 都是 $ℝ$ 上的奇函数，则 $h(x) = f(g(x))：$","options":[{"id":"A","text":"Even"},{"id":"B","text":"Neither odd nor even"},{"id":"C","text":"Odd"},{"id":"D","text":"Depends on the particular functions"}],"answer":"C","explanation":"$h(−x) = f(g(−x)) = f(−g(x)) = −f(g(x)) = −h(x)$, so the composition of two odd functions is odd.","explanationZh":"$h(−x) = f(g(−x)) = f(−g(x)) = −f(g(x)) = −h(x)，故两个奇函数的复合仍为奇函数。$","tags":["hard","parity","composite"],"status":"published"},{"id":"math-parity-028","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":5,"stem":"For $a > 0, a ≠ 1$, the function $f(x) = x(a^x + 1)/(a^x − 1) is$:","stemZh":"$设 a > 0，a ≠ 1，函数 f(x) = x(a^x + 1)/(a^x − 1)$ 是：","options":[{"id":"A","text":"Odd"},{"id":"B","text":"Neither odd nor even"},{"id":"C","text":"Depends on the value of a"},{"id":"D","text":"Even"}],"answer":"D","explanation":"The domain is $ℝ \\ {0}$. The factor $(a^x + 1)/(a^x − 1)$ is odd (replacing x by −x multiplies it by $−1)$, and x is odd, so the product of two odd factors is $even: f(−x) = f(x)$.","explanationZh":"定义域为 $ℝ \\ {0}。因子 (a^x + 1)/(a^x − 1)$ 为奇函数，x 也是奇函数，奇函数之积为偶函数，故 $f(−x) = f(x)。$","tags":["hard","parity","exponential"],"status":"published"},{"id":"math-parity-029","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":4,"stem":"The function $f(x) = lg((1 + x)/(1 − x)) is$:","stemZh":"函数 $f(x) = lg((1 + x)/(1 − x))$ 是：","options":[{"id":"A","text":"Even"},{"id":"B","text":"Neither odd nor even"},{"id":"C","text":"Odd"},{"id":"D","text":"Both odd and even"}],"answer":"C","explanation":"The domain is $(−1, 1)$, symmetric about 0, and $f(−x) = lg((1 − x)/(1 + x)) = −lg((1 + x)/(1 − x)) = −f(x)$, so f is odd.","explanationZh":"定义域为 $(−1, 1)，关于原点对称；f(−x) = lg((1 − x)/(1 + x)) = −lg((1 + x)/(1 − x)) = −f(x)，故为奇函数。$","tags":["hard","parity","logarithm"],"status":"published"},{"id":"math-parity-030","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"parity","type":"single-choice","difficulty":5,"stem":"For which value of a is $f(x) = x^{3} + (a − 1)x^{2} + 2x + (a − 1)$ an odd function?","stemZh":"$当 a 取何值时，f(x) = x^{3} + (a − 1)x^{2} + 2x + (a − 1)$ 为奇函数？","options":[{"id":"A","text":"$a = 0$"},{"id":"B","text":"$a = 1$"},{"id":"C","text":"$a = 2$"},{"id":"D","text":"No such value of a exists"}],"answer":"B","explanation":"An odd polynomial can contain only odd powers. The quadratic term and the constant term both vanish exactly when $a − 1 = 0$, i.e. $a = 1$, leaving $f(x) = x^{3} + 2x$.","explanationZh":"奇函数只含奇次项。二次项系数 a − 1 与常数项 a − 1 同时为 $0，即 a = 1，此时 f(x) = x^{3} + 2x$ 为奇函数。","tags":["hard","parity","parameter"],"status":"published"},{"id":"math-inverse-001","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":1,"stem":"The inverse of $f(x) = 2x + 1 is$:","stemZh":"函数 $f(x) = 2x + 1$ 的反函数是：","options":[{"id":"A","text":"$(x + 1)/2$"},{"id":"B","text":"$(x − 1)/2$"},{"id":"C","text":"2x − 1"},{"id":"D","text":"$(1 − x)/2$"}],"answer":"B","explanation":"Let $y = 2x + 1$. Solving for x gives $x = (y − 1)/2$, so $f^{-1}(x) = (x − 1)/2$.","explanationZh":"$令 y = 2x + 1，解出 x = (y − 1)/2，故 f^{-1}(x) = (x − 1)/2。$","tags":["easy","inverse","linear"],"status":"published"},{"id":"math-inverse-002","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":1,"stem":"The inverse of $f(x) = 3x − 2 is$:","stemZh":"函数 $f(x) = 3x − 2$ 的反函数是：","options":[{"id":"A","text":"$(x − 2)/3$"},{"id":"B","text":"3x + 2"},{"id":"C","text":"$(x + 2)/3$"},{"id":"D","text":"x/3 − 2"}],"answer":"C","explanation":"From $y = 3x − 2$ we get $x = (y + 2)/3$, hence $f^{-1}(x) = (x + 2)/3$.","explanationZh":"$由 y = 3x − 2 得 x = (y + 2)/3，故 f^{-1}(x) = (x + 2)/3。$","tags":["easy","inverse","linear"],"status":"published"},{"id":"math-inverse-003","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":2,"stem":"The inverse of $f(x) = x^{3} is$:","stemZh":"函数 $f(x) = x^{3}$ 的反函数是：","options":[{"id":"A","text":"$x^{2}$"},{"id":"B","text":"$\\sqrt{x}$"},{"id":"C","text":"$\\dfrac{1}{x^{3}}$"},{"id":"D","text":"$\\sqrt[3]{x}$"}],"answer":"D","explanation":"$y = x^{3}$ is one-to-one on $\\mathbb{R}$ with range $\\mathbb{R}$, and solving $y = x^{3}$ gives $x = \\sqrt[3]{y}$, so $f^{-1}(x) = \\sqrt[3]{x}$.","explanationZh":"$y = x^{3} 在 \\mathbb{R}$ 上一一对应且值域为 $\\mathbb{R}，由 y = x^{3} 得 x = \\sqrt[3]{y}，故 f^{-1}(x) = \\sqrt[3]{x}。$","tags":["easy","inverse","power"],"status":"published"},{"id":"math-inverse-004","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":2,"stem":"The inverse of $f(x) = x^{2}$ with domain $x ≥ 0 is$:","stemZh":"函数 $f(x) = x^{2}（定义域 x ≥ 0）的反函数是：$","options":[{"id":"A","text":"$\\sqrt{x}$, with domain $x ≥ 0$"},{"id":"B","text":"$−\\sqrt{x}$, with domain $x ≥ 0$"},{"id":"C","text":"$x^{2}$, with domain $x ≥ 0$"},{"id":"D","text":"$\\pm\\sqrt{x}$, with domain $x ≥ 0$"}],"answer":"A","explanation":"On $x ≥ 0$ the function is one-to-one with range $[0, +∞)$; solving $y = x^{2}$ gives $x = \\sqrt{y} (the$ positive $root)$, so $f^{-1}(x) = \\sqrt{x}$ on $[0, +∞)$.","explanationZh":"$在 x ≥ 0$ 上函数一一对应、值域为 $[0, +∞)；由 y = x^{2} 得 x = \\sqrt{y}（取正根），故 f^{-1}(x) = \\sqrt{x}，定义域 [0, +∞)。$","tags":["easy","inverse","quadratic"],"status":"published"},{"id":"math-inverse-005","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":1,"stem":"The inverse of $f(x) = 2^x is$:","stemZh":"函数 $f(x) = 2^x$ 的反函数是：","options":[{"id":"A","text":"$(1/2)^x$"},{"id":"B","text":"$2^(−x)$"},{"id":"C","text":"$\\log_{_{2}}x$, with domain $x > 0$"},{"id":"D","text":"$x^{2}$"}],"answer":"C","explanation":"Exponential and logarithmic functions with the same base are $inverses: 2^x$ and $\\log_{_{2}}x$. The domain of $f^{-1}$ is the range of f, namely $(0, +∞)$.","explanationZh":"$同底的指数函数与对数函数互为反函数：2^x 与 \\log_{_{2}}x。f^{-1}$ 的定义域即 f 的值域 $(0, +∞)。$","tags":["easy","inverse","exponential"],"status":"published"},{"id":"math-inverse-006","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":2,"stem":"The inverse of $f(x) = \\log_{_{2}}x is$:","stemZh":"函数 $f(x) = \\log_{_{2}}x$ 的反函数是：","options":[{"id":"A","text":"$2^x$, with domain $x ∈ ℝ$"},{"id":"B","text":"$\\log_{_{2}}x$"},{"id":"C","text":"$x^{2}$"},{"id":"D","text":"$(1/2)^x$"}],"answer":"A","explanation":"Solving $y = \\log_{_{2}}x$ gives $x = 2^y$, so $f^{-1}(x) = 2^x$; its domain is the range of $\\log_{_{2}}x$, which is all of $ℝ$.","explanationZh":"$由 y = \\log_{_{2}}x 得 x = 2^y，故 f^{-1}(x) = 2^x，其定义域为 \\log_{_{2}}x$ 的值域 $ℝ。$","tags":["easy","inverse","logarithm"],"status":"published"},{"id":"math-inverse-007","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":1,"stem":"The graphs of $y = f(x)$ and $y = f^{-1}(x)$ are symmetric about:","stemZh":"函数 $y = f(x) 与 y = f^{-1}(x)$ 的图象关于什么对称？","options":[{"id":"A","text":"The x-axis"},{"id":"B","text":"The line $y = x$"},{"id":"C","text":"The y-axis"},{"id":"D","text":"The origin"}],"answer":"B","explanation":"If $(a, b)$ lies on $y = f(x)$ then $(b, a)$ lies on $y = f^{-1}(x)$; swapping coordinates is reflection in the line $y =$ x.","explanationZh":"$若 (a, b) 在 y = f(x)$ 上，则 $(b, a) 在 y = f^{-1}(x)$ 上；交换横纵坐标即关于直线 $y = x$ 对称。","tags":["easy","inverse","graph"],"status":"published"},{"id":"math-inverse-008","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":2,"stem":"The inverse of $f(x) = 1/x (x ≠ 0) is$:","stemZh":"函数 $f(x) = 1/x（x ≠ 0）的反函数是：$","options":[{"id":"A","text":"−1/x"},{"id":"B","text":"x"},{"id":"C","text":"$x^{2}$"},{"id":"D","text":"1/x, with domain $x ≠ 0$"}],"answer":"D","explanation":"Solving $y = 1/x$ gives $x = 1/y$, so the function is its own inverse; its domain $(x ≠ 0)$ is also its range.","explanationZh":"$由 y = 1/x 得 x = 1/y，故该函数自为反函数，其定义域（x ≠ 0）也是值域。$","tags":["easy","inverse","rational"],"status":"published"},{"id":"math-inverse-009","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":1,"stem":"If the point $(2, 3)$ lies on the graph of $y = f(x)$, which point lies on the graph of $y = f^{-1}(x)$?","stemZh":"若点 $(2, 3) 在 y = f(x)$ 的图象上，则哪个点在 $y = f^{-1}(x)$ 的图象上？","options":[{"id":"A","text":"$(3, 2)$"},{"id":"B","text":"$(2, 3)$"},{"id":"C","text":"$(−2, −3)$"},{"id":"D","text":"$(−3, −2)$"}],"answer":"A","explanation":"$f(2) = 3$ means $f^{-1}(3) = 2$, so the point $(3, 2) —$ the reflection of $(2, 3)$ in $y = x —$ lies on the inverse graph.","explanationZh":"$f(2) = 3$ 等价于 $f^{-1}(3) = 2，故点 (3, 2)（即 (2, 3)$ 关于 $y = x$ 的对称点）在反函数图象上。","tags":["easy","inverse","graph"],"status":"published"},{"id":"math-inverse-010","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":2,"stem":"The inverse of $f(x) = x^{3} + 1 is$:","stemZh":"函数 $f(x) = x^{3} + 1$ 的反函数是：","options":[{"id":"A","text":"$\\sqrt[3]{x} - 1$"},{"id":"B","text":"$\\sqrt[3]{x + 1}$"},{"id":"C","text":"$\\sqrt[3]{x - 1}$"},{"id":"D","text":"$x^{3} - 1$"}],"answer":"C","explanation":"From $y = x^{3} + 1$ we get $x^{3} = y - 1$, hence $x = \\sqrt[3]{y - 1}$ and $f^{-1}(x) = \\sqrt[3]{x - 1}$.","explanationZh":"$由 y = x^{3} + 1 得 x^{3} = y - 1，故 x = \\sqrt[3]{y - 1}，即 f^{-1}(x) = \\sqrt[3]{x - 1}。$","tags":["easy","inverse","power"],"status":"published"},{"id":"math-inverse-011","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":3,"stem":"The inverse of $f(x) = (x + 1)/(x − 1) is$:","stemZh":"函数 $f(x) = (x + 1)/(x − 1)$ 的反函数是：","options":[{"id":"A","text":"$(x + 1)/(x − 1)$"},{"id":"B","text":"$(x − 1)/(x + 1)$"},{"id":"C","text":"$(1 − x)/(x + 1)$"},{"id":"D","text":"1/x"}],"answer":"A","explanation":"Let $y = (x + 1)/(x − 1)$. Then $y(x − 1) = x + 1$, so $x(y − 1) = y + 1$ and $x = (y + 1)/(y − 1)$. The function is its own inverse.","explanationZh":"$令 y = (x + 1)/(x − 1)，则 y(x − 1) = x + 1，得 x(y − 1) = y + 1，即 x = (y + 1)/(y − 1)，原函数自为反函数。$","tags":["medium","inverse","rational"],"status":"published"},{"id":"math-inverse-012","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":3,"stem":"The inverse of $f(x) = 2/(x − 1) is$:","stemZh":"函数 $f(x) = 2/(x − 1)$ 的反函数是：","options":[{"id":"A","text":"2/x − 1"},{"id":"B","text":"2/x + 1, with domain $x ≠ 0$"},{"id":"C","text":"x/2 + 1"},{"id":"D","text":"$2/(x + 1)$"}],"answer":"B","explanation":"From $y = 2/(x − 1)$ we get $x − 1 = 2/y$, so $x = 2/y + 1$ and $f^{-1}(x) = 2/x + 1$. The domain of $f^{-1}$ is the range of f, i.e. $x ≠ 0$.","explanationZh":"$由 y = 2/(x − 1) 得 x − 1 = 2/y，故 x = 2/y + 1，即 f^{-1}(x) = 2/x + 1；其定义域为 f$ 的值域，即 $x ≠ 0。$","tags":["medium","inverse","rational"],"status":"published"},{"id":"math-inverse-013","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":3,"stem":"The inverse of $f(x) = 3^{x − 1} is$:","stemZh":"函数 $f(x) = 3^(x − 1)$ 的反函数是：","options":[{"id":"A","text":"$\\log_{3}(x + 1)$"},{"id":"B","text":"$\\\\log_{3}x − 1$"},{"id":"C","text":"$\\log_{3}x + 1$, with domain $x > 0$"},{"id":"D","text":"$3^{x + 1}$"}],"answer":"C","explanation":"From $y = 3^(x − 1)$ we get $x − 1 = \\log_{3}y$, so $x = \\log_{3}y + 1$, i.e. $f^{-1}(x) = \\log_{3}x + 1$ on $(0, +∞)$.","explanationZh":"$由 y = 3^(x − 1) 得 x − 1 = \\log_{3}y，故 x = \\log_{3}y + 1，即 f^{-1}(x) = \\log_{3}x + 1，定义域 (0, +∞)。$","tags":["medium","inverse","exponential"],"status":"published"},{"id":"math-inverse-014","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":3,"stem":"The inverse of $f(x) = \\log_{_{2}}(x − 1) is$:","stemZh":"函数 $f(x) = \\log_{_{2}}(x − 1)$ 的反函数是：","options":[{"id":"A","text":"$2^x − 1$"},{"id":"B","text":"$\\log_{_{2}}x + 1$"},{"id":"C","text":"$\\log_{_{2}}(x + 1)$"},{"id":"D","text":"$2^x + 1$, with domain $x ∈ ℝ$"}],"answer":"D","explanation":"From $y = \\log_{_{2}}(x − 1)$ we get $x − 1 = 2^y$, so $x = 2^y + 1$ and $f^{-1}(x) = 2^x + 1$. Its domain is the range of f, namely $ℝ$.","explanationZh":"$由 y = \\log_{_{2}}(x − 1) 得 x − 1 = 2^y，故 x = 2^y + 1，即 f^{-1}(x) = 2^x + 1；其定义域为 f$ 的值域 $ℝ。$","tags":["medium","inverse","logarithm"],"status":"published"},{"id":"math-inverse-015","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":3,"stem":"The inverse of $f(x) = x^{2} − 2x$ with domain $x ≥ 1 is$:","stemZh":"函数 $f(x) = x^{2} − 2x（定义域 x ≥ 1）的反函数是：$","options":[{"id":"A","text":"$1 + \\sqrt{x + 1}$, with domain $x ≥ −1$"},{"id":"B","text":"$1 − \\sqrt{x + 1}$, with domain $x ≥ −1$"},{"id":"C","text":"$\\sqrt{x − 1} + 1$"},{"id":"D","text":"$\\sqrt{x + 1} − 1$"}],"answer":"A","explanation":"$f(x) = (x − 1)^{2} − 1$ has range $[−1, +∞)$. Solving $(x − 1)^{2} = y + 1$ with $x ≥ 1$ gives $x = 1 + \\sqrt{y + 1}$, so $f^{-1}(x) = 1 + \\sqrt{x + 1}$ on $[−1, +∞)$.","explanationZh":"$f(x) = (x − 1)^{2} − 1$ 的值域为 $[−1, +∞)。由 (x − 1)^{2} = y + 1 且 x ≥ 1 得 x = 1 + \\sqrt{y + 1}，故 f^{-1}(x) = 1 + \\sqrt{x + 1}，定义域 [−1, +∞)。$","tags":["medium","inverse","quadratic"],"status":"published"},{"id":"math-inverse-016","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":3,"stem":"The inverse of $f(x) = \\sqrt{x − 2} is$:","stemZh":"函数 $f(x) = \\sqrt{x − 2}$ 的反函数是：","options":[{"id":"A","text":"$x^{2} − 2$"},{"id":"B","text":"$(x + 2)^{2}$"},{"id":"C","text":"$x^{2} + 2$, with domain $x ≥ 0$"},{"id":"D","text":"$\\sqrt{x} + 2$"}],"answer":"C","explanation":"The range of f is $[0, +∞)$, so the inverse has domain $x ≥ 0$. From $y = \\sqrt{x − 2}$ we get $x = y^{2} + 2$, hence $f^{-1}(x) = x^{2} + 2 (x ≥ 0)$.","explanationZh":"f 的值域为 $[0, +∞)，故反函数定义域为 x ≥ 0。由 y = \\sqrt{x − 2} 得 x = y^{2} + 2，即 f^{-1}(x) = x^{2} + 2（x ≥ 0）。$","tags":["medium","inverse","radical"],"status":"published"},{"id":"math-inverse-017","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":3,"stem":"The inverse of $f(x) = 2^x + 1 is$:","stemZh":"函数 $f(x) = 2^x + 1$ 的反函数是：","options":[{"id":"A","text":"$\\log_{_{2}}x − 1$"},{"id":"B","text":"$\\log_{_{2}}(x + 1)$"},{"id":"C","text":"$2^{x − 1}$"},{"id":"D","text":"$\\log_{_{2}}(x − 1)$, with domain $x > 1$"}],"answer":"D","explanation":"The range of f is $(1, +∞)$, so $f^{-1}$ has domain $x > 1$. From $y = 2^x + 1$ we get $2^x = y − 1$, so $f^{-1}(x) = \\log_{_{2}}(x − 1)$.","explanationZh":"f 的值域为 $(1, +∞)，故反函数定义域为 x > 1。由 y = 2^x + 1 得 2^x = y − 1，故 f^{-1}(x) = \\log_{_{2}}(x − 1)。$","tags":["medium","inverse","exponential"],"status":"published"},{"id":"math-inverse-018","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":3,"stem":"Which of the following functions has no inverse function on the stated domain?","stemZh":"下列函数中，在给定定义域上不存在反函数的是：","options":[{"id":"A","text":"$f(x) = x^{3}, x ∈ ℝ$"},{"id":"B","text":"$f(x) = 2x − 1, x ∈ ℝ$"},{"id":"C","text":"$f(x) = 1/x, x ≠ 0$"},{"id":"D","text":"$f(x) = x^{2}, x ∈ ℝ$"}],"answer":"D","explanation":"An inverse exists only when the function is one-to-one. $f(x) = x^{2}$ on $ℝ$ fails $this: f(−1) = f(1) = 1$, so two inputs share one output.","explanationZh":"$只有一一对应的函数才有反函数。f(x) = x^{2} 在 ℝ 上不是一一对应：f(−1) = f(1) = 1，两个自变量对应同一个函数值。$","tags":["medium","inverse","one-to-one"],"status":"published"},{"id":"math-inverse-019","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":3,"stem":"Let $f(x) = 2^x$ with domain $[0, 2]$. What is the domain of $f^{-1}$?","stemZh":"$设 f(x) = 2^x$ 的定义域为 $[0, 2]，则 f^{-1}$ 的定义域是：","options":[{"id":"A","text":"$[0, 2]$"},{"id":"B","text":"$[1, 4]$"},{"id":"C","text":"$(0, +∞)$"},{"id":"D","text":"$[0, 4]$"}],"answer":"B","explanation":"The domain of $f^{-1}$ is the range of f. Since $2^x increases$, its values on $[0, 2]$ run from $2^{0} = 1$ to $2^{2} = 4$, so the domain of $f^{-1}$ is $[1, 4]$.","explanationZh":"$f^{-1}$ 的定义域等于 $f 的值域。2^x$ 递增，在 $[0, 2]$ 上的取值从 $2^{0} = 1 到 2^{2} = 4，故 f^{-1}$ 的定义域为 $[1, 4]。$","tags":["medium","inverse","domain-range"],"status":"published"},{"id":"math-inverse-020","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":3,"stem":"The inverse of $f(x) = (2x + 1)/(x − 3) is$:","stemZh":"函数 $f(x) = (2x + 1)/(x − 3)$ 的反函数是：","options":[{"id":"A","text":"$(x − 2)/(3x + 1)$"},{"id":"B","text":"$(3x − 1)/(x + 2)$"},{"id":"C","text":"$(x + 3)/(x − 2)$"},{"id":"D","text":"$(3x + 1)/(x − 2)$, with domain $x ≠ 2$"}],"answer":"D","explanation":"Let $y = (2x + 1)/(x − 3)$. Then $y(x − 3) = 2x + 1$, so $x(y − 2) = 3y + 1$ and $x = (3y + 1)/(y − 2)$, hence $f^{-1}(x) = (3x + 1)/(x − 2)$.","explanationZh":"$令 y = (2x + 1)/(x − 3)，则 y(x − 3) = 2x + 1，得 x(y − 2) = 3y + 1，即 x = (3y + 1)/(y − 2)，故 f^{-1}(x) = (3x + 1)/(x − 2)。$","tags":["medium","inverse","rational"],"status":"published"},{"id":"math-inverse-021","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":4,"stem":"The inverse of $f(x) = (x − 1)/(x + 2) is$:","stemZh":"函数 $f(x) = (x − 1)/(x + 2)$ 的反函数是：","options":[{"id":"A","text":"$(x + 2)/(x − 1)$"},{"id":"B","text":"$(2x + 1)/(1 − x)$, with domain $x ≠ 1$"},{"id":"C","text":"$(1 − 2x)/(x − 1)$"},{"id":"D","text":"$(x − 1)/(2x + 1)$"}],"answer":"B","explanation":"From $y = (x − 1)/(x + 2): y(x + 2) = x − 1$, so $x(y − 1) = −1 − 2y$ and $x = (2y + 1)/(1 − y)$. Hence $f^{-1}(x) = (2x + 1)/(1 − x)$, whose domain $x ≠ 1$ is the range of f.","explanationZh":"$由 y = (x − 1)/(x + 2) 得 y(x + 2) = x − 1，即 x(y − 1) = −1 − 2y，故 x = (2y + 1)/(1 − y)，f^{-1}(x) = (2x + 1)/(1 − x)，定义域 x ≠ 1 即 f$ 的值域。","tags":["hard","inverse","rational"],"status":"published"},{"id":"math-inverse-022","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":5,"stem":"The inverse of $f(x) = 2^x/(2^x + 1) is$:","stemZh":"函数 $f(x) = 2^x/(2^x + 1)$ 的反函数是：","options":[{"id":"A","text":"$\\log_{_{2}}((1 − x)/x)$, with domain $(0, 1)$"},{"id":"B","text":"$\\log_{_{2}}(1 − x)$, with domain $(0, 1)$"},{"id":"C","text":"$\\log_{_{2}}(x/(1 − x))$, with domain $(0, 1)$"},{"id":"D","text":"$\\log_{_{2}}(x/(1 + x))$, with domain $(0, +∞)$"}],"answer":"C","explanation":"Let $y = 2^x/(2^x + 1)$. Then $y·2^x + y = 2^x$, so $2^x(1 − y) = y$ and $2^x = y/(1 − y) > 0$, giving $0 < y < 1$. Hence $f^{-1}(x) = \\log_{_{2}}(x/(1 − x))$ on $(0, 1)$.","explanationZh":"$令 y = 2^x/(2^x + 1)，则 y·2^x + y = 2^x，即 2^x(1 − y) =$ y，得 $2^x = y/(1 − y) > 0，故 0 < y < 1，f^{-1}(x) = \\log_{_{2}}(x/(1 − x))，定义域 (0, 1)。$","tags":["hard","inverse","exponential"],"status":"published"},{"id":"math-inverse-023","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":4,"stem":"The inverse of $f(x) = (x − 1)^{2}$ with domain $x ≤ 1 is$:","stemZh":"函数 $f(x) = (x − 1)^{2}（定义域 x ≤ 1）的反函数是：$","options":[{"id":"A","text":"$1 + \\sqrt{x}$, with domain $x ≥ 0$"},{"id":"B","text":"$\\sqrt{x} − 1$, with domain $x ≥ 0$"},{"id":"C","text":"$(x + 1)^{2}$"},{"id":"D","text":"$1 − \\sqrt{x}$, with domain $x ≥ 0$"}],"answer":"D","explanation":"The range is $[0, +∞)$, so the inverse has domain $x ≥ 0$. Since $x ≤ 1$ we have $x − 1 = −\\sqrt{y}$, giving $x = 1 − \\sqrt{y}$, i.e. $f^{-1}(x) = 1 − \\sqrt{x}$.","explanationZh":"值域为 $[0, +∞)，故反函数定义域为 x ≥ 0。因 x ≤ 1，有 x − 1 = −\\sqrt{y}，得 x = 1 − \\sqrt{y}，即 f^{-1}(x) = 1 − \\sqrt{x}。$","tags":["hard","inverse","quadratic"],"status":"published"},{"id":"math-inverse-024","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":4,"stem":"The inverse of $f(x) = \\log_{_{2}}x + 1 is$:","stemZh":"函数 $f(x) = \\log_{_{2}}x + 1$ 的反函数是：","options":[{"id":"A","text":"$2^{x − 1}$, with domain $x ∈ ℝ$"},{"id":"B","text":"$2^x + 1$"},{"id":"C","text":"$2^{x + 1}$"},{"id":"D","text":"$2^x − 1$"}],"answer":"A","explanation":"From $y = \\log_{_{2}}x + 1$ we get $\\log_{_{2}}x = y − 1$, so $x = 2^(y − 1)$, hence $f^{-1}(x) = 2^{x − 1}$; the domain is the range of f, i.e. $ℝ$.","explanationZh":"$由 y = \\log_{_{2}}x + 1 得 \\log_{_{2}}x = y − 1，故 x = 2^(y − 1)，f^{-1}(x) = 2^(x − 1)；定义域为 f$ 的值域 $ℝ。$","tags":["hard","inverse","logarithm"],"status":"published"},{"id":"math-inverse-025","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":5,"stem":"The inverse of $f(x) = x^{2} + 2x$ with domain $x ≥ −1 is$:","stemZh":"函数 $f(x) = x^{2} + 2x（定义域 x ≥ −1）的反函数是：$","options":[{"id":"A","text":"$1 + \\sqrt{x + 1}$, with domain $x ≥ −1$"},{"id":"B","text":"$−1 + \\sqrt{x + 1}$, with domain $x ≥ −1$"},{"id":"C","text":"$−1 − \\sqrt{x + 1}$, with domain $x ≥ −1$"},{"id":"D","text":"$1 − \\sqrt{x − 1}$"}],"answer":"B","explanation":"$f(x) = (x + 1)^{2} − 1$ has range $[−1, +∞)$. With $x ≥ −1$ we get $x + 1 = \\sqrt{y + 1}$, so $x = −1 + \\sqrt{y + 1}$ and $f^{-1}(x) = −1 + \\sqrt{x + 1}$ on $[−1, +∞)$.","explanationZh":"$f(x) = (x + 1)^{2} − 1$ 的值域为 $[−1, +∞)。由 x ≥ −1 得 x + 1 = \\sqrt{y + 1}，故 x = −1 + \\sqrt{y + 1}，f^{-1}(x) = −1 + \\sqrt{x + 1}，定义域 [−1, +∞)。$","tags":["hard","inverse","quadratic"],"status":"published"},{"id":"math-inverse-026","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":4,"stem":"The inverse of $f(x) = \\sqrt{x + 1} − 2 is$:","stemZh":"函数 $f(x) = \\sqrt{x + 1} − 2$ 的反函数是：","options":[{"id":"A","text":"$(x + 2)^{2} + 1$"},{"id":"B","text":"$(x − 2)^{2} − 1$"},{"id":"C","text":"$(x + 2)^{2} − 1$, with domain $x ≥ −2$"},{"id":"D","text":"$x^{2} + 3$"}],"answer":"C","explanation":"The range of f is $[−2, +∞)$, so $f^{-1}$ has domain $x ≥ −2$. From $y + 2 = \\sqrt{x + 1}$ we get $x = (y + 2)^{2} − 1$, hence $f^{-1}(x) = (x + 2)^{2} − 1$.","explanationZh":"f 的值域为 $[−2, +∞)，故反函数定义域 x ≥ −2。由 y + 2 = \\sqrt{x + 1} 得 x = (y + 2)^{2} − 1，即 f^{-1}(x) = (x + 2)^{2} − 1。$","tags":["hard","inverse","radical"],"status":"published"},{"id":"math-inverse-027","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":5,"stem":"The inverse of $f(x) = (2^x − 1)/(2^x + 1) is$:","stemZh":"函数 $f(x) = (2^x − 1)/(2^x + 1)$ 的反函数是：","options":[{"id":"A","text":"$\\log_{_{2}}((x − 1)/(x + 1))$"},{"id":"B","text":"$\\log_{_{2}}((1 − x)/(1 + x))$, with domain $(−1, 1)$"},{"id":"C","text":"$\\log_{_{2}}((x + 1)/(x − 1))$"},{"id":"D","text":"$\\log_{_{2}}((1 + x)/(1 − x))$, with domain $(−1, 1)$"}],"answer":"D","explanation":"Let $y = (2^x − 1)/(2^x + 1)$. Then $y·2^x + y = 2^x − 1$, so $2^x(1 − y) = 1 + y$ and $2^x = (1 + y)/(1 − y) > 0$, which forces $−1 < y < 1$. Hence $f^{-1}(x) = \\log_{_{2}}((1 + x)/(1 − x))$ on $(−1, 1)$.","explanationZh":"$令 y = (2^x − 1)/(2^x + 1)，则 y·2^x + y = 2^x − 1，即 2^x(1 − y) = 1 +$ y，得 $2^x = (1 + y)/(1 − y) > 0，故 −1 < y < 1，f^{-1}(x) = \\log_{_{2}}((1 + x)/(1 − x))，定义域 (−1, 1)。$","tags":["hard","inverse","exponential"],"status":"published"},{"id":"math-inverse-028","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":4,"stem":"The inverse of $f(x) = 3^{2x − 1} is$:","stemZh":"函数 $f(x) = 3^(2x − 1)$ 的反函数是：","options":[{"id":"A","text":"$(1 + \\log_{3}x)/2$, with domain $x > 0$"},{"id":"B","text":"$(\\log_{3}x − 1)/2$, with domain $x > 0$"},{"id":"C","text":"$2\\log_{3}x − 1$"},{"id":"D","text":"$(\\log_{3}x)/2 + 1$"}],"answer":"A","explanation":"From $y = 3^(2x − 1)$ we get $2x − 1 = \\log_{3}y$, so $x = (1 + \\log_{3}y)/2$, i.e. $f^{-1}(x) = (1 + \\log_{3}x)/2$ on $(0, +∞)$.","explanationZh":"$由 y = 3^(2x − 1) 得 2x − 1 = \\log_{3}y，故 x = (1 + \\log_{3}y)/2，即 f^{-1}(x) = (1 + \\log_{3}x)/2，定义域 (0, +∞)。$","tags":["hard","inverse","exponential"],"status":"published"},{"id":"math-inverse-029","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":5,"stem":"The inverse of $f(x) = \\log_{_{2}}(x^{2} + 1)$ with domain $x ≥ 0 is$:","stemZh":"函数 $f(x) = \\log_{_{2}}(x^{2} + 1)（定义域 x ≥ 0）的反函数是：$","options":[{"id":"A","text":"$\\sqrt{2^x + 1}$, with domain $x ≥ 0$"},{"id":"B","text":"$\\sqrt{2^x − 1}$, with domain $x ≥ 0$"},{"id":"C","text":"$\\sqrt{2^x} − 1$"},{"id":"D","text":"$2^(\\sqrt{x}) − 1$"}],"answer":"B","explanation":"On $x ≥ 0$ the function is one-to-one with range $[0, +∞)$. From $y = \\log_{2}(x^{2} + 1)$ we get $x^{2} = 2^y − 1$, and $x ≥ 0$ gives $x = \\sqrt{2^y − 1}$, so $f^{-1}(x) = \\sqrt{2^x − 1}$ on $[0, +∞)$.","explanationZh":"$在 x ≥ 0$ 上函数一一对应、值域为 $[0, +∞)。由 y = \\log_{2}(x^{2} + 1) 得 x^{2} = 2^y − 1，结合 x ≥ 0 得 x = \\sqrt{2^y − 1}，故 f^{-1}(x) = \\sqrt{2^x − 1}，定义域 [0, +∞)。$","tags":["hard","inverse","logarithm"],"status":"published"},{"id":"math-inverse-030","subject":"math","moduleId":"functions","unitId":"concepts","topicId":"inverse","type":"single-choice","difficulty":4,"stem":"If $f^{-1}(x) = (x − 1)/2$, then f(x) is:","stemZh":"$若 f^{-1}(x) = (x − 1)/2，则 f(x)$ 为：","options":[{"id":"A","text":"2x + 1"},{"id":"B","text":"$(x + 1)/2$"},{"id":"C","text":"2x − 1"},{"id":"D","text":"x/2 + 1"}],"answer":"A","explanation":"The inverse of $f^{-1}$ is f itself. Solving $y = (x − 1)/2$ for x gives $x = 2y + 1$, so $f(x) = 2x + 1. (Check: f^{-1}(f(x)) = ((2x + 1) − 1)/2 = x.)$","explanationZh":"$f^{-1}$ 的反函数就是 f。由 $y = (x − 1)/2$ 解出 $x = 2y + 1，故 f(x) = 2x + 1。（验证：f^{-1}(f(x)) = ((2x + 1) − 1)/2 =$ x。）","tags":["hard","inverse","linear"],"status":"published"},{"id":"math-power-001","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":1,"stem":"Which of the following is a power function?","stemZh":"下列函数中，是幂函数的是？","options":[{"id":"A","text":"$y = 2^{x}$"},{"id":"B","text":"$y = x^{3}$"},{"id":"C","text":"$y = 3x − 1$"},{"id":"D","text":"$y = \\log_{_{2}}x$"}],"answer":"B","explanation":"A power function has the form $y = x^{a}$: coefficient exactly 1 and the variable in the base. $y = 2^{x}$ is $exponential, y = 3x − 1$ is linear, and $y = \\log_{_{2}}x$ is logarithmic.","explanationZh":"幂函数的形式为 $y = x^{a}：系数必须为 1，自变量在底数位置。y = 2^{x}$ 是指数函数，y $= 3x − 1$ 是一次函数，y $= \\log_{_{2}}x$ 是对数函数。","tags":["easy","power","definition"],"status":"published"},{"id":"math-power-002","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":1,"stem":"Simplify $x^{2}·x^{5}$.","stemZh":"化简 $x^{2}·x^{5}。$","options":[{"id":"A","text":"$x^{10}$"},{"id":"B","text":"$2x^{7}$"},{"id":"C","text":"$x^{7}$"},{"id":"D","text":"$x^{3}$"}],"answer":"C","explanation":"With the same base, keep the base and add the $exponents: x^{2}·x^{5} = x^{2+5} = x^{7}$. Exponents are not multiplied.","explanationZh":"$同底数幂相乘，底数不变、指数相加：x^{2}·x^{5} = x^(2+5) = x^{7}。指数不能相乘。$","tags":["easy","power","exponent-rules"],"status":"published"},{"id":"math-power-003","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":2,"stem":"Simplify $(x^{3})^{2}$.","stemZh":"化简 $(x^{3})^{2}。$","options":[{"id":"A","text":"$x^{6}$"},{"id":"B","text":"$x^{9}$"},{"id":"C","text":"$x^{5}$"},{"id":"D","text":"$2x^{6}$"}],"answer":"A","explanation":"A power raised to a power multiplies the $exponents: (x^{3})^{2} = x^{3·2} = x^{6}$.","explanationZh":"$幂的乘方，底数不变、指数相乘：(x^{3})^{2} = x^(3·2) = x^{6}。$","tags":["easy","power","exponent-rules"],"status":"published"},{"id":"math-power-004","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":1,"stem":"For $x ≠ 0$, what does $x^{0}$ equal?","stemZh":"$当 x ≠ 0 时，x^{0}$ 等于多少？","options":[{"id":"A","text":"0"},{"id":"B","text":"x"},{"id":"C","text":"1/x"},{"id":"D","text":"1"}],"answer":"D","explanation":"Any non-zero number raised to the power 0 equals $1: x^{0} = 1 (x ≠ 0)$.","explanationZh":"任何非零数的 0 次幂都等于 $1：x^{0} = 1（x ≠ 0）。$","tags":["easy","power","exponent-rules"],"status":"published"},{"id":"math-power-005","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":2,"stem":"Evaluate $2^{-3}$.","stemZh":"计算 $2^{-3}$ 的值。","options":[{"id":"A","text":"1/8"},{"id":"B","text":"−8"},{"id":"C","text":"−1/8"},{"id":"D","text":"8"}],"answer":"A","explanation":"$2^{-3} = 1/2^{3} = 1/8. A$ negative exponent gives a reciprocal; it does not make the number negative.","explanationZh":"$2^{-3} = 1/2^{3} = 1/8。负指数表示取倒数，并不表示结果是负数。$","tags":["easy","power","exponent-rules"],"status":"published"},{"id":"math-power-006","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":2,"stem":"Evaluate $8^{2/3}$.","stemZh":"计算 $8^(2/3)$ 的值。","options":[{"id":"A","text":"2"},{"id":"B","text":"6"},{"id":"C","text":"4"},{"id":"D","text":"16"}],"answer":"C","explanation":"$8^(2/3) = (\\sqrt[3]{}8)^{2} = 2^{2} = 4$.","explanationZh":"$8^(2/3) = (\\sqrt[3]{}8)^{2} = 2^{2} = 4。$","tags":["easy","power","fractional-exponent"],"status":"published"},{"id":"math-power-007","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":1,"stem":"What is the domain of $y = \\sqrt{x}$?","stemZh":"函数 $y = \\sqrt{x}$ 的定义域是？","options":[{"id":"A","text":"$(0,+∞)$"},{"id":"B","text":"$[0,+∞)$"},{"id":"C","text":"$ℝ$"},{"id":"D","text":"$(−∞,0]$"}],"answer":"B","explanation":"$y = \\sqrt{x} = x^(1/2)$ requires $x ≥ 0$, so the domain is $[0,+∞)$.","explanationZh":"$y = \\sqrt{x} = x^(1/2)$ 要求 $x ≥ 0，所以定义域为 [0,+∞)。$","tags":["easy","power","domain"],"status":"published"},{"id":"math-power-008","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":2,"stem":"Which point lies on the graph of $y = x^{3}$?","stemZh":"下列哪个点在函数 $y = x^{3}$ 的图象上？","options":[{"id":"A","text":"$(−1, 1)$"},{"id":"B","text":"$(2, 6)$"},{"id":"C","text":"$(1, −1)$"},{"id":"D","text":"$(−2, −8)$"}],"answer":"D","explanation":"$(−2)^{3} = −8$, so $(−2, −8)$ satisfies $y = x^{3}$. Also $2^{3} = 8$, not 6.","explanationZh":"$(−2)^{3} = −8，所以 (−2, −8)$ 满足 $y = x^{3}；另外 2^{3} = 8$ 而不是 6。","tags":["easy","power","graph"],"status":"published"},{"id":"math-power-009","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":1,"stem":"Every power function $y = x^{a}$ passes through which point?","stemZh":"所有幂函数 $y = x^{a}$ 的图象都经过哪个点？","options":[{"id":"A","text":"$(1,1)$"},{"id":"B","text":"$(0,1)$"},{"id":"C","text":"$(1,0)$"},{"id":"D","text":"$(0,0)$"}],"answer":"A","explanation":"$1^{a} = 1$ for every real a, so every power function passes through $(1,1)$. When $a ≤ 0$ the graph does not pass through $(0,0)$.","explanationZh":"对任意实数 a 都有 $1^{a} = 1，所以所有幂函数都过 (1,1)；当 a ≤ 0$ 时图象不过 $(0,0)。$","tags":["easy","power","graph"],"status":"published"},{"id":"math-power-010","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":2,"stem":"On which interval is $y = x^{2}$ decreasing?","stemZh":"函数 $y = x^{2}$ 在哪个区间上单调递减？","options":[{"id":"A","text":"$[0,+∞)$"},{"id":"B","text":"$ℝ$"},{"id":"C","text":"$(−∞,0]$"},{"id":"D","text":"$(1,+∞)$"}],"answer":"C","explanation":"$y = x^{2}$ decreases on $(−∞,0]$ and increases on $[0,+∞); x = 0$ is its minimum point.","explanationZh":"$y = x^{2} 在 (−∞,0]$ 上递减，在 $[0,+∞)$ 上递增，x $= 0$ 是最小值点。","tags":["easy","power","monotonicity"],"status":"published"},{"id":"math-power-011","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":3,"stem":"What is the domain of $y = x^{−1/2}$?","stemZh":"函数 $y = x^{−1/2}$ 的定义域是？","options":[{"id":"A","text":"$[0,+∞)$"},{"id":"B","text":"$ℝ$"},{"id":"C","text":"$(−∞,0)∪(0,+∞)$"},{"id":"D","text":"$(0,+∞)$"}],"answer":"D","explanation":"$y = x^{−1/2} = 1/\\sqrt{x}$. The square root needs $x ≥ 0$ and the denominator needs $x ≠ 0$, so together $x > 0$.","explanationZh":"$y = x^{−1/2} = 1/\\sqrt{x}。偶次根式要求 x ≥ 0，分母要求 x ≠ 0，综合得 x > 0。$","tags":["medium","power","domain"],"status":"published"},{"id":"math-power-012","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":3,"stem":"The power function $f(x) = x^{a}$ passes through $(4, 2)$. Find $f(9)$.","stemZh":"幂函数 $f(x) = x^{a}$ 的图象经过点 $(4, 2)，求 f(9)。$","options":[{"id":"A","text":"9"},{"id":"B","text":"3"},{"id":"C","text":"81"},{"id":"D","text":"$\\sqrt{3}$"}],"answer":"B","explanation":"$4^{a} = 2$ gives $a = 1/2$, so $f(x) = \\sqrt{x}$ and $f(9) = 3$.","explanationZh":"$由 4^{a} = 2 得 a = 1/2，所以 f(x) = \\sqrt{x}，f(9) = 3。$","tags":["medium","power","find-exponent"],"status":"published"},{"id":"math-power-013","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":3,"stem":"The power function $f(x) = x^{a}$ passes through $(2, 1/4)$. Find a.","stemZh":"幂函数 $f(x) = x^{a}$ 的图象经过点 $(2, 1/4)，求$ a。","options":[{"id":"A","text":"−2"},{"id":"B","text":"2"},{"id":"C","text":"−1/2"},{"id":"D","text":"1/2"}],"answer":"A","explanation":"$2^{a} = 1/4 = 2^{-2}$, hence $a = −2$.","explanationZh":"$2^{a} = 1/4 = 2^{-2}，所以 a = −2。$","tags":["medium","power","find-exponent"],"status":"published"},{"id":"math-power-014","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":3,"stem":"Which function has domain $[0,+∞)$ and is increasing on that domain?","stemZh":"下列函数中定义域为 $[0,+∞)$ 且在该定义域上单调递增的是？","options":[{"id":"A","text":"$y = x^{-1}$"},{"id":"B","text":"$y = x^{−1/2}$"},{"id":"C","text":"$y = x^{1/2}$"},{"id":"D","text":"$y = x^{-2}$"}],"answer":"C","explanation":"$y = x^{1/2} = \\sqrt{x}$ has domain $[0,+∞)$ and increases there. The two negative-exponent functions are undefined at $x = 0$, and $x^{-1}$ has domain $x ≠ 0$.","explanationZh":"$y = x^(1/2) = \\sqrt{x}$ 的定义域为 $[0,+∞)$ 且在其上递增；两个负指数函数在 $x = 0$ 处无意义。","tags":["medium","power","domain"],"status":"published"},{"id":"math-power-015","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":3,"stem":"For $y = x^{-2}$, which statement about monotonicity is correct?","stemZh":"关于 $y = x^{-2}$ 的单调性，下列说法正确的是？","options":[{"id":"A","text":"decreasing on $(−∞,0)$ and decreasing on $(0,+∞)$"},{"id":"B","text":"decreasing on $(−∞,0)$ and increasing on $(0,+∞)$"},{"id":"C","text":"increasing on $(−∞,0)$ and increasing on $(0,+∞)$"},{"id":"D","text":"increasing on $(−∞,0)$ and decreasing on $(0,+∞)$"}],"answer":"D","explanation":"$y = 1/x^{2}$ is even. On $(−∞,0)$ it rises from 0 to $+∞$ as x increases; on $(0,+∞)$ it falls. It is not monotone across $x = 0$.","explanationZh":"$y = 1/x^{2}$ 是偶函数：在 $(−∞,0)$ 上由 0 增到 $+∞，在 (0,+∞)$ 上递减；不能跨越 $x = 0$ 讨论单调性。","tags":["medium","power","monotonicity"],"status":"published"},{"id":"math-power-016","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":3,"stem":"Simplify $(x^{1/2}·x^{2/3})^{6}$ for $x > 0$.","stemZh":"化简 $(x^(1/2)·x^(2/3))^{6}（x > 0）。$","options":[{"id":"A","text":"$x^{7}$"},{"id":"B","text":"$x^{5}$"},{"id":"C","text":"$x^{7/6}$"},{"id":"D","text":"$x^{4}$"}],"answer":"A","explanation":"First $x^{1/2}·x^{2/3} = x^{7/6}$; then $(x^{7/6})^{6} = x^{7}$.","explanationZh":"先得 $x^(1/2)·x^(2/3) = x^{7/6}，再乘方：(x^(7/6))^{6} = x^{7}。$","tags":["medium","power","exponent-rules"],"status":"published"},{"id":"math-power-017","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":3,"stem":"Which ordering of the three numbers is correct?","stemZh":"下列三个数的大小关系正确的是？","options":[{"id":"A","text":"$2^{3/4} < 2^{2/3} < 2^{1/2}$"},{"id":"B","text":"$2^{1/2} < 2^{2/3} < 2^{3/4}$"},{"id":"C","text":"$2^{2/3} < 2^{1/2} < 2^{3/4}$"},{"id":"D","text":"$2^{1/2} < 2^{3/4} < 2^{2/3}$"}],"answer":"B","explanation":"The base 2 is greater than 1, so $y = 2^{a}$ increases with the exponent. Since $1/2 < 2/3 < 3/4$, we get $2^{1/2} < 2^{2/3} < 2^{3/4}$.","explanationZh":"底数 $2 > 1，幂值随指数增大而增大。因为 1/2 < 2/3 < 3/4，所以 2^{1/2} < 2^{2/3} < 2^{3/4}。$","tags":["medium","power","comparison"],"status":"published"},{"id":"math-power-018","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":3,"stem":"A power function $f(x) = x^{a}$ is even and decreasing on $(0,+∞)$. Which value of a works?","stemZh":"幂函数 $f(x) = x^{a}$ 是偶函数且在 $(0,+∞)$ 上递减，a 可以取下列哪个值？","options":[{"id":"A","text":"3"},{"id":"B","text":"1/2"},{"id":"C","text":"−2"},{"id":"D","text":"2"}],"answer":"C","explanation":"$a = −2$ gives the even function $y = 1/x^{2}$, which decreases on $(0,+∞). a = 2$ and $a = 1/2$ increase on $(0,+∞)$, and $a = 3$ gives an odd function.","explanationZh":"$a = −2 时 y = 1/x^{2}$ 是偶函数且在 $(0,+∞)$ 上递减；a $= 2、a = 1/2$ 时递增；a $= 3$ 时为奇函数。","tags":["medium","power","monotonicity"],"status":"published"},{"id":"math-power-019","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":3,"stem":"The graph of $y = x^{1/3}$ is symmetric about:","stemZh":"函数 $y = x^(1/3)$ 的图象关于什么对称？","options":[{"id":"A","text":"the y-axis"},{"id":"B","text":"the x-axis"},{"id":"C","text":"the line $y = x$"},{"id":"D","text":"the origin"}],"answer":"D","explanation":"$y = \\sqrt[3]{}x$ is an odd function defined on $ℝ$, so its graph is symmetric about the origin.","explanationZh":"$y = \\sqrt[3]{}x$ 是定义在 $ℝ$ 上的奇函数，图象关于原点对称。","tags":["medium","power","graph"],"status":"published"},{"id":"math-power-020","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":3,"stem":"The power function $f(x) = x^{a}$ passes through $(2, 8)$. Find $f(3)$.","stemZh":"幂函数 $f(x) = x^{a}$ 的图象经过点 $(2, 8)，求 f(3)。$","options":[{"id":"A","text":"9"},{"id":"B","text":"27"},{"id":"C","text":"8"},{"id":"D","text":"81"}],"answer":"B","explanation":"$2^{a} = 8 = 2^{3}$ gives $a = 3$, so $f(3) = 3^{3} = 27$.","explanationZh":"$由 2^{a} = 8 = 2^{3} 得 a = 3，所以 f(3) = 3^{3} = 27。$","tags":["medium","power","find-exponent"],"status":"published"},{"id":"math-power-021","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":4,"stem":"Which statement about power functions is correct?","stemZh":"下列关于幂函数的说法正确的是？","options":[{"id":"A","text":"$y = x^{1/2}$ is increasing on $ℝ$"},{"id":"B","text":"$y = x^{-1}$ is decreasing on its whole domain"},{"id":"C","text":"$y = x^{2/3}$ is an even function"},{"id":"D","text":"$y = x^{3/2}$ has domain $ℝ$"}],"answer":"C","explanation":"$y = x^(2/3) = (\\sqrt[3]{}x)^{2}$ is defined on $ℝ$ and satisfies $f(−x) = f(x)$, so it is even. A is false because the domain is $[0,+∞); B$ is false since 1/x is not decreasing across $x = 0 (f(−1) = −1 < f(1) = 1); D$ is false because the domain is $[0,+∞)$.","explanationZh":"$y = x^(2/3) = (\\sqrt[3]{}x)^{2} 在 ℝ$ 上有定义且 $f(−x) = f(x)，是偶函数。A$ 错在定义域为 $[0,+∞)；B$ 错在 1/x 不能跨越 $x = 0 说递减（f(−1) = −1 < f(1) = 1）；D$ 错在定义域为 $[0,+∞)。$","tags":["hard","power","domain"],"status":"published"},{"id":"math-power-022","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":5,"stem":"What is the domain of $y = x^{−2/3}$?","stemZh":"函数 $y = x^(−2/3)$ 的定义域是？","options":[{"id":"A","text":"$(−∞,0)∪(0,+∞)$"},{"id":"B","text":"$(0,+∞)$"},{"id":"C","text":"$[0,+∞)$"},{"id":"D","text":"$ℝ$"}],"answer":"A","explanation":"$y = x^(−2/3) = 1/(\\sqrt[3]{}x)^{2}$. The cube root is defined for every real x, so only $x = 0$ has to be excluded; the domain is $x ≠ 0. (Restricting$ to $x > 0$ is a common error — that restriction only applies when the denominator of the exponent is even.)","explanationZh":"$y = x^(−2/3) = 1/(\\sqrt[3]{}x)^{2}。立方根对一切实数有意义，只需排除 x = 0，故定义域为 x ≠ 0。（误答 x > 0$ 是常见错误——只有当指数分母为偶数时才需 $x > 0。）$","tags":["hard","power","domain"],"status":"published"},{"id":"math-power-023","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":4,"stem":"Solve $x^{2/3} = 4$.","stemZh":"解方程 $x^(2/3) = 4。$","options":[{"id":"A","text":"{8}"},{"id":"B","text":"{−8}"},{"id":"C","text":"{4}"},{"id":"D","text":"{−8, 8}"}],"answer":"D","explanation":"$x^(2/3) = (\\sqrt[3]{}x)^{2} = 4$ gives $\\sqrt[3]{}x = \\pm2$, so $x = 8$ or $x = −8$. Both $check: (−8)^(2/3) = (−2)^{2} = 4$. Missing the negative root is the usual error.","explanationZh":"$x^(2/3) = (\\sqrt[3]{}x)^{2} = 4，得 \\sqrt[3]{}x = \\pm2，所以 x = 8 或 x = −8。检验：(−8)^(2/3) = (−2)^{2} = 4。漏掉负根是常见错误。$","tags":["hard","power","equations"],"status":"published"},{"id":"math-power-024","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":4,"stem":"A power function $f(x) = x^{a}$ is decreasing on $(0,+∞)$ and $f(1/2) = 4$. Find $f(2)$.","stemZh":"幂函数 $f(x) = x^{a} 在 (0,+∞)$ 上递减，且 $f(1/2) = 4，求 f(2)。$","options":[{"id":"A","text":"4"},{"id":"B","text":"1/4"},{"id":"C","text":"−4"},{"id":"D","text":"1/2"}],"answer":"B","explanation":"$(1/2)^{a} = 4$ gives $a = −2$, so $f(2) = 2^{-2} = 1/4$.","explanationZh":"$由 (1/2)^{a} = 4 得 a = −2，所以 f(2) = 2^{-2} = 1/4。$","tags":["hard","power","monotonicity"],"status":"published"},{"id":"math-power-025","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":5,"stem":"$f(x) = (m^{2} − m − 1)x^{m^{2} − 2m − 3}$ is a power function that is decreasing on $(0,+∞)$. Find m.","stemZh":"$f(x) = (m^{2} − m − 1)x^(m^{2} − 2m − 3)$ 是幂函数且在 $(0,+∞)$ 上递减，求 m。","options":[{"id":"A","text":"−1"},{"id":"B","text":"3"},{"id":"C","text":"2"},{"id":"D","text":"−2"}],"answer":"C","explanation":"A power function requires coefficient $1: m^{2} − m − 1 = 1$, i.e. $m^{2} − m − 2 = 0$, so $m = 2$ or $m = −1$. For $m = 2$ the exponent is $4 − 4 − 3 = −3 < 0$, which is decreasing on $(0,+∞)$; for $m = −1$ the exponent is 0, giving the constant $y = 1$. Hence $m = 2$.","explanationZh":"幂函数要求系数为 $1：m^{2} − m − 1 = 1，即 m^{2} − m − 2 = 0，得 m = 2 或 m = −1。m = 2$ 时指数为 $−3 < 0，在 (0,+∞)$ 上递减；m $= −1$ 时指数为 $0，为常函数。故 m = 2。$","tags":["hard","power","definition"],"status":"published"},{"id":"math-power-026","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":4,"stem":"Order $a = 2^{1/3}, b = 3^{1/2}, c = 2^{1/2}$.","stemZh":"比较 $a = 2^(1/3)、b = 3^(1/2)、c = 2^(1/2)$ 的大小。","options":[{"id":"A","text":"$a < c < b$"},{"id":"B","text":"$b < c < a$"},{"id":"C","text":"$c 1$ with $1/3 < 1/2$ gives $a <$ c. Same positive exponent 1/2 with bases $2 < 3$ gives $c <$ b. Hence $a < c <$ b.","explanationZh":"同底 $2 > 1，指数 1/3 < 1/2，故 a <$ c；同为正指数 $1/2，底数 2 < 3，故 c <$ b。所以 $a < c <$ b。","tags":["hard","power","comparison"],"status":"published"},{"id":"math-power-027","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":5,"stem":"A power function $y = x^{a}$ passes through $(3, 1/9)$. What is its domain?","stemZh":"幂函数 $y = x^{a}$ 的图象经过点 $(3, 1/9)，它的定义域是？$","options":[{"id":"A","text":"$[0,+∞)$"},{"id":"B","text":"$(0,+∞)$"},{"id":"C","text":"$ℝ$"},{"id":"D","text":"$(−∞,0)∪(0,+∞)$"}],"answer":"D","explanation":"$3^{a} = 1/9 = 3^{-2}$ gives $a = −2$, so $y = x^{-2} = 1/x^{2}$, whose domain is $x ≠ 0$.","explanationZh":"$由 3^{a} = 1/9 = 3^{-2} 得 a = −2，即 y = x^{-2} = 1/x^{2}，定义域为 x ≠ 0。$","tags":["hard","power","domain"],"status":"published"},{"id":"math-power-028","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":4,"stem":"Order $a = (1/3)^{1/2}, b = (1/2)^{1/2}, c = (1/2)^{1/3}$.","stemZh":"比较 $a = (1/3)^(1/2)、b = (1/2)^(1/2)、c = (1/2)^(1/3)$ 的大小。","options":[{"id":"A","text":"$c < b < a$"},{"id":"B","text":"$a < b < c$"},{"id":"C","text":"$b < a < c$"},{"id":"D","text":"$a < c < b$"}],"answer":"B","explanation":"Positive exponent 1/2 with bases $1/3 < 1/2$ gives $a <$ b. Base 1/2 lies in $(0,1)$, so a larger exponent gives a smaller $value: 1/2 > 1/3$ gives $b <$ c. Hence $a 1/3 故 b <$ c。所以 $a < b <$ c。","tags":["hard","power","comparison"],"status":"published"},{"id":"math-power-029","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":5,"stem":"A power function $f(x) = x^{a}$ is increasing on $(0,+∞)$ and $f(2) = \\sqrt{2}$. Find $f(1/2)$.","stemZh":"幂函数 $f(x) = x^{a} 在 (0,+∞)$ 上递增，且 $f(2) = \\sqrt{2}，求 f(1/2)。$","options":[{"id":"A","text":"$\\sqrt{2}$"},{"id":"B","text":"2"},{"id":"C","text":"$\\sqrt{2}/2$"},{"id":"D","text":"1/4"}],"answer":"C","explanation":"$2^{a} = \\sqrt{2} = 2^(1/2)$ gives $a = 1/2 (positive$, consistent with increasing). Thus $f(1/2) = (1/2)^(1/2) = 1/\\sqrt{2} = \\sqrt{2}/2$.","explanationZh":"$2^{a} = \\sqrt{2} = 2^(1/2) 得 a = 1/2（为正，与递增一致）。所以 f(1/2) = (1/2)^(1/2) = 1/\\sqrt{2} = \\sqrt{2}/2。$","tags":["hard","power","monotonicity"],"status":"published"},{"id":"math-power-030","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"power","type":"single-choice","difficulty":4,"stem":"Which exponent a gives $y = x^{a}$ domain $ℝ$ and an odd function?","stemZh":"下列哪个指数 $a 使 y = x^{a}$ 的定义域为 $ℝ$ 且为奇函数？","options":[{"id":"A","text":"1/3"},{"id":"B","text":"1/2"},{"id":"C","text":"−1/2"},{"id":"D","text":"2"}],"answer":"A","explanation":"$y = x^(1/3) = \\sqrt[3]{}x$ is defined on $ℝ$ and is odd. $y = x^{2}$ is $even; y = x^(1/2)$ has domain $[0,+∞); y = x^(−1/2)$ has domain $(0,+∞)$.","explanationZh":"$y = x^(1/3) = \\sqrt[3]{}x 在 ℝ$ 上有定义且为奇函数；y $= x^{2}$ 是偶函数；y $= x^(1/2)$ 定义域为 $[0,+∞)；y = x^(−1/2)$ 定义域为 $(0,+∞)。$","tags":["hard","power","domain"],"status":"published"},{"id":"math-exponential-001","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":1,"stem":"For $y = a^{x}$ to be an exponential function, the base a must satisfy:","stemZh":"要使 $y = a^{x}$ 是指数函数，底数 a 应满足：","options":[{"id":"A","text":"$a > 0$ and $a ≠ 1$"},{"id":"B","text":"$a > 0$"},{"id":"C","text":"$a ≠ 1$"},{"id":"D","text":"$a > 1$"}],"answer":"A","explanation":"By definition the base must be positive and not equal to 1. If $a = 1$ the function becomes the constant $y = 1$, and $a ≤ 0$ is not defined for all real x.","explanationZh":"由定义，底数必须大于 0 且不等于 $1。a = 1$ 时函数为常函数 $y = 1；a ≤ 0$ 时不能对一切实数 x 有定义。","tags":["easy","exponential","definition"],"status":"published"},{"id":"math-exponential-002","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":1,"stem":"Every exponential function $y = a^{x} (a > 0, a ≠ 1)$ passes through which point?","stemZh":"所有指数函数 $y = a^{x}（a > 0，a ≠ 1）的图象都经过哪个点？$","options":[{"id":"A","text":"$(1,0)$"},{"id":"B","text":"$(0,0)$"},{"id":"C","text":"$(1,1)$"},{"id":"D","text":"$(0,1)$"}],"answer":"D","explanation":"$a^{0} = 1$ for every admissible base, so the graph always passes through $(0,1)$.","explanationZh":"对任意合格底数都有 $a^{0} = 1，所以图象恒过点 (0,1)。$","tags":["easy","exponential","graph"],"status":"published"},{"id":"math-exponential-003","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":1,"stem":"What is the range of $y = 2^{x}$?","stemZh":"函数 $y = 2^{x}$ 的值域是？","options":[{"id":"A","text":"$ℝ$"},{"id":"B","text":"$(0,+∞)$"},{"id":"C","text":"$[0,+∞)$"},{"id":"D","text":"$[1,+∞)$"}],"answer":"B","explanation":"$2^{x} > 0$ for every real x and every positive value is attained, so the range is $(0,+∞)$. The value 0 is only approached as $x → −∞$.","explanationZh":"对任意实数 x 都有 $2^{x} > 0，且能取遍所有正数，故值域为 (0,+∞)；0$ 只在 $x → −∞$ 时逼近。","tags":["easy","exponential","range"],"status":"published"},{"id":"math-exponential-004","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":2,"stem":"What is the domain of $y = 3^{x}$?","stemZh":"函数 $y = 3^{x}$ 的定义域是？","options":[{"id":"A","text":"$(0,+∞)$"},{"id":"B","text":"$[0,+∞)$"},{"id":"C","text":"$ℝ$"},{"id":"D","text":"$(−∞,0)$"}],"answer":"C","explanation":"An exponential function is defined for every real exponent x, so the domain is $ℝ$.","explanationZh":"指数函数对任意实数指数 x 都有意义，所以定义域为 $ℝ。$","tags":["easy","exponential","domain"],"status":"published"},{"id":"math-exponential-005","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":1,"stem":"Which of the following functions is decreasing on $ℝ$?","stemZh":"下列函数中在 $ℝ$ 上单调递减的是？","options":[{"id":"A","text":"$y = (1/2)^{x}$"},{"id":"B","text":"$y = 2^{x}$"},{"id":"C","text":"$y = 3^{x}$"},{"id":"D","text":"$y = e^{x}$"}],"answer":"A","explanation":"$y = a^{x}$ decreases when $0 < a < 1$. The bases 2, 3 and e are all greater than 1, so those functions increase.","explanationZh":"$当 0 < a < 1 时 y = a^{x} 递减。2、3、e$ 都大于 1，对应的函数均递增。","tags":["easy","exponential","monotonicity"],"status":"published"},{"id":"math-exponential-006","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":2,"stem":"Solve $2^{x} = 8$.","stemZh":"解方程 $2^{x} = 8。$","options":[{"id":"A","text":"2"},{"id":"B","text":"4"},{"id":"C","text":"8"},{"id":"D","text":"3"}],"answer":"D","explanation":"$8 = 2^{3}$, so $2^{x} = 2^{3}$ and $x = 3$.","explanationZh":"$8 = 2^{3}，所以 2^{x} = 2^{3}，得 x = 3。$","tags":["easy","exponential","equations"],"status":"published"},{"id":"math-exponential-007","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":1,"stem":"Which function is increasing on $ℝ$?","stemZh":"下列函数中在 $ℝ$ 上单调递增的是？","options":[{"id":"A","text":"$y = (1/3)^{x}$"},{"id":"B","text":"$y = 0.5^{x}$"},{"id":"C","text":"$y = 2^{x}$"},{"id":"D","text":"$y = (1/e)^{x}$"}],"answer":"C","explanation":"$y = a^{x}$ increases when $a > 1$. The other three bases $(1/3, 0.5, 1/e)$ all lie in $(0,1)$.","explanationZh":"$当 a > 1 时 y = a^{x}$ 递增。其余三个底数 1/3、0.5、1/e 都在 $(0,1)$ 内。","tags":["easy","exponential","monotonicity"],"status":"published"},{"id":"math-exponential-008","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":2,"stem":"Evaluate $2^{2}·2^{5}$.","stemZh":"计算 $2^{2}·2^{5}。$","options":[{"id":"A","text":"$2^{10}$"},{"id":"B","text":"128"},{"id":"C","text":"64"},{"id":"D","text":"256"}],"answer":"B","explanation":"$2^{2}·2^{5} = 2^{2+5} = 2^{7} = 128$. Exponents are added, not multiplied.","explanationZh":"$2^{2}·2^{5} = 2^(2+5) = 2^{7} = 128。指数相加而非相乘。$","tags":["easy","exponential","exponent-rules"],"status":"published"},{"id":"math-exponential-009","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":1,"stem":"Which of the following is an exponential function?","stemZh":"下列函数中是指数函数的是？","options":[{"id":"A","text":"$y = 3^{x}$"},{"id":"B","text":"$y = x^{3}$"},{"id":"C","text":"$y = 3x$"},{"id":"D","text":"$y = \\log_{3}x$"}],"answer":"A","explanation":"An exponential function has the variable in the $exponent, y = a^{x}$ with $a > 0$ and $a ≠ 1. y = x^{3}$ is a power function and $y = \\log_{3}x$ is a logarithmic function.","explanationZh":"指数函数的自变量在指数位置，形式为 $y = a^{x}（a > 0，a ≠ 1）。y = x^{3}$ 是幂函数，y $= \\log_{3}x$ 是对数函数。","tags":["easy","exponential","definition"],"status":"published"},{"id":"math-exponential-010","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":2,"stem":"If $f(x) = 2^{x}$, what is $f(3)$?","stemZh":"$若 f(x) = 2^{x}，则 f(3)$ 等于多少？","options":[{"id":"A","text":"6"},{"id":"B","text":"9"},{"id":"C","text":"4"},{"id":"D","text":"8"}],"answer":"D","explanation":"$f(3) = 2^{3} = 8$. Note that $2·3 = 6$ is the product, not the power.","explanationZh":"$f(3) = 2^{3} = 8。注意 2·3 = 6$ 是乘法而不是乘方。","tags":["easy","exponential","evaluation"],"status":"published"},{"id":"math-exponential-011","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":3,"stem":"Solve $(1/2)^{x} = 8$.","stemZh":"解方程 $(1/2)^{x} = 8。$","options":[{"id":"A","text":"3"},{"id":"B","text":"−3"},{"id":"C","text":"1/8"},{"id":"D","text":"−1/3"}],"answer":"B","explanation":"$(1/2)^{x} = 2^{−x}$ and $8 = 2^{3}$, so −x $= 3$ and $x = −3$.","explanationZh":"$(1/2)^{x} = 2^{−x}，8 = 2^{3}，所以$ −x $= 3，x = −3。$","tags":["medium","exponential","equations"],"status":"published"},{"id":"math-exponential-012","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":3,"stem":"Which of the following numbers is the largest?","stemZh":"下列数中最大的是？","options":[{"id":"A","text":"$(1/2)^{3}$"},{"id":"B","text":"$(1/2)^{-1}$"},{"id":"C","text":"$(1/2)^{-2}$"},{"id":"D","text":"$(1/2)^{0}$"}],"answer":"C","explanation":"The values are 1/8, 2, 4 and 1. Since the base 1/2 lies in $(0,1), y = (1/2)^{x}$ is decreasing, so the smallest exponent $(−2)$ gives the largest value, 4.","explanationZh":"四个值分别为 $1/8、2、4、1。底数 1/2 ∈ (0,1)，y = (1/2)^{x} 递减，因此指数最小（−2）时值最大，为 4。$","tags":["medium","exponential","comparison"],"status":"published"},{"id":"math-exponential-013","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":3,"stem":"What is the range of $y = 2^{x} + 1$?","stemZh":"函数 $y = 2^{x} + 1$ 的值域是？","options":[{"id":"A","text":"$(1,+∞)$"},{"id":"B","text":"$(0,+∞)$"},{"id":"C","text":"$[1,+∞)$"},{"id":"D","text":"$ℝ$"}],"answer":"A","explanation":"$2^{x} > 0$ gives $2^{x} + 1 > 1$, and every value greater than 1 is attained. The value 1 is only approached as $x → −∞$, so the interval is open.","explanationZh":"$由 2^{x} > 0 得 2^{x} + 1 > 1，且大于 1 的值都能取到；1$ 只在 $x → −∞$ 时逼近，故区间为开区间。","tags":["medium","exponential","range"],"status":"published"},{"id":"math-exponential-014","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":3,"stem":"Which point lies on the graph of $y = 2^{x−1}$?","stemZh":"下列哪个点在函数 $y = 2^(x−1)$ 的图象上？","options":[{"id":"A","text":"$(1,0)$"},{"id":"B","text":"$(0,1)$"},{"id":"C","text":"$(1,2)$"},{"id":"D","text":"$(1,1)$"}],"answer":"D","explanation":"When $x = 1, y = 2^{0} = 1$, so $(1,1)$ is on the graph. The graph of $y = 2^{x−1}$ is $y = 2^{x}$ shifted right by 1, which moves $(0,1)$ to $(1,1)$.","explanationZh":"$x = 1 时 y = 2^{0} = 1，所以 (1,1)$ 在图象上。y $= 2^(x−1)$ 的图象由 $y = 2^{x}$ 向右平移 1 个单位得到，点 $(0,1)$ 变为 $(1,1)。$","tags":["medium","exponential","graph"],"status":"published"},{"id":"math-exponential-015","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":3,"stem":"Solve $9^{x} = 27$.","stemZh":"解方程 $9^{x} = 27。$","options":[{"id":"A","text":"2/3"},{"id":"B","text":"3/2"},{"id":"C","text":"3"},{"id":"D","text":"2"}],"answer":"B","explanation":"Write both sides in base $3: 9^{x} = 3^{2x}$ and $27 = 3^{3}$, so $2x = 3$ and $x = 3/2$.","explanationZh":"两边化为同底 $3：9^{x} = 3^(2x)，27 = 3^{3}，所以 2x = 3，x = 3/2。$","tags":["medium","exponential","equations"],"status":"published"},{"id":"math-exponential-016","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":3,"stem":"Solve $4^{x} = 1/8$.","stemZh":"解方程 $4^{x} = 1/8。$","options":[{"id":"A","text":"−2/3"},{"id":"B","text":"3/2"},{"id":"C","text":"−3/2"},{"id":"D","text":"3"}],"answer":"C","explanation":"$4^{x} = 2^{2x}$ and $1/8 = 2^{-3}$, so $2x = −3$ and $x = −3/2$.","explanationZh":"$4^{x} = 2^(2x)，1/8 = 2^{-3}，所以 2x = −3，x = −3/2。$","tags":["medium","exponential","equations"],"status":"published"},{"id":"math-exponential-017","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":3,"stem":"If $(1/3)^{x} > (1/3)^{2}, then$:","stemZh":"$若 (1/3)^{x} > (1/3)^{2}，则：$","options":[{"id":"A","text":"$x < 2$"},{"id":"B","text":"$x > 2$"},{"id":"C","text":"$x > 1/9$"},{"id":"D","text":"$x < 1/9$"}],"answer":"A","explanation":"The base 1/3 lies in $(0,1)$, so $y = (1/3)^{x}$ is decreasing on $ℝ: a$ larger function value means a smaller exponent, hence $x < 2$.","explanationZh":"底数 $1/3 ∈ (0,1)，y = (1/3)^{x} 在 ℝ$ 上递减：函数值越大则指数越小，所以 $x < 2。$","tags":["medium","exponential","monotonicity"],"status":"published"},{"id":"math-exponential-018","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":3,"stem":"If $2^{a} = 3$ and $2^{b} = 5$, what is $2^{a+b}$?","stemZh":"$若 2^{a} = 3，2^{b} = 5，则 2^(a+b)$ 等于多少？","options":[{"id":"A","text":"8"},{"id":"B","text":"10"},{"id":"C","text":"$2^{15}$"},{"id":"D","text":"15"}],"answer":"D","explanation":"$2^{a+b} = 2^{a}·2^{b} = 3·5 = 15$.","explanationZh":"$2^(a+b) = 2^{a}·2^{b} = 3·5 = 15。$","tags":["medium","exponential","exponent-rules"],"status":"published"},{"id":"math-exponential-019","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":3,"stem":"What is the domain of $y = 1/(2^{x} − 1)$?","stemZh":"函数 $y = 1/(2^{x} − 1)$ 的定义域是？","options":[{"id":"A","text":"$x > 0$"},{"id":"B","text":"$x < 0$"},{"id":"C","text":"$x ≠ 0$"},{"id":"D","text":"$ℝ$"}],"answer":"C","explanation":"The denominator cannot $vanish: 2^{x} − 1 ≠ 0$, i.e. $2^{x} ≠ 1$, so $x ≠ 0$.","explanationZh":"分母不能为 $0：2^{x} − 1 ≠ 0，即 2^{x} ≠ 1，所以 x ≠ 0。$","tags":["medium","exponential","domain"],"status":"published"},{"id":"math-exponential-020","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":3,"stem":"A bacteria culture starts at 100 and doubles every hour. What is the population after 3 hours?","stemZh":"某种细菌初始数量为 100，每过 1 小时翻一倍。3 小时后的数量是多少？","options":[{"id":"A","text":"600"},{"id":"B","text":"800"},{"id":"C","text":"300"},{"id":"D","text":"1600"}],"answer":"B","explanation":"The model is $P(t) = 100·2^{t}$, so $P(3) = 100·2^{3} = 800$.","explanationZh":"模型为 $P(t) = 100·2^{t}，所以 P(3) = 100·2^{3} = 800。$","tags":["medium","exponential","word-problem"],"status":"published"},{"id":"math-exponential-021","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":4,"stem":"Solve $2^{2x} − 5·2^{x} + 4 = 0$.","stemZh":"解方程 $2^(2x) − 5·2^{x} + 4 = 0。$","options":[{"id":"A","text":"{0}"},{"id":"B","text":"{2}"},{"id":"C","text":"{1,4}"},{"id":"D","text":"{0,2}"}],"answer":"D","explanation":"Let $t = 2^{x} > 0$. Then $t^{2} − 5t + 4 = 0$ gives $t = 1$ or $t = 4$, so $x = 0$ or $x = 2. (C\\text{ lists the t-values, not the }x-values.)$","explanationZh":"$令 t = 2^{x} > 0，则 t^{2} − 5t + 4 = 0，得 t = 1 或 t = 4，于是 x = 0 或 x = 2。（C$ 给的是 t 的值而不是 x 的值。）","tags":["hard","exponential","equations"],"status":"published"},{"id":"math-exponential-022","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":5,"stem":"If $0 < a < 1$, which ordering is correct?","stemZh":"$若 0 < a < 1，则下列大小关系正确的是？$","options":[{"id":"A","text":"$a^{-2} > a^{-1} > 1$"},{"id":"B","text":"$a^{-2} < a^{-1} < 1$"},{"id":"C","text":"$1 > a^{-1} > a^{-2}$"},{"id":"D","text":"$a^{-1} > a^{-2} > 1$"}],"answer":"A","explanation":"With $0 < a < 1$ the function $y = a^{x}$ is decreasing. Since $−2 < −1 < 0$, we get $a^{-2} > a^{-1} > a^{0} = 1$. For example $a = 1/2$ gives $4 > 2 > 1$.","explanationZh":"$0 < a < 1 时 y = a^{x}$ 递减。因为 $−2 < −1 < 0，所以 a^{-2} > a^{-1} > a^{0} = 1。例如 a = 1/2 时 4 > 2 > 1。$","tags":["hard","exponential","comparison"],"status":"published"},{"id":"math-exponential-023","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":4,"stem":"Compare $a = 2^0.6, b = 4^0.3$ and $c = (1/2)^{−0.5}$.","stemZh":"比较 $a = 2^0.6、b = 4^0.3、c = (1/2)^(−0.5)$ 的大小。","options":[{"id":"A","text":"$a < b < c$"},{"id":"B","text":"$c 1$ and $0.5 < 0.6$, we get $c < a =$ b.","explanationZh":"$b = 4^0.3 = (2^{2})^0.3 = 2^0.6 =$ a，c $= (1/2)^(−0.5) = 2^0.5。底数 2 > 1 且 0.5 < 0.6，所以 c < a =$ b。","tags":["hard","exponential","comparison"],"status":"published"},{"id":"math-exponential-024","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":4,"stem":"An exponential function $f(x) = a^{x}$ satisfies $f(1) = 4$. Find $f(−2)$.","stemZh":"指数函数 $f(x) = a^{x}$ 满足 $f(1) = 4，求 f(−2)。$","options":[{"id":"A","text":"1/8"},{"id":"B","text":"1/16"},{"id":"C","text":"−16"},{"id":"D","text":"16"}],"answer":"B","explanation":"$f(1) = a = 4$, so $f(−2) = 4^{-2} = 1/16$.","explanationZh":"$f(1) = a = 4，所以 f(−2) = 4^{-2} = 1/16。$","tags":["hard","exponential","evaluation"],"status":"published"},{"id":"math-exponential-025","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":5,"stem":"If $3^{x} = 4$ and $3^{y} = 12$, find $3^{2x−y}$.","stemZh":"$若 3^{x} = 4，3^{y} = 12，求 3^(2x−y)。$","options":[{"id":"A","text":"3/4"},{"id":"B","text":"4"},{"id":"C","text":"16"},{"id":"D","text":"4/3"}],"answer":"D","explanation":"$3^{2x} = (3^{x})^{2} = 16$, so $3^{2x−y} = 3^{2x}/3^{y} = 16/12 = 4/3$.","explanationZh":"$3^(2x) = (3^{x})^{2} = 16，所以 3^(2x−y) = 3^(2x)/3^{y} = 16/12 = 4/3。$","tags":["hard","exponential","exponent-rules"],"status":"published"},{"id":"math-exponential-026","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":4,"stem":"What is the range of $y = 1/(2^{x} + 1)$?","stemZh":"函数 $y = 1/(2^{x} + 1)$ 的值域是？","options":[{"id":"A","text":"$(0,1)$"},{"id":"B","text":"$(0,+∞)$"},{"id":"C","text":"$(1,+∞)$"},{"id":"D","text":"$ℝ$"}],"answer":"A","explanation":"$2^{x} > 0$ gives $2^{x} + 1 > 1$, hence $0 < 1/(2^{x} + 1) < 1$. As $x → +∞$ the value tends to 0 and as $x → −∞$ it tends to 1, but neither endpoint is reached.","explanationZh":"$由 2^{x} > 0 得 2^{x} + 1 > 1，故 0 < 1/(2^{x} + 1) < 1。x → +∞$ 时趋于 $0，x → −∞$ 时趋于 1，但两个端点都取不到。","tags":["hard","exponential","range"],"status":"published"},{"id":"math-exponential-027","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":5,"stem":"Solve $4^{x} − 6·2^{x} + 8 = 0$.","stemZh":"解方程 $4^{x} − 6·2^{x} + 8 = 0。$","options":[{"id":"A","text":"{1}"},{"id":"B","text":"{2}"},{"id":"C","text":"{1,2}"},{"id":"D","text":"{2,4}"}],"answer":"C","explanation":"Let $t = 2^{x} > 0$. Then $t^{2} − 6t + 8 = 0$ gives $t = 2$ or $t = 4$, so $x = 1$ or $x = 2$. Both roots are positive, so both are admissible.","explanationZh":"$令 t = 2^{x} > 0，则 t^{2} − 6t + 8 = 0，得 t = 2 或 t = 4，所以 x = 1 或 x = 2。两根均为正，都符合要求。$","tags":["hard","exponential","equations"],"status":"published"},{"id":"math-exponential-028","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":4,"stem":"A sample has mass $m(t) = 100·(1/2)^{t/3}$ grams after t days. What is the mass after 6 days?","stemZh":"某样品的质量满足 $m(t) = 100·(1/2)^(t/3)（克），t 为天数。6$ 天后质量是多少？","options":[{"id":"A","text":"50"},{"id":"B","text":"25"},{"id":"C","text":"12.5"},{"id":"D","text":"200"}],"answer":"B","explanation":"$m(6) = 100·(1/2)^{6/3} = 100·(1/2)^{2} = 100·1/4 = 25$ g.","explanationZh":"$m(6) = 100·(1/2)^(6/3) = 100·(1/2)^{2} = 100·1/4 = 25$ 克。","tags":["hard","exponential","word-problem"],"status":"published"},{"id":"math-exponential-029","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":5,"stem":"The graphs of $y = 2^{x}$ and $y = (1/2)^{x}$ are symmetric about:","stemZh":"函数 $y = 2^{x} 与 y = (1/2)^{x}$ 的图象关于什么对称？","options":[{"id":"A","text":"the origin"},{"id":"B","text":"the line $y = x$"},{"id":"C","text":"the x-axis"},{"id":"D","text":"the y-axis"}],"answer":"D","explanation":"$(1/2)^{x} = 2^{−x}$, so replacing x by −x turns one graph into the other, which is a reflection in the y-axis.","explanationZh":"$(1/2)^{x} = 2^{−x}，把 x$ 换成 −x 即可由一个图象得到另一个，这是关于 y 轴对称。","tags":["hard","exponential","graph"],"status":"published"},{"id":"math-exponential-030","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"exponential","type":"single-choice","difficulty":4,"stem":"If $f(x) = a^{x}$ is decreasing on $ℝ$ and $f(2) = 1/9$, find a.","stemZh":"$若 f(x) = a^{x} 在 ℝ$ 上递减且 $f(2) = 1/9，求$ a。","options":[{"id":"A","text":"1/3"},{"id":"B","text":"−1/3"},{"id":"C","text":"3"},{"id":"D","text":"1/9"}],"answer":"A","explanation":"Decreasing means $0 < a < 1$. From $a^{2} = 1/9$ and $a > 0$ we get $a = 1/3$; the base can never be negative, and $a = 3$ would give an increasing function.","explanationZh":"递减说明 $0 < a < 1。由 a^{2} = 1/9 且 a > 0 得 a = 1/3；底数不能为负，a = 3$ 时函数递增。","tags":["hard","exponential","monotonicity"],"status":"published"},{"id":"math-logarithmic-001","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":1,"stem":"Evaluate $\\log_{_{2}}8$.","stemZh":"计算 $\\log_{_{2}}8$ 的值。","options":[{"id":"A","text":"2"},{"id":"B","text":"4"},{"id":"C","text":"3"},{"id":"D","text":"8"}],"answer":"C","explanation":"$2^{3} = 8$, so $\\log_{_{2}}8 = 3$.","explanationZh":"因为 $2^{3} = 8，所以 \\log_{_{2}}8 = 3。$","tags":["easy","logarithmic","definition"],"status":"published"},{"id":"math-logarithmic-002","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":1,"stem":"Evaluate $\\log_{a}1 (a > 0, a ≠ 1)$.","stemZh":"计算 $\\log_{a}1（a > 0，a ≠ 1）。$","options":[{"id":"A","text":"0"},{"id":"B","text":"1"},{"id":"C","text":"a"},{"id":"D","text":"−1"}],"answer":"A","explanation":"$a^{0} = 1$ for every admissible base, so $\\log_{a}1 = 0$.","explanationZh":"对任意合格底数都有 $a^{0} = 1，所以 \\log_{a}1 = 0。$","tags":["easy","logarithmic","\\log-rules"],"status":"published"},{"id":"math-logarithmic-003","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":2,"stem":"What is the domain of $y = \\log_{_{2}}x$?","stemZh":"函数 $y = \\log_{_{2}}x$ 的定义域是？","options":[{"id":"A","text":"$ℝ$"},{"id":"B","text":"$[0,+∞)$"},{"id":"C","text":"$(−∞,0)∪(0,+∞)$"},{"id":"D","text":"$(0,+∞)$"}],"answer":"D","explanation":"A logarithm is only defined for positive arguments, so $x > 0$.","explanationZh":"对数的真数必须大于 $0，所以 x > 0。$","tags":["easy","logarithmic","domain"],"status":"published"},{"id":"math-logarithmic-004","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":1,"stem":"Evaluate $\\log_{3}3$.","stemZh":"计算 $\\log_{3}3$ 的值。","options":[{"id":"A","text":"0"},{"id":"B","text":"1"},{"id":"C","text":"3"},{"id":"D","text":"9"}],"answer":"B","explanation":"$\\log_{a}a = 1$ for every admissible base, so $\\log_{3}3 = 1$.","explanationZh":"对任意合格底数都有 $\\log_{a}a = 1，所以 \\log_{3}3 = 1。$","tags":["easy","logarithmic","\\log-rules"],"status":"published"},{"id":"math-logarithmic-005","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":2,"stem":"Every logarithmic function $y = \\log_{a}x$ passes through which point?","stemZh":"所有对数函数 $y = \\log_{a}x$ 的图象都经过哪个点？","options":[{"id":"A","text":"$(0,1)$"},{"id":"B","text":"$(0,0)$"},{"id":"C","text":"$(1,0)$"},{"id":"D","text":"$(1,1)$"}],"answer":"C","explanation":"$\\log_{a}1 = 0$, so the point $(1,0)$ lies on every logarithmic graph.","explanationZh":"因为 $\\log_{a}1 = 0，所以点 (1,0)$ 在所有对数函数图象上。","tags":["easy","logarithmic","graph"],"status":"published"},{"id":"math-logarithmic-006","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":1,"stem":"What is the range of $y = \\log_{_{2}}x$?","stemZh":"函数 $y = \\log_{_{2}}x$ 的值域是？","options":[{"id":"A","text":"$ℝ$"},{"id":"B","text":"$(0,+∞)$"},{"id":"C","text":"$[0,+∞)$"},{"id":"D","text":"$(1,+∞)$"}],"answer":"A","explanation":"As x runs over $(0,+∞), \\log_{_{2}}x$ takes every real value, so the range is $ℝ$.","explanationZh":"当 x 取遍 $(0,+∞) 时，\\log_{_{2}}x$ 取遍所有实数，所以值域为 $ℝ。$","tags":["easy","logarithmic","range"],"status":"published"},{"id":"math-logarithmic-007","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":2,"stem":"Evaluate $\\log_{_{2}}(4·8)$.","stemZh":"计算 $\\log_{_{2}}(4·8)$ 的值。","options":[{"id":"A","text":"3"},{"id":"B","text":"2"},{"id":"C","text":"6"},{"id":"D","text":"5"}],"answer":"D","explanation":"$\\log_{_{2}}(4·8) = \\log_{_{2}}32 = 5$, or equivalently $\\log_{_{2}}4 + \\log_{_{2}}8 = 2 + 3 = 5$. Note that $\\log_{_{2}}(4·8)$ is not $\\log_{_{2}}4 · \\log_{_{2}}8 = 6$.","explanationZh":"$\\log_{_{2}}(4·8) = \\log_{_{2}}32 = 5，或 \\log_{_{2}}4 + \\log_{_{2}}8 = 2 + 3 = 5。注意 \\log_{_{2}}(4·8)$ 不等于 $\\log_{_{2}}4 · \\log_{_{2}}8 = 6。$","tags":["easy","logarithmic","\\log-rules"],"status":"published"},{"id":"math-logarithmic-008","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":1,"stem":"What is the inverse function of $y = 2^{x}$?","stemZh":"函数 $y = 2^{x}$ 的反函数是？","options":[{"id":"A","text":"$y = \\log_{_{2}}(1/x)$"},{"id":"B","text":"$y = \\log_{_{2}}x$"},{"id":"C","text":"$y = x^{2}$"},{"id":"D","text":"$y = (1/2)^{x}$"}],"answer":"B","explanation":"The exponential $y = a^{x}$ and the logarithm $y = \\log_{a}x$ are inverse functions of each other.","explanationZh":"指数函数 $y = a^{x}$ 与对数函数 $y = \\log_{a}x$ 互为反函数。","tags":["easy","logarithmic","inverse"],"status":"published"},{"id":"math-logarithmic-009","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":2,"stem":"Evaluate $\\log_{_{2}}(1/4)$.","stemZh":"计算 $\\log_{_{2}}(1/4)$ 的值。","options":[{"id":"A","text":"2"},{"id":"B","text":"1/2"},{"id":"C","text":"−2"},{"id":"D","text":"−1/2"}],"answer":"C","explanation":"$1/4 = 2^{-2}$, so $\\log_{_{2}}(1/4) = −2$.","explanationZh":"$1/4 = 2^{-2}，所以 \\log_{_{2}}(1/4) = −2。$","tags":["easy","logarithmic","definition"],"status":"published"},{"id":"math-logarithmic-010","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":1,"stem":"For $\\log_{a}b$ to be defined, which conditions are required?","stemZh":"要使 $\\log_{a}b$ 有意义，需要满足的条件是？","options":[{"id":"A","text":"$a > 0, a ≠ 1, b > 0$"},{"id":"B","text":"$a > 0, b > 0$"},{"id":"C","text":"$a ≠ 1, b > 0$"},{"id":"D","text":"$a > 0, a ≠ 1$"}],"answer":"A","explanation":"The base must be positive and different from 1, and the argument must be positive. All three conditions are needed.","explanationZh":"底数必须大于 0 且不等于 1，真数必须大于 0，三个条件缺一不可。","tags":["easy","logarithmic","definition"],"status":"published"},{"id":"math-logarithmic-011","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":3,"stem":"Evaluate $\\log_{_{2}}(32/4)$.","stemZh":"计算 $\\log_{_{2}}(32/4)$ 的值。","options":[{"id":"A","text":"5"},{"id":"B","text":"2"},{"id":"C","text":"8"},{"id":"D","text":"3"}],"answer":"D","explanation":"$\\log_{_{2}}(32/4) = \\log_{_{2}}8 = 3$, or $\\log_{_{2}}32 − \\log_{_{2}}4 = 5 − 2 = 3$. The quotient rule subtracts logs; it does not divide them.","explanationZh":"$\\log_{_{2}}(32/4) = \\log_{_{2}}8 = 3，或 \\log_{_{2}}32 − \\log_{_{2}}4 = 5 − 2 = 3。商的对数等于对数相减，不是相除。$","tags":["medium","logarithmic","\\log-rules"],"status":"published"},{"id":"math-logarithmic-012","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":3,"stem":"Evaluate $\\log_{_{2}}(4^{3})$.","stemZh":"计算 $\\log_{_{2}}(4^{3})$ 的值。","options":[{"id":"A","text":"8"},{"id":"B","text":"6"},{"id":"C","text":"12"},{"id":"D","text":"9"}],"answer":"B","explanation":"The power rule gives $\\log_{_{2}}(4^{3}) = 3·\\log_{_{2}}4 = 3·2 = 6$. It is not $(\\log_{_{2}}4)^{3} = 8$.","explanationZh":"由幂的运算法则 $\\log_{_{2}}(4^{3}) = 3·\\log_{_{2}}4 = 3·2 = 6，而不是 (\\log_{_{2}}4)^{3} = 8。$","tags":["medium","logarithmic","\\log-rules"],"status":"published"},{"id":"math-logarithmic-013","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":3,"stem":"Evaluate $\\log_{_{8}}16$.","stemZh":"计算 $\\log_{_{8}}16$ 的值。","options":[{"id":"A","text":"2"},{"id":"B","text":"3/4"},{"id":"C","text":"4/3"},{"id":"D","text":"1/2"}],"answer":"C","explanation":"Using the change-of-base $formula, \\log_{_{8}}16 = \\log_{_{2}}16 / \\log_{_{2}}8 = 4/3$. Writing it the other way round gives the common wrong answer 3/4.","explanationZh":"由换底公式 $\\log_{_{8}}16 = \\log_{_{2}}16 / \\log_{_{2}}8 = 4/3。分子分母颠倒会得到常见错误答案 3/4。$","tags":["medium","logarithmic","change-of-base"],"status":"published"},{"id":"math-logarithmic-014","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":3,"stem":"What is the domain of $y = \\log_{_{2}}(x − 3)$?","stemZh":"函数 $y = \\log_{_{2}}(x − 3)$ 的定义域是？","options":[{"id":"A","text":"$(3,+∞)$"},{"id":"B","text":"$[3,+∞)$"},{"id":"C","text":"$(−∞,3)$"},{"id":"D","text":"$x ≠ 3$"}],"answer":"A","explanation":"The argument must be strictly $positive: x − 3 > 0$, so $x > 3$.","explanationZh":"真数必须严格大于 $0：x − 3 > 0，所以 x > 3。$","tags":["medium","logarithmic","domain"],"status":"published"},{"id":"math-logarithmic-015","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":3,"stem":"Solve $\\log_{3}x = 2$.","stemZh":"解方程 $\\log_{3}x = 2。$","options":[{"id":"A","text":"6"},{"id":"B","text":"8"},{"id":"C","text":"3"},{"id":"D","text":"9"}],"answer":"D","explanation":"$\\log_{3}x = 2$ means $x = 3^{2} = 9. (3 \\cdot 2 = 6$ is not the conversion.)","explanationZh":"$\\log_{3}x = 2 即 x = 3^{2} = 9。（3·2 = 6$ 是错误的转化方式。）","tags":["medium","logarithmic","equations"],"status":"published"},{"id":"math-logarithmic-016","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":3,"stem":"Solve $\\log_{3}(x + 1) = 2$.","stemZh":"解方程 $\\log_{3}(x + 1) = 2。$","options":[{"id":"A","text":"5"},{"id":"B","text":"8"},{"id":"C","text":"9"},{"id":"D","text":"10"}],"answer":"B","explanation":"$x + 1 = 3^{2} = 9$, so $x = 8. Check: x + 1 = 9 > 0$, so the solution is admissible.","explanationZh":"$x + 1 = 3^{2} = 9，所以 x = 8。检验：x + 1 = 9 > 0，符合定义域要求。$","tags":["medium","logarithmic","equations"],"status":"published"},{"id":"math-logarithmic-017","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":3,"stem":"If $\\log_{_{2}}x < 1, then$:","stemZh":"$若 \\log_{_{2}}x < 1，则：$","options":[{"id":"A","text":"$x < 2$"},{"id":"B","text":"$x > 2$"},{"id":"C","text":"$0 < x < 2$"},{"id":"D","text":"$x < 1$"}],"answer":"C","explanation":"First the domain requires $x > 0$. The base $2 > 1$ makes $y = \\log_{_{2}}x increasing$, so $\\log_{_{2}}x < 1 = \\log_{_{2}}2$ gives $x < 2. Together: 0 < x < 2$.","explanationZh":"首先定义域要求 $x > 0；底数 2 > 1 时 y = \\log_{_{2}}x$ 递增，由 $\\log_{_{2}}x < 1 = \\log_{_{2}}2 得 x < 2。综合得 0 < x < 2。$","tags":["medium","logarithmic","inequalities"],"status":"published"},{"id":"math-logarithmic-018","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":3,"stem":"Evaluate $\\log_{_{2}}5 + \\log_{_{2}}(1/5)$.","stemZh":"计算 $\\log_{_{2}}5 + \\log_{_{2}}(1/5)$ 的值。","options":[{"id":"A","text":"0"},{"id":"B","text":"5"},{"id":"C","text":"1"},{"id":"D","text":"−1"}],"answer":"A","explanation":"$\\log_{_{2}}5 + \\log_{_{2}}(1/5) = \\log_{_{2}}(5·1/5) = \\log_{_{2}}1 = 0$.","explanationZh":"$\\log_{_{2}}5 + \\log_{_{2}}(1/5) = \\log_{_{2}}(5·1/5) = \\log_{_{2}}1 = 0。$","tags":["medium","logarithmic","\\log-rules"],"status":"published"},{"id":"math-logarithmic-019","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":3,"stem":"Evaluate lg 4 + lg 25.","stemZh":"计算 lg 4 + lg 25 的值。","options":[{"id":"A","text":"100"},{"id":"B","text":"2"},{"id":"C","text":"29"},{"id":"D","text":"10"}],"answer":"B","explanation":"lg is the common logarithm $(base 10)$, and lg 4 + lg $25 = lg(4·25) =$ lg $100 = 2$. Note lg 4 + lg $25 ≠$ lg 29.","explanationZh":"lg 表示以 10 为底的常用对数，lg 4 + lg $25 = lg(4·25) =$ lg $100 = 2。注意$ lg 4 + lg $25 ≠$ lg 29。","tags":["medium","logarithmic","\\log-rules"],"status":"published"},{"id":"math-logarithmic-020","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":3,"stem":"Evaluate $e^(\\ln 5)$.","stemZh":"计算 $e^(\\ln 5)$ 的值。","options":[{"id":"A","text":"$\\ln 5$"},{"id":"B","text":"$e^{5}$"},{"id":"C","text":"1"},{"id":"D","text":"5"}],"answer":"D","explanation":"$y = e^{x}$ and $y = \\ln x$ are inverse functions, so $e^(\\ln 5) = 5$.","explanationZh":"$y = e^{x} 与 y = \\ln x$ 互为反函数，所以 $e^(\\ln 5) = 5。$","tags":["medium","logarithmic","natural-\\log"],"status":"published"},{"id":"math-logarithmic-021","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":4,"stem":"Solve $\\log_{_{2}}x + \\log_{_{2}}(x − 2) = 3$.","stemZh":"解方程 $\\log_{_{2}}x + \\log_{_{2}}(x − 2) = 3。$","options":[{"id":"A","text":"{4}"},{"id":"B","text":"{−2, 4}"},{"id":"C","text":"{−2}"},{"id":"D","text":"{2}"}],"answer":"A","explanation":"Combining gives $\\log_{_{2}}(x(x−2)) = 3$, so $x^{2} − 2x = 8$, i.e. $x = 4$ or $x = −2$. The domain requires $x > 0$ and $x − 2 > 0$, i.e. $x > 2$, so only $x = 4$ survives.","explanationZh":"合并得 $\\log_{_{2}}(x(x−2)) = 3，即 x^{2} − 2x = 8，解得 x = 4 或 x = −2。定义域要求 x > 0 且 x − 2 > 0，即 x > 2，所以只有 x = 4$ 符合。","tags":["hard","logarithmic","equations"],"status":"published"},{"id":"math-logarithmic-022","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":5,"stem":"For $x, y > 0$, which statement is correct?","stemZh":"$当 x, y > 0$ 时，下列等式正确的是？","options":[{"id":"A","text":"$\\log_{_{2}}(x + y) = \\log_{_{2}}x + \\log_{_{2}}y$"},{"id":"B","text":"$\\log_{_{2}}(xy) = \\log_{_{2}}x · \\log_{_{2}}y$"},{"id":"C","text":"$\\log_{_{2}}(x/y) = \\log_{_{2}}x − \\log_{_{2}}y$"},{"id":"D","text":"$(\\log_{_{2}}x)^{2} = 2\\log_{_{2}}x$"}],"answer":"C","explanation":"The quotient rule is $\\log_{a}(x/y) = \\log_{a}x − \\log_{a}y$. The product rule adds logs of $factors, \\log_{a}(xy) = \\log_{a}x + \\log_{a}y$; there is no rule for $\\log_{a}(x + y)$, and $(\\log_{_{2}}x)^{2}$ is not $2\\log_{_{2}}x$.","explanationZh":"商的运算法则为 $\\log_{a}(x/y) = \\log_{a}x − \\log_{a}y。积的法则是 \\log_{a}(xy) = \\log_{a}x + \\log_{a}y；\\log_{a}(x + y) 没有运算法则，(\\log_{_{2}}x)^{2}$ 也不等于 $2\\log_{_{2}}x。$","tags":["hard","logarithmic","\\log-rules"],"status":"published"},{"id":"math-logarithmic-023","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":4,"stem":"Order $a = \\log_{_{2}}3, b = \\log_{3}2$ and $c = \\log_{_{2}}(1/3)$.","stemZh":"比较 $a = \\log_{_{2}}3、b = \\log_{3}2、c = \\log_{_{2}}(1/3)$ 的大小。","options":[{"id":"A","text":"$a \\log_{_{2}}2 = 1$, so $a > 1. \\log_{3}1 < \\log_{3}2 < \\log_{3}3$ gives $0 \\log_{_{2}}2 = 1，所以 a > 1；由 \\log_{3}1 < \\log_{3}2 < \\log_{3}3 得 0 < b < 1；c = \\log_{_{2}}(1/3) = −\\log_{_{2}}3 < 0。故 c < b <$ a。","tags":["hard","logarithmic","comparison"],"status":"published"},{"id":"math-logarithmic-024","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":5,"stem":"Solve $\\log_(1/2)x > 1$.","stemZh":"解不等式 $\\log_(1/2)x > 1。$","options":[{"id":"A","text":"$x > 1/2$"},{"id":"B","text":"$0 < x < 1$"},{"id":"C","text":"$x < 1/2$"},{"id":"D","text":"$0 < x < 1/2$"}],"answer":"D","explanation":"The domain gives $x > 0$. The base 1/2 lies in $(0,1)$, so the function is $decreasing: \\log_(1/2)x > 1 = \\log_(1/2)(1/2)$ implies $x < 1/2. Together: 0 < x < 1/2$.","explanationZh":"定义域要求 $x > 0。底数 1/2 ∈ (0,1)，函数递减：由 \\log_(1/2)x > 1 = \\log_(1/2)(1/2) 得 x < 1/2。综合得 0 < x < 1/2。$","tags":["hard","logarithmic","inequalities"],"status":"published"},{"id":"math-logarithmic-025","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":4,"stem":"If lg $2 = a$ and lg $3 = b$, express lg 12 in terms of a and b.","stemZh":"若 lg $2 =$ a，lg $3 =$ b，用 a、b 表示 lg 12。","options":[{"id":"A","text":"2a + b"},{"id":"B","text":"a + b"},{"id":"C","text":"2ab"},{"id":"D","text":"$a^{2} + b$"}],"answer":"A","explanation":"lg $12 = lg(4·3) =$ lg $2^{2} +$ lg $3 = 2lg 2 +$ lg $3 = 2a +$ b.","explanationZh":"lg $12 = lg(4·3) =$ lg $2^{2} +$ lg $3 = 2lg 2 +$ lg $3 = 2a +$ b。","tags":["hard","logarithmic","\\log-rules"],"status":"published"},{"id":"math-logarithmic-026","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":5,"stem":"Evaluate $\\log_{_{2}}3 · \\log_{3}4 · \\log_{_{4}}8$.","stemZh":"计算 $\\log_{_{2}}3 · \\log_{3}4 · \\log_{_{4}}8$ 的值。","options":[{"id":"A","text":"1"},{"id":"B","text":"2"},{"id":"C","text":"3"},{"id":"D","text":"4"}],"answer":"C","explanation":"Change everything to a common $base: (lg3/lg2)·(lg4/lg3)·(lg8/lg4) = lg8/lg2 = \\log_{_{2}}8 = 3$. The product telescopes.","explanationZh":"$统一换底：(lg3/lg2)·(lg4/lg3)·(lg8/lg4) = lg8/lg2 = \\log_{_{2}}8 = 3，中间项全部约掉。$","tags":["hard","logarithmic","change-of-base"],"status":"published"},{"id":"math-logarithmic-027","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":4,"stem":"Solve $\\log_{3}(x^{2} − 1) = 1$.","stemZh":"解方程 $\\log_{3}(x^{2} − 1) = 1。$","options":[{"id":"A","text":"{2}"},{"id":"B","text":"{−2, 2}"},{"id":"C","text":"{−2}"},{"id":"D","text":"{4}"}],"answer":"B","explanation":"$x^{2} − 1 = 3$ gives $x^{2} = 4$, so $x = 2$ or $x = −2$. Both satisfy $x^{2} − 1 = 3 > 0$, so both are valid; dropping the negative root is the usual error.","explanationZh":"$x^{2} − 1 = 3 得 x^{2} = 4，所以 x = 2 或 x = −2。两者都满足 x^{2} − 1 = 3 > 0，均为有效解；漏掉负根是常见错误。$","tags":["hard","logarithmic","equations"],"status":"published"},{"id":"math-logarithmic-028","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":5,"stem":"What is the range of $y = \\log_{_{2}}(x^{2} + 1)$?","stemZh":"函数 $y = \\log_{_{2}}(x^{2} + 1)$ 的值域是？","options":[{"id":"A","text":"$(0,+∞)$"},{"id":"B","text":"$ℝ$"},{"id":"C","text":"$[1,+∞)$"},{"id":"D","text":"$[0,+∞)$"}],"answer":"D","explanation":"$x^{2} + 1 ≥ 1$ for every real x, and the base $2 > 1$ makes $\\log_{_{2}} increasing$, so $\\log_{_{2}}(x^{2} + 1) ≥ \\log_{_{2}}1 = 0$. The value 0 is attained at $x = 0$, so the endpoint is included.","explanationZh":"对任意实数 x 都有 $x^{2} + 1 ≥ 1，且底数 2 > 1 时 \\log_{_{2}}$ 递增，故 $\\log_{_{2}}(x^{2} + 1) ≥ \\log_{_{2}}1 = 0。x = 0$ 时取到 0，所以端点取得到。","tags":["hard","logarithmic","range"],"status":"published"},{"id":"math-logarithmic-029","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":4,"stem":"If $\\log_{a}(1/2) > \\log_{a}(1/3)$, what is the range of a?","stemZh":"$若 \\log_{a}(1/2) > \\log_{a}(1/3)，则 a$ 的取值范围是？","options":[{"id":"A","text":"$a > 1$"},{"id":"B","text":"$0 < a < 1$"},{"id":"C","text":"$a > 0, a ≠ 1$"},{"id":"D","text":"$0 < a < 1/3$"}],"answer":"A","explanation":"Since $1/2 > 1/3$ and the larger argument has the larger $logarithm, y = \\log_{a}x$ must be increasing, which happens exactly when $a > 1. (For 0 < a < 1$ the inequality would reverse.)","explanationZh":"因为 $1/2 > 1/3，而较大的真数对应较大的对数值，说明 y = \\log_{a}x$ 递增，这只有当 $a > 1$ 时成立。（若 $0 < a < 1，不等号方向相反。）$","tags":["hard","logarithmic","monotonicity"],"status":"published"},{"id":"math-logarithmic-030","subject":"math","moduleId":"functions","unitId":"elementary","topicId":"logarithmic","type":"single-choice","difficulty":5,"stem":"Solve $\\log_{_{2}}x + \\log_{_{4}}x = 3$.","stemZh":"解方程 $\\log_{_{2}}x + \\log_{_{4}}x = 3。$","options":[{"id":"A","text":"2"},{"id":"B","text":"8"},{"id":"C","text":"4"},{"id":"D","text":"16"}],"answer":"C","explanation":"First unify the $bases: \\log_{_{4}}x = \\log_{_{2}}x / \\log_{_{2}}4 = (\\log_{_{2}}x)/2$. Let $t = \\log_{_{2}}x$. Then $t + t/2 = 3$, so $t = 2$ and $x = 2^{2} = 4$.","explanationZh":"$先统一底数：\\log_{_{4}}x = \\log_{_{2}}x / \\log_{_{2}}4 = (\\log_{_{2}}x)/2。令 t = \\log_{_{2}}x，则 t + t/2 = 3，得 t = 2，x = 2^{2} = 4。$","tags":["hard","logarithmic","equations"],"status":"published"},{"id":"math-angles-001","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":1,"stem":"Convert 60° to radians.","stemZh":"把 60° 化成弧度。","options":[{"id":"A","text":"$π/3$"},{"id":"B","text":"$π/6$"},{"id":"C","text":"$π/4$"},{"id":"D","text":"$2π/3$"}],"answer":"A","explanation":"Multiply by $π/180: 60 \\times π/180 = π/3$.","explanationZh":"利用 $1° = π/180，60 \\times π/180 = π/3。$","tags":["easy","radian","conversion"],"status":"published"},{"id":"math-angles-002","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":1,"stem":"Convert $π/2$ radians to degrees.","stemZh":"$把 π/2$ 弧度化成角度。","options":[{"id":"A","text":"30°"},{"id":"B","text":"45°"},{"id":"C","text":"60°"},{"id":"D","text":"90°"}],"answer":"D","explanation":"Multiply by $180/π: (π/2) \\times 180/π = 90°$.","explanationZh":"1 弧度 $= 180°/π，(π/2) \\times 180°/π = 90°。$","tags":["easy","radian","conversion"],"status":"published"},{"id":"math-angles-003","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":1,"stem":"How many radians does one full revolution measure?","stemZh":"一周角等于多少弧度？","options":[{"id":"A","text":"$π$"},{"id":"B","text":"$2π$"},{"id":"C","text":"$3π$"},{"id":"D","text":"$4π$"}],"answer":"B","explanation":"A full turn is 360°, and $360° = 2π$ radians.","explanationZh":"一周为 $360°，等于 2π$ 弧度。","tags":["easy","radian"],"status":"published"},{"id":"math-angles-004","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":1,"stem":"A circle has radius 2. A central angle of 3 radians cuts off an arc of length","stemZh":"半径为 2 的圆中，圆心角 3 弧度所对的弧长为","options":[{"id":"A","text":"3/2"},{"id":"B","text":"5"},{"id":"C","text":"6"},{"id":"D","text":"12"}],"answer":"C","explanation":"Arc length $l = |α|r = 3 \\times 2 = 6$; the angle must be in radians for this formula.","explanationZh":"弧长 $l = |α|r = 3 \\times 2 = 6；使用该公式时角必须用弧度表示。$","tags":["easy","arc-length"],"status":"published"},{"id":"math-angles-005","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":2,"stem":"The area of a sector with radius 3 and central angle 2 radians is","stemZh":"半径为 3、圆心角为 2 弧度的扇形面积为","options":[{"id":"A","text":"9"},{"id":"B","text":"6"},{"id":"C","text":"3"},{"id":"D","text":"18"}],"answer":"A","explanation":"$S = ½|α|r^{2} = ½ \\times 2 \\times 9 = 9. (The$ value 6 is the arc length, not the area.)","explanationZh":"$S = ½|α|r^{2} = ½ \\times 2 \\times 9 = 9。（6$ 是弧长，注意与面积区分。）","tags":["easy","sector"],"status":"published"},{"id":"math-angles-006","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":2,"stem":"Which of the following angles is coterminal with 30°?","stemZh":"下列哪个角与 30° 终边相同？","options":[{"id":"A","text":"150°"},{"id":"B","text":"210°"},{"id":"C","text":"330°"},{"id":"D","text":"390°"}],"answer":"D","explanation":"Coterminal angles differ by a whole number of $turns: 30° + 360° = 390°$.","explanationZh":"$终边相同的角相差周角的整数倍：30° + 360° = 390°。$","tags":["easy","coterminal"],"status":"published"},{"id":"math-angles-007","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":2,"stem":"In standard position, the terminal side of the angle $−π/3$ lies in","stemZh":"标准位置下，角 $−π/3$ 的终边在","options":[{"id":"A","text":"Quadrant I"},{"id":"B","text":"Quadrant IV"},{"id":"C","text":"Quadrant III"},{"id":"D","text":"Quadrant II"}],"answer":"B","explanation":"A negative angle rotates $clockwise; −π/3 = −60°$, so its terminal side is in Quadrant IV.","explanationZh":"$负角按顺时针方向旋转，−π/3 = −60°，终边落在第四象限。$","tags":["easy","quadrant","negative-angle"],"status":"published"},{"id":"math-angles-008","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":2,"stem":"Which of the following is a quadrantal angle (its terminal side lies on a coordinate axis)?","stemZh":"下列哪个是轴线角（终边在坐标轴上）？","options":[{"id":"A","text":"$π/6$"},{"id":"B","text":"$π/4$"},{"id":"C","text":"$π/2$"},{"id":"D","text":"$5π/6$"}],"answer":"C","explanation":"$π/2 = 90°$, whose terminal side lies on the positive y-axis.","explanationZh":"$π/2 = 90°，终边在 y$ 轴的正半轴上，是轴线角。","tags":["easy","quadrantal"],"status":"published"},{"id":"math-angles-009","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":2,"stem":"Convert 300° to radians.","stemZh":"把 300° 化成弧度。","options":[{"id":"A","text":"$5π/3$"},{"id":"B","text":"$5π/6$"},{"id":"C","text":"$4π/3$"},{"id":"D","text":"$7π/6$"}],"answer":"A","explanation":"$300 \\times π/180 = 5π/3$.","explanationZh":"$300 \\times π/180 = 5π/3。$","tags":["easy","radian","conversion"],"status":"published"},{"id":"math-angles-010","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":2,"stem":"In standard position, an angle produced by a clockwise rotation is","stemZh":"标准位置下，由顺时针旋转得到的角是","options":[{"id":"A","text":"positive"},{"id":"B","text":"zero"},{"id":"C","text":"negative"},{"id":"D","text":"always 360°"}],"answer":"C","explanation":"By convention a counterclockwise rotation gives a positive angle and a clockwise rotation gives a negative one.","explanationZh":"规定逆时针旋转为正角，顺时针旋转为负角。","tags":["easy","sign","rotation"],"status":"published"},{"id":"math-angles-011","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":3,"stem":"For an angle $α$ measured in radians, the set of all angles coterminal with $α$ is","stemZh":"若角 $α$ 用弧度表示，则与 $α$ 终边相同的所有角组成的集合是","options":[{"id":"A","text":"${α +\\text{ kπ }| k∈ℤ}$"},{"id":"B","text":"${α + 360°k | k∈ℤ}$"},{"id":"C","text":"${−α + 2kπ | k∈ℤ}$"},{"id":"D","text":"${α + 2kπ | k∈ℤ}$"}],"answer":"D","explanation":"Coterminal angles differ by whole turns, i.e. by $2kπ$. Option B mixes the degree measure with a radian angle.","explanationZh":"终边相同的角相差周角的整数倍，即 $2kπ；B$ 把角度制与弧度制混用了。","tags":["medium","coterminal","radian"],"status":"published"},{"id":"math-angles-012","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":3,"stem":"Which quadrant contains the terminal side of an angle of 4 radians?","stemZh":"4 弧度的角的终边在第几象限？","options":[{"id":"A","text":"Quadrant II"},{"id":"B","text":"Quadrant III"},{"id":"C","text":"Quadrant IV"},{"id":"D","text":"Quadrant I"}],"answer":"B","explanation":"$π ≈ 3.14 < 4 < 3π/2 ≈ 4.71$, so the terminal side lies in Quadrant III.","explanationZh":"$π ≈ 3.14 < 4 < 3π/2 ≈ 4.71，故终边在第三象限。$","tags":["medium","quadrant","radian"],"status":"published"},{"id":"math-angles-013","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":3,"stem":"The angle −2025° is coterminal with","stemZh":"角 −2025° 与下列哪个角终边相同？","options":[{"id":"A","text":"45°"},{"id":"B","text":"225°"},{"id":"C","text":"135°"},{"id":"D","text":"315°"}],"answer":"C","explanation":"Add whole $turns: −2025° + 6 \\times 360° = −2025° + 2160° = 135°$.","explanationZh":"$加上周角的整数倍：−2025° + 6 \\times 360° = 135°。$","tags":["medium","coterminal"],"status":"published"},{"id":"math-angles-014","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":3,"stem":"A sector has radius 6 and arc length $4π$. Its central angle in radians is","stemZh":"半径为 $6、弧长为 4π$ 的扇形，其圆心角（弧度）为","options":[{"id":"A","text":"$2π/3$"},{"id":"B","text":"$4π/3$"},{"id":"C","text":"$3π/4$"},{"id":"D","text":"$π/3$"}],"answer":"A","explanation":"From $l = |α|r$ we get $α = l/r = 4π/6 = 2π/3$.","explanationZh":"$由 l = |α|r 得 α = l/r = 4π/6 = 2π/3。$","tags":["medium","arc-length","sector"],"status":"published"},{"id":"math-angles-015","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":3,"stem":"A sector of radius r has arc length l. Its area is","stemZh":"半径为 r、弧长为 l 的扇形面积为","options":[{"id":"A","text":"lr"},{"id":"B","text":"$lr^{2}$"},{"id":"C","text":"$½l^{2}r$"},{"id":"D","text":"½lr"}],"answer":"D","explanation":"$S = ½|α|r^{2}$ and $l = |α|r$, so $S =$ ½lr — the same shape as the triangle area formula.","explanationZh":"$S = ½|α|r^{2}，又 l = |α|r，故 S =$ ½lr，形式类似三角形面积公式。","tags":["medium","sector","arc-length"],"status":"published"},{"id":"math-angles-016","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":3,"stem":"The terminal side of −1000° lies in","stemZh":"角 −1000° 的终边在","options":[{"id":"A","text":"Quadrant IV"},{"id":"B","text":"Quadrant I"},{"id":"C","text":"Quadrant II"},{"id":"D","text":"Quadrant III"}],"answer":"B","explanation":"$−1000° + 3 \\times 360° = 80°$, and 80° lies in Quadrant I.","explanationZh":"$−1000° + 3 \\times 360° = 80°，在第一象限。$","tags":["medium","coterminal","quadrant"],"status":"published"},{"id":"math-angles-017","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":3,"stem":"Which of the following angles has its terminal side in Quadrant III?","stemZh":"下列哪个角的终边在第三象限？","options":[{"id":"A","text":"$−π/3$"},{"id":"B","text":"$5π/3$"},{"id":"C","text":"$4π/3$"},{"id":"D","text":"$−π/6$"}],"answer":"C","explanation":"$4π/3 = 240°$ lies in Quadrant III, while $−π/3, −π/6$ and $5π/3$ all end in Quadrant IV.","explanationZh":"$4π/3 = 240° 在第三象限；−π/3、−π/6、5π/3$ 的终边都在第四象限。","tags":["medium","quadrant"],"status":"published"},{"id":"math-angles-018","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":3,"stem":"The angle $7π/4$ is coterminal with","stemZh":"$角 7π/4$ 与下列哪个角终边相同？","options":[{"id":"A","text":"$−π/4$"},{"id":"B","text":"$π/4$"},{"id":"C","text":"$3π/4$"},{"id":"D","text":"$−3π/4$"}],"answer":"A","explanation":"Subtract one full $turn: 7π/4 − 2π = −π/4$.","explanationZh":"$减去一周：7π/4 − 2π = −π/4。$","tags":["medium","coterminal"],"status":"published"},{"id":"math-angles-019","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":3,"stem":"A wheel of radius 0.5 m turns through an angle of 4 radians. How far does a point on its rim travel?","stemZh":"半径为 0.5 m 的轮子转过 4 弧度，轮缘上一点经过的路程为","options":[{"id":"A","text":"8 m"},{"id":"B","text":"1 m"},{"id":"C","text":"4 m"},{"id":"D","text":"2 m"}],"answer":"D","explanation":"$l = |α|r = 4 \\times 0.5 = 2 m$; the circumference $2πr$ only applies to a full revolution.","explanationZh":"$l = |α|r = 4 \\times 0.5 = 2$ m；只有转满一周才用周长 $2πr。$","tags":["medium","arc-length","application"],"status":"published"},{"id":"math-angles-020","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":3,"stem":"Convert $5π/12$ radians to degrees.","stemZh":"$把 5π/12$ 弧度化成角度。","options":[{"id":"A","text":"105°"},{"id":"B","text":"75°"},{"id":"C","text":"60°"},{"id":"D","text":"150°"}],"answer":"B","explanation":"$(5π/12) \\times 180°/π = 5 \\times 15° = 75°$.","explanationZh":"$(5π/12) \\times 180°/π = 5 \\times 15° = 75°。$","tags":["medium","radian","conversion"],"status":"published"},{"id":"math-angles-021","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":4,"stem":"Which set describes all angles whose terminal side lies on the negative x-axis?","stemZh":"终边在 x 轴负半轴上的所有角组成的集合是","options":[{"id":"A","text":"${kπ | k∈ℤ}$"},{"id":"B","text":"${π/2 + 2kπ | k∈ℤ}$"},{"id":"C","text":"${π + 2kπ | k∈ℤ}$"},{"id":"D","text":"${3π/2 + 2kπ | k∈ℤ}$"}],"answer":"C","explanation":"The negative x-axis corresponds to $180° = π$ plus whole $turns: π + 2kπ$. Option A covers both halves of the x-axis and D is the negative y-axis.","explanationZh":"x 轴负半轴对应 $180° = π，再加整周：π + 2kπ。A$ 覆盖 x 轴的两条半轴，D 是 y 轴负半轴。","tags":["hard","quadrantal","coterminal"],"status":"published"},{"id":"math-angles-022","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":5,"stem":"If $α$ is a Quadrant II angle, where can the terminal side of $α/2$ lie?","stemZh":"$若 α$ 是第二象限角，则 $α/2$ 的终边可能在","options":[{"id":"A","text":"Quadrant I or Quadrant III"},{"id":"B","text":"Quadrant I or Quadrant II"},{"id":"C","text":"Quadrant II or Quadrant IV"},{"id":"D","text":"Quadrant I only"}],"answer":"A","explanation":"$α ∈ (π/2 + 2kπ, π + 2kπ)$, so $α/2 ∈ (π/4 +\\text{ kπ, }π/2 + kπ)$: Quadrant I when k is even and Quadrant III when k is odd.","explanationZh":"$α ∈ (π/2 + 2kπ, π + 2kπ)，故 α/2 ∈ (π/4 + kπ, π/2 + kπ)：k$ 为偶数时在第一象限，k 为奇数时在第三象限。","tags":["hard","quadrant","half-angle"],"status":"published"},{"id":"math-angles-023","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":4,"stem":"The angle $−17π/4$ is coterminal with","stemZh":"$角 −17π/4$ 与下列哪个角终边相同？","options":[{"id":"A","text":"$π/4$"},{"id":"B","text":"$3π/4$"},{"id":"C","text":"$5π/4$"},{"id":"D","text":"$7π/4$"}],"answer":"D","explanation":"Add $2π = 8π/4 repeatedly: −17π/4 + 24π/4 = 7π/4$.","explanationZh":"反复加 $2π = 8π/4：−17π/4 + 24π/4 = 7π/4。$","tags":["hard","coterminal"],"status":"published"},{"id":"math-angles-024","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":5,"stem":"A sector has perimeter 8 and area 4. Its central angle, in radians, is","stemZh":"一个扇形的周长为 8、面积为 4，则其圆心角（弧度）为","options":[{"id":"A","text":"1"},{"id":"B","text":"2"},{"id":"C","text":"4"},{"id":"D","text":"$π/2$"}],"answer":"B","explanation":"Let the radius be r and the arc length l. Then $2r + l = 8$ and ½lr $= 4$, which give $r = 2$ and $l = 4$; hence $α = l/r = 2$.","explanationZh":"设半径为 r、弧长为 l，则 $2r + l = 8，½lr = 4，解得 r = 2，l = 4，故 α = l/r = 2。$","tags":["hard","sector","arc-length"],"status":"published"},{"id":"math-angles-025","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":4,"stem":"Let $S = {θ | θ = π/6 + kπ/2, k∈ℤ}$. Which of the following angles belongs to S?","stemZh":"$设 S = {θ | θ = π/6 + kπ/2, k∈ℤ}，下列哪个角属于$ S？","options":[{"id":"A","text":"$π/3$"},{"id":"B","text":"$5π/6$"},{"id":"C","text":"$2π/3$"},{"id":"D","text":"$4π/3$"}],"answer":"C","explanation":"With $k = 1$ we get $π/6 + π/2 = 2π/3$. The other options would require $k = 1/3, 4/3$ and 7/3, which are not integers.","explanationZh":"$取 k = 1 得 π/6 + π/2 = 2π/3。其余选项对应 k = 1/3、4/3、7/3，都不是整数。$","tags":["hard","coterminal","angle-set"],"status":"published"},{"id":"math-angles-026","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":4,"stem":"The minute hand of a clock is 6 cm long. How far does its tip travel in 20 minutes?","stemZh":"时钟分针长 6 cm，20 分钟内分针尖端走过的路程为","options":[{"id":"A","text":"$4π$ cm"},{"id":"B","text":"$2π$ cm"},{"id":"C","text":"$8π$ cm"},{"id":"D","text":"$12π$ cm"}],"answer":"A","explanation":"In 20 minutes the hand turns $120° = 2π/3 radians$, so $l = |α|r = (2π/3) \\times 6 = 4π$ cm.","explanationZh":"20 分钟转过 $120° = 2π/3$ 弧度，l $= |α|r = (2π/3) \\times 6 = 4π$ cm。","tags":["hard","arc-length","application"],"status":"published"},{"id":"math-angles-027","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":5,"stem":"If the terminal side of $α$ lies in Quadrant III, which of the following must have its terminal side in Quadrant II?","stemZh":"$若 α$ 的终边在第三象限，则下列哪个角的终边一定在第二象限？","options":[{"id":"A","text":"$α + π$"},{"id":"B","text":"$α + π/2$"},{"id":"C","text":"$α − π$"},{"id":"D","text":"$−α$"}],"answer":"D","explanation":"Take $α$ between 180° and 270°. Then $α + π$ and $α − π$ end in Quadrant $I, α + π/2$ ends in Quadrant IV, and $−α$ lands between 90° and 180°, i.e. Quadrant II.","explanationZh":"$设 180° < α < 270°，则 α + π 与 α − π 的终边在第一象限，α + π/2$ 在第四象限，而 $−α$ 落在 90° 到 180° 之间，即第二象限。","tags":["hard","quadrant","negative-angle"],"status":"published"},{"id":"math-angles-028","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":4,"stem":"How many angles $θ$ with $−2π ≤ θ ≤ 2π$ have their terminal side on the y-axis?","stemZh":"在区间 $−2π ≤ θ ≤ 2π$ 内，终边在 y 轴上的角 $θ$ 有多少个？","options":[{"id":"A","text":"2"},{"id":"B","text":"4"},{"id":"C","text":"3"},{"id":"D","text":"5"}],"answer":"B","explanation":"These angles are $θ = π/2 + kπ$; in the interval they are $−3π/2, −π/2, π/2$ and $3π/2$, so there are $4 (the$ negative ones are easy to miss).","explanationZh":"这些角满足 $θ = π/2 + kπ，在区间内为 −3π/2、−π/2、π/2、3π/2，共 4$ 个（负值容易被漏掉）。","tags":["hard","quadrantal","counting"],"status":"published"},{"id":"math-angles-029","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":4,"stem":"Given the angle $7π/5$, which of the following can NOT be coterminal with it?","stemZh":"已知角 $7π/5，下列哪个角不可能与它终边相同？$","options":[{"id":"A","text":"$−3π/5$"},{"id":"B","text":"$17π/5$"},{"id":"C","text":"$12π/5$"},{"id":"D","text":"$27π/5$"}],"answer":"C","explanation":"Coterminal angles differ by $2kπ = 10kπ/5$. The differences are $−10π/5, 10π/5, 5π/5$ and $20π/5$; only $12π/5$ differs by $π$ instead of $2π$.","explanationZh":"终边相同需相差 $2kπ = 10kπ/5：差值分别为 −10π/5、10π/5、5π/5、20π/5；只有 12π/5$ 相差 $π$ 而不是 $2π。$","tags":["hard","coterminal","error-distinction"],"status":"published"},{"id":"math-angles-030","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"angles","type":"single-choice","difficulty":4,"stem":"Which set of inequalities describes all angles whose terminal side lies in Quadrant IV?","stemZh":"终边在第四象限的所有角可表示为","options":[{"id":"A","text":"${θ | 3π/2 + 2kπ < θ < 2π + 2kπ, k∈ℤ}$"},{"id":"B","text":"${θ | π/2 + 2kπ < θ < π + 2kπ, k∈ℤ}$"},{"id":"C","text":"${θ | π + 2kπ < θ < 3π/2 + 2kπ, k∈ℤ}$"},{"id":"D","text":"${θ | 2kπ < θ < π/2 + 2kπ, k∈ℤ}$"}],"answer":"A","explanation":"Quadrant IV runs from $270° = 3π/2$ to $360° = 2π$, modulo whole turns. Option B is Quadrant II, C is Quadrant III and D is Quadrant I.","explanationZh":"第四象限为 $270° = 3π/2 到 360° = 2π，再加整周。B$ 是第二象限，C 是第三象限，D 是第一象限。","tags":["hard","quadrant","angle-set"],"status":"published"},{"id":"math-trig-functions-001","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":1,"stem":"What is $\\sin(π/6)$?","stemZh":"$\\sin(π/6)$ 的值为","options":[{"id":"A","text":"$\\sqrt{3}/2$"},{"id":"B","text":"1/2"},{"id":"C","text":"$\\sqrt{2}/2$"},{"id":"D","text":"−1/2"}],"answer":"B","explanation":"$π/6 = 30°$, and $\\sin 30° = 1/2$.","explanationZh":"$π/6 = 30°，\\sin 30° = 1/2。$","tags":["easy","special-angle","value"],"status":"published"},{"id":"math-trig-functions-002","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":1,"stem":"What is $\\cos(π/2)$?","stemZh":"$\\cos(π/2)$ 的值为","options":[{"id":"A","text":"0"},{"id":"B","text":"1"},{"id":"C","text":"−1"},{"id":"D","text":"1/2"}],"answer":"A","explanation":"At $π/2$ the point on the unit circle is $(0, 1)$, so $\\cos(π/2) = 0$.","explanationZh":"$π/2$ 对应单位圆上的点 $(0, 1)，故 \\cos(π/2) = 0。$","tags":["easy","unit-circle","value"],"status":"published"},{"id":"math-trig-functions-003","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":1,"stem":"What is $\\tan(π/4)$?","stemZh":"$\\tan(π/4)$ 的值为","options":[{"id":"A","text":"$\\sqrt{3}$"},{"id":"B","text":"$\\sqrt{3}/3$"},{"id":"C","text":"−1"},{"id":"D","text":"1"}],"answer":"D","explanation":"$\\tan(π/4) = \\sin(π/4)/\\cos(π/4) = (\\sqrt{2}/2)/(\\sqrt{2}/2) = 1$.","explanationZh":"$\\tan(π/4) = \\sin(π/4)/\\cos(π/4) = (\\sqrt{2}/2)/(\\sqrt{2}/2) = 1。$","tags":["easy","special-angle","tangent"],"status":"published"},{"id":"math-trig-functions-004","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":1,"stem":"For every real $x, \\sin^{2}x + \\cos^{2}x$ equals","stemZh":"对任意实数 $x，\\sin^{2}x + \\cos^{2}x$ 等于","options":[{"id":"A","text":"0"},{"id":"B","text":"2"},{"id":"C","text":"1"},{"id":"D","text":"$\\sin 2x$"}],"answer":"C","explanation":"This is the Pythagorean identity $\\sin^{2}x + \\cos^{2}x = 1$, valid for every real x.","explanationZh":"这是平方关系 $\\sin^{2}x + \\cos^{2}x = 1，对任意实数 x$ 都成立。","tags":["easy","identity"],"status":"published"},{"id":"math-trig-functions-005","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":1,"stem":"What is the range of $y = \\sin$ x?","stemZh":"函数 $y = \\sin x$ 的值域是","options":[{"id":"A","text":"$(0, 1)$"},{"id":"B","text":"$[−1, 1]$"},{"id":"C","text":"$[0, 1]$"},{"id":"D","text":"$(−∞, +∞)$"}],"answer":"B","explanation":"The sine function takes every value from −1 to 1 inclusive, so the endpoints are included.","explanationZh":"正弦函数取遍 −1 到 1 的一切值，端点可以取到，故为 $[−1, 1]。$","tags":["easy","range"],"status":"published"},{"id":"math-trig-functions-006","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":2,"stem":"What is the least positive period of $y = \\sin$ x?","stemZh":"函数 $y = \\sin x$ 的最小正周期是","options":[{"id":"A","text":"$π/2$"},{"id":"B","text":"$π$"},{"id":"C","text":"$4π$"},{"id":"D","text":"$2π$"}],"answer":"D","explanation":"$\\sin(x + 2π) = \\sin x$ for all x, and no smaller positive number works.","explanationZh":"对任意 $x 有 \\sin(x + 2π) = \\sin$ x，且没有更小的正周期，故为 $2π。$","tags":["easy","period"],"status":"published"},{"id":"math-trig-functions-007","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":2,"stem":"What is the least positive period of $y = \\tan$ x?","stemZh":"函数 $y = \\tan x$ 的最小正周期是","options":[{"id":"A","text":"$π$"},{"id":"B","text":"$2π$"},{"id":"C","text":"$π/2$"},{"id":"D","text":"$4π$"}],"answer":"A","explanation":"$\\tan(x + π) = \\tan x$ whenever both sides are defined, so the period is $π$, not $2π$.","explanationZh":"$\\tan(x + π) = \\tan$ x（在定义域内），故最小正周期是 $π$ 而不是 $2π。$","tags":["easy","period","tangent"],"status":"published"},{"id":"math-trig-functions-008","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":2,"stem":"If $θ$ is an angle in Quadrant III, which statement is true?","stemZh":"$若 θ$ 是第三象限角，则下列结论正确的是","options":[{"id":"A","text":"$\\\\sin θ > 0$"},{"id":"B","text":"$\\\\cos θ > 0$"},{"id":"C","text":"$\\tan θ > 0$"},{"id":"D","text":"$\\tan θ < 0$"}],"answer":"C","explanation":"In Quadrant III both $\\sin θ$ and $\\cos θ$ are negative, so their quotient $\\tan θ = \\sin θ/\\cos θ$ is positive.","explanationZh":"第三象限中 $\\sin θ < 0 且 \\cos θ < 0，故 \\tan θ = \\sin θ/\\cos θ > 0。$","tags":["easy","sign","quadrant"],"status":"published"},{"id":"math-trig-functions-009","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":2,"stem":"What is the domain of $y = \\tan$ x?","stemZh":"函数 $y = \\tan x$ 的定义域是","options":[{"id":"A","text":"$ℝ$"},{"id":"B","text":"${x | x ≠ π/2 +\\text{ kπ, }k∈ℤ}$"},{"id":"C","text":"${x | x ≠\\text{ kπ, }k∈ℤ}$"},{"id":"D","text":"${x | x ≠ 2kπ, k∈ℤ}$"}],"answer":"B","explanation":"Since $\\tan x = \\sin x/\\cos x$, it is undefined whenever $\\cos x = 0$, that is at $x = π/2 + kπ$.","explanationZh":"$\\tan x = \\sin x/\\cos$ x，当 $\\cos x = 0（即 x = π/2 + kπ）时无意义。$","tags":["easy","domain","tangent"],"status":"published"},{"id":"math-trig-functions-010","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":2,"stem":"What is the amplitude of $y = 3 \\sin(2x)$?","stemZh":"函数 $y = 3 \\sin(2x)$ 的振幅是","options":[{"id":"A","text":"2"},{"id":"B","text":"6"},{"id":"C","text":"1/3"},{"id":"D","text":"3"}],"answer":"D","explanation":"In $y = A \\sin(ωx + φ)$ the amplitude is $|A| = 3$; the coefficient 2 controls the period, not the amplitude.","explanationZh":"$y = A \\sin(ωx + φ)$ 的振幅为 $|A| = 3；系数 2$ 决定周期，与振幅无关。","tags":["easy","amplitude","graph"],"status":"published"},{"id":"math-trig-functions-011","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":3,"stem":"What is $\\sin(2π/3)$?","stemZh":"$\\sin(2π/3)$ 的值为","options":[{"id":"A","text":"$\\sqrt{3}/2$"},{"id":"B","text":"$−\\sqrt{3}/2$"},{"id":"C","text":"1/2"},{"id":"D","text":"−1/2"}],"answer":"A","explanation":"$\\sin(2π/3) = \\sin(π − π/3) = \\sin(π/3) = \\sqrt{3}/2$; sine stays positive in Quadrant II.","explanationZh":"$\\sin(2π/3) = \\sin(π − π/3) = \\sin(π/3) = \\sqrt{3}/2；第二象限正弦为正。$","tags":["medium","reduction-formula","special-angle"],"status":"published"},{"id":"math-trig-functions-012","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":3,"stem":"What is $\\cos(7π/6)$?","stemZh":"$\\cos(7π/6)$ 的值为","options":[{"id":"A","text":"$\\sqrt{3}/2$"},{"id":"B","text":"1/2"},{"id":"C","text":"$−\\sqrt{3}/2$"},{"id":"D","text":"−1/2"}],"answer":"C","explanation":"$7π/6 = π + π/6$ lies in Quadrant III, so $\\cos(7π/6) = −\\cos(π/6) = −\\sqrt{3}/2$.","explanationZh":"$7π/6 = π + π/6$ 在第三象限，故 $\\cos(7π/6) = −\\cos(π/6) = −\\sqrt{3}/2。$","tags":["medium","reduction-formula","special-angle"],"status":"published"},{"id":"math-trig-functions-013","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":3,"stem":"What is $\\tan(2π/3)$?","stemZh":"$\\tan(2π/3)$ 的值为","options":[{"id":"A","text":"$\\sqrt{3}$"},{"id":"B","text":"$−\\sqrt{3}$"},{"id":"C","text":"$−\\sqrt{3}/3$"},{"id":"D","text":"$\\sqrt{3}/3$"}],"answer":"B","explanation":"$\\tan(2π/3) = \\tan(π − π/3) = −\\tan(π/3) = −\\sqrt{3}$; tangent is negative in Quadrant II.","explanationZh":"$\\tan(2π/3) = \\tan(π − π/3) = −\\tan(π/3) = −\\sqrt{3}；第二象限正切为负。$","tags":["medium","reduction-formula","tangent"],"status":"published"},{"id":"math-trig-functions-014","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":3,"stem":"If $\\sin θ = 3/5$ and $θ$ is in Quadrant II, then $\\cos θ$ equals","stemZh":"$若 \\sin θ = 3/5 且 θ$ 在第二象限，则 $\\cos θ$ 等于","options":[{"id":"A","text":"4/5"},{"id":"B","text":"−3/5"},{"id":"C","text":"3/5"},{"id":"D","text":"−4/5"}],"answer":"D","explanation":"$\\cos^{2}θ = 1 − 9/25 = 16/25$, so $|\\cos θ| = 4/5$; cosine is negative in Quadrant II, hence $\\cos θ = −4/5$.","explanationZh":"$\\cos^{2}θ = 1 − 9/25 = 16/25，|\\cos θ| = 4/5；第二象限余弦为负，故 \\cos θ = −4/5。$","tags":["medium","identity","quadrant-sign"],"status":"published"},{"id":"math-trig-functions-015","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":3,"stem":"Simplify $\\sin(π − α)$.","stemZh":"化简 $\\sin(π − α)。$","options":[{"id":"A","text":"$\\sin α$"},{"id":"B","text":"$−\\sin α$"},{"id":"C","text":"$\\cos α$"},{"id":"D","text":"$−\\cos α$"}],"answer":"A","explanation":"By the reduction formula $\\sin(π − α) = \\sin α$: supplementary angles have equal sines.","explanationZh":"由诱导公式 $\\sin(π − α) = \\sin α，互补角的正弦值相等。$","tags":["medium","reduction-formula"],"status":"published"},{"id":"math-trig-functions-016","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":3,"stem":"Simplify $\\cos(π/2 + α)$.","stemZh":"化简 $\\cos(π/2 + α)。$","options":[{"id":"A","text":"$\\sin α$"},{"id":"B","text":"$\\cos α$"},{"id":"C","text":"$−\\sin α$"},{"id":"D","text":"$−\\cos α$"}],"answer":"C","explanation":"$\\cos(π/2 + α) = −\\sin α$: the function changes to its co-function and the sign is taken from Quadrant II.","explanationZh":"$\\cos(π/2 + α) = −\\sin α：奇变偶不变，符号看象限（第二象限余弦为负）。$","tags":["medium","reduction-formula"],"status":"published"},{"id":"math-trig-functions-017","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":3,"stem":"What is the least positive period of $y = \\sin(2x)$?","stemZh":"函数 $y = \\sin(2x)$ 的最小正周期是","options":[{"id":"A","text":"$2π$"},{"id":"B","text":"$π$"},{"id":"C","text":"$π/2$"},{"id":"D","text":"$4π$"}],"answer":"B","explanation":"For $y = \\sin(ωx)$ the period is $2π/|ω| = 2π/2 = π$.","explanationZh":"$y = \\sin(ωx)$ 的周期为 $2π/|ω| = 2π/2 = π。$","tags":["medium","period","graph"],"status":"published"},{"id":"math-trig-functions-018","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":3,"stem":"What is the amplitude of $y = −2 \\sin(3x)$?","stemZh":"函数 $y = −2 \\sin(3x)$ 的振幅是","options":[{"id":"A","text":"−2"},{"id":"B","text":"3"},{"id":"C","text":"6"},{"id":"D","text":"2"}],"answer":"D","explanation":"The amplitude is $|A| = |−2| = 2$; an amplitude is never negative, and 3 is the angular frequency.","explanationZh":"振幅为 $|A| = |−2| = 2，振幅不能为负；3$ 是角频率。","tags":["medium","amplitude","graph"],"status":"published"},{"id":"math-trig-functions-019","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":3,"stem":"On which of the following intervals is $y = \\sin x$ increasing?","stemZh":"函数 $y = \\sin x$ 在下列哪个区间上单调递增？","options":[{"id":"A","text":"$[−π/2, π/2]$"},{"id":"B","text":"$[0, π]$"},{"id":"C","text":"$[π/2, 3π/2]$"},{"id":"D","text":"$[−π, −π/2]$"}],"answer":"A","explanation":"On $[−π/2, π/2]$ the sine curve rises from −1 to 1; on $[π/2, 3π/2]$ it falls from 1 to −1.","explanationZh":"$在 [−π/2, π/2]$ 上正弦曲线由 −1 升到 $1；在 [π/2, 3π/2]$ 上由 1 降到 −1。","tags":["medium","graph","monotonicity"],"status":"published"},{"id":"math-trig-functions-020","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":3,"stem":"If $\\sin θ \\cos θ < 0$, then the terminal side of $θ$ lies in","stemZh":"$若 \\sin θ \\cos θ < 0，则 θ$ 的终边在","options":[{"id":"A","text":"Quadrant I or Quadrant II"},{"id":"B","text":"Quadrant I or Quadrant III"},{"id":"C","text":"Quadrant II or Quadrant IV"},{"id":"D","text":"Quadrant III or Quadrant IV"}],"answer":"C","explanation":"The product is negative exactly when $\\sin θ$ and $\\cos θ$ have opposite signs, which happens in Quadrants II and IV.","explanationZh":"乘积为负说明 $\\sin θ 与 \\cos θ$ 异号，这只在第二象限和第四象限发生。","tags":["medium","sign","quadrant"],"status":"published"},{"id":"math-trig-functions-021","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":4,"stem":"If $\\tan α = 2$, then $(\\sin α + 2\\cos α)/(3\\sin α − \\cos α)$ equals","stemZh":"$若 \\tan α = 2，则 (\\sin α + 2\\cos α)/(3\\sin α − \\cos α)$ 等于","options":[{"id":"A","text":"5/4"},{"id":"B","text":"4/5"},{"id":"C","text":"4/7"},{"id":"D","text":"2/5"}],"answer":"B","explanation":"Divide numerator and denominator by $\\cos α: (\\tan α + 2)/(3\\tan α − 1) = (2 + 2)/(6 − 1) = 4/5$.","explanationZh":"分子分母同除以 $\\cos α：(\\tan α + 2)/(3\\tan α − 1) = (2 + 2)/(6 − 1) = 4/5。$","tags":["hard","identity","tangent"],"status":"published"},{"id":"math-trig-functions-022","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":4,"stem":"If $\\sin α + \\cos α = 1/2$, then $\\sin α \\cos α$ equals","stemZh":"$若 \\sin α + \\cos α = 1/2，则 \\sin α \\cos α$ 等于","options":[{"id":"A","text":"3/8"},{"id":"B","text":"3/4"},{"id":"C","text":"−3/4"},{"id":"D","text":"−3/8"}],"answer":"D","explanation":"Square both $sides: \\sin^{2}α + 2 \\sin α \\cos α + \\cos^{2}α = 1/4$, so $1 + 2 \\sin α \\cos α = 1/4$ and $\\sin α \\cos α = −3/8$.","explanationZh":"$两边平方：\\sin^{2}α + 2 \\sin α \\cos α + \\cos^{2}α = 1/4，即 1 + 2 \\sin α \\cos α = 1/4，故 \\sin α \\cos α = −3/8。$","tags":["hard","identity"],"status":"published"},{"id":"math-trig-functions-023","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":4,"stem":"Simplify $[\\sin(π + α) · \\cos(π/2 − α)] / [\\cos(π − α) · \\sin(π/2 + α)]$.","stemZh":"化简 $[\\sin(π + α) · \\cos(π/2 − α)] / [\\cos(π − α) · \\sin(π/2 + α)]。$","options":[{"id":"A","text":"$\\tan^{2}α$"},{"id":"B","text":"$−\\tan^{2}α$"},{"id":"C","text":"1"},{"id":"D","text":"−1"}],"answer":"A","explanation":"$\\sin(π + α) = −\\sin α, \\cos(π/2 − α) = \\sin α, \\cos(π − α) = −\\cos α$ and $\\sin(π/2 + α) = \\cos α$, so the quotient is $(−\\sin^{2}α)/(−\\cos^{2}α) = \\tan^{2}α$.","explanationZh":"$\\sin(π + α) = −\\sin α，\\cos(π/2 − α) = \\sin α，\\cos(π − α) = −\\cos α，\\sin(π/2 + α) = \\cos α，故原式 = (−\\sin^{2}α)/(−\\cos^{2}α) = \\tan^{2}α。$","tags":["hard","reduction-formula","identity"],"status":"published"},{"id":"math-trig-functions-024","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":4,"stem":"What is the least positive period of $f(x) = \\sin x \\cos$ x?","stemZh":"函数 $f(x) = \\sin x \\cos x$ 的最小正周期是","options":[{"id":"A","text":"$4π$"},{"id":"B","text":"$2π$"},{"id":"C","text":"$π$"},{"id":"D","text":"$π/2$"}],"answer":"C","explanation":"$\\sin x \\cos x = (1/2) \\sin 2x$, whose least positive period is $2π/2 = π$.","explanationZh":"$\\sin x \\cos x = ½ \\sin 2x，其最小正周期为 2π/2 = π。$","tags":["hard","period","identity"],"status":"published"},{"id":"math-trig-functions-025","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":5,"stem":"If $\\sin(π/6 + α) = 1/3$, then $\\cos(π/3 − α)$ equals","stemZh":"$若 \\sin(π/6 + α) = 1/3，则 \\cos(π/3 − α)$ 等于","options":[{"id":"A","text":"$2\\sqrt{2}/3$"},{"id":"B","text":"1/3"},{"id":"C","text":"−1/3"},{"id":"D","text":"$−2\\sqrt{2}/3$"}],"answer":"B","explanation":"Since $(π/6 + α) + (π/3 − α) = π/2$, the two angles are complementary, so $\\cos(π/3 − α) = \\sin(π/6 + α) = 1/3$.","explanationZh":"因为 $(π/6 + α) + (π/3 − α) = π/2，两角互余，故 \\cos(π/3 − α) = \\sin(π/6 + α) = 1/3。$","tags":["hard","reduction-formula","complementary"],"status":"published"},{"id":"math-trig-functions-026","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":4,"stem":"What is the domain of $f(x) = \\sqrt{\\sin x}$?","stemZh":"函数 $f(x) = \\sqrt{\\sin x}$ 的定义域是","options":[{"id":"A","text":"${x | −π/2 + 2kπ ≤ x ≤ π/2 + 2kπ, k∈ℤ}$"},{"id":"B","text":"${x | 0 ≤ x ≤ π}$"},{"id":"C","text":"${x | π + 2kπ ≤ x ≤ 2π + 2kπ, k∈ℤ}$"},{"id":"D","text":"${x | 2kπ ≤ x ≤ π + 2kπ, k∈ℤ}$"}],"answer":"D","explanation":"We need $\\sin x ≥ 0$, which holds on $[2kπ, π + 2kπ]$. Option B is incomplete because it lists only one of the infinitely many intervals.","explanationZh":"$需 \\sin x ≥ 0，即 x ∈ [2kπ, π + 2kπ]；B$ 只写出了其中一个区间，不完整。","tags":["hard","domain","graph"],"status":"published"},{"id":"math-trig-functions-027","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":4,"stem":"Which of the following identities is correct?","stemZh":"下列诱导公式中正确的是","options":[{"id":"A","text":"$\\sin(π/2 + x) = \\cos x$"},{"id":"B","text":"$\\cos(π/2 + x) = \\cos x$"},{"id":"C","text":"$\\sin(π − x) = −\\sin x$"},{"id":"D","text":"$\\tan(π + x) = −\\tan x$"}],"answer":"A","explanation":"$\\sin(π/2 + x) = \\cos$ x. The others should read $\\cos(π/2 + x) = −\\sin x, \\sin(π − x) = \\sin x$ and $\\tan(π + x) = \\tan$ x.","explanationZh":"$\\sin(π/2 + x) = \\cos$ x。其余应为 $\\cos(π/2 + x) = −\\sin x，\\sin(π − x) = \\sin x，\\tan(π + x) = \\tan$ x。","tags":["hard","reduction-formula","error-distinction"],"status":"published"},{"id":"math-trig-functions-028","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":5,"stem":"If $\\sin α = 3/5$ and $α ∈ (π/2, π)$, then $\\sin(α − π/3)$ equals","stemZh":"$若 \\sin α = 3/5 且 α ∈ (π/2, π)，则 \\sin(α − π/3)$ 等于","options":[{"id":"A","text":"$(3 − 4\\sqrt{3})/10$"},{"id":"B","text":"$(4\\sqrt{3} − 3)/10$"},{"id":"C","text":"$(3 + 4\\sqrt{3})/10$"},{"id":"D","text":"$(−3 + 4\\sqrt{3})/10$"}],"answer":"C","explanation":"In Quadrant II $\\cos α = −4/5$, so $\\sin(α − π/3) = \\sin α \\cos(π/3) − \\cos α \\sin(π/3) = (3/5)(1/2) − (−4/5)(\\sqrt{3}/2) = (3 + 4\\sqrt{3})/10$.","explanationZh":"第二象限 $\\cos α = −4/5，故 \\sin(α − π/3) = \\sin α \\cos(π/3) − \\cos α \\sin(π/3) = (3/5)(1/2) − (−4/5)(\\sqrt{3}/2) = (3 + 4\\sqrt{3})/10。$","tags":["hard","quadrant-sign","addition-formula"],"status":"published"},{"id":"math-trig-functions-029","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":4,"stem":"The graph of $y = \\sin(2x + π/3)$ is obtained from the graph of $y = \\sin 2x$ by","stemZh":"$由 y = \\sin 2x$ 的图像得到 $y = \\sin(2x + π/3)$ 的图像，需要","options":[{"id":"A","text":"shifting it right by $π/3$"},{"id":"B","text":"shifting it left by $π/6$"},{"id":"C","text":"shifting it right by $π/6$"},{"id":"D","text":"shifting it left by $π/3$"}],"answer":"B","explanation":"Write $\\sin(2x + π/3) = \\sin(2(x + π/6))$: the shift is $π/6$ to the left, not $π/3$, because the coefficient of x must be factored out first.","explanationZh":"$\\sin(2x + π/3) = \\sin(2(x + π/6))，应向左平移 π/6$ 而不是 $π/3；必须先提取 x$ 的系数。","tags":["hard","graph","phase-shift"],"status":"published"},{"id":"math-trig-functions-030","subject":"math","moduleId":"functions","unitId":"trigonometry","topicId":"trig-functions","type":"single-choice","difficulty":5,"stem":"If $\\tan α = −3$ and $α ∈ (π/2, π)$, then $\\sin α$ equals","stemZh":"$若 \\tan α = −3 且 α ∈ (π/2, π)，则 \\sin α$ 等于","options":[{"id":"A","text":"$−3\\sqrt{10}/10$"},{"id":"B","text":"$\\sqrt{10}/10$"},{"id":"C","text":"$−\\sqrt{10}/10$"},{"id":"D","text":"$3\\sqrt{10}/10$"}],"answer":"D","explanation":"In Quadrant II sine is positive and cosine is negative. With $\\tan α = −3$ we get $\\cos^{2}α = 1/10$, so $\\cos α = −\\sqrt{10}/10$ and $\\sin α = |\\tan α| · |\\cos α| = 3\\sqrt{10}/10$.","explanationZh":"第二象限正弦为正、余弦为负。由 $\\tan α = −3 得 \\cos^{2}α = 1/10，故 \\cos α = −\\sqrt{10}/10，\\sin α = |\\tan α| · |\\cos α| = 3\\sqrt{10}/10。$","tags":["hard","quadrant-sign","tangent"],"status":"published"},{"id":"math-derivatives-001","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":1,"stem":"If $f(x) = x^{5}$, then $f′(x) =$ ?","stemZh":"$若 f(x) = x^{5}，则 f′(x) = ？$","options":[{"id":"A","text":"$5x^{4}$"},{"id":"B","text":"$x^{4}$"},{"id":"C","text":"$5x^{5}$"},{"id":"D","text":"$4x^{5}$"}],"answer":"A","explanation":"By the power rule $(x^{n})′ = n·x^{n-1}$, with $n = 5$ we get $(x^{5})′ = 5x^{4}$. The exponent drops by 1, it does not stay 5.","explanationZh":"由幂函数求导公式 $(x^{n})′ = n·x^{n-1}，取 n = 5 得 (x^{5})′ = 5x^{4}。指数要减 1，不能保持不变。$","tags":["easy","power-rule","derivative-formula"],"status":"published"},{"id":"math-derivatives-002","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":1,"stem":"If $f(x) = −3$ for every real x, then $f′(x) =$ ?","stemZh":"若对任意实数 x 都有 $f(x) = −3，则 f′(x) = ？$","options":[{"id":"A","text":"−3"},{"id":"B","text":"0"},{"id":"C","text":"1"},{"id":"D","text":"x"}],"answer":"B","explanation":"A constant function has a horizontal graph, so its rate of change is $0 everywhere: (c)′ = 0$.","explanationZh":"常数函数的图象是一条水平直线，变化率恒为 $0，即 (c)′ = 0。$","tags":["easy","constant-derivative","derivative-formula"],"status":"published"},{"id":"math-derivatives-003","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":1,"stem":"The derivative of $f(x) = e^x$ is","stemZh":"函数 $f(x) = e^x$ 的导数是","options":[{"id":"A","text":"$x·e^{x−1}$"},{"id":"B","text":"$e^x·\\ln x$"},{"id":"C","text":"$e^x$"},{"id":"D","text":"1"}],"answer":"C","explanation":"The natural exponential function is its own $derivative: (e^x)′ = e^x$. The power rule does not apply here, since the variable is in the exponent.","explanationZh":"$自然指数函数的导数等于它本身：(e^x)′ = e^x。这里变量在指数位置，不能用幂函数求导公式。$","tags":["easy","exponential-derivative","derivative-formula"],"status":"published"},{"id":"math-derivatives-004","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":1,"stem":"If $f(x) = \\ln x$ with $x > 0$, then $f′(x) =$ ?","stemZh":"$若 f(x) = \\ln$ x（x $> 0），则 f′(x) = ？$","options":[{"id":"A","text":"$x·\\ln x$"},{"id":"B","text":"$e^x$"},{"id":"C","text":"$\\ln x$"},{"id":"D","text":"1/x"}],"answer":"D","explanation":"The derivative of the natural logarithm is $(\\ln x)′ = 1/x$ for $x > 0$. It is not differentiable from the exponential rule.","explanationZh":"自然对数的导数为 $(\\ln x)′ = 1/x（x > 0），不要与指数函数的求导公式混淆。$","tags":["easy","logarithm-derivative","derivative-formula"],"status":"published"},{"id":"math-derivatives-005","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":1,"stem":"If $f(x) = x^{2}$, then $f′(3) =$ ?","stemZh":"$若 f(x) = x^{2}，则 f′(3) = ？$","options":[{"id":"A","text":"3"},{"id":"B","text":"6"},{"id":"C","text":"9"},{"id":"D","text":"2"}],"answer":"B","explanation":"First $f′(x) = 2x$, then substitute $x = 3: f′(3) = 2·3 = 6$. Note that $f(3) = 9$ is the function value, not the derivative.","explanationZh":"先求导得 $f′(x) = 2x，再代入 x = 3：f′(3) = 2\\times3 = 6。注意 f(3) = 9$ 是函数值，不是导数值。","tags":["easy","derivative-value","power-rule"],"status":"published"},{"id":"math-derivatives-006","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":2,"stem":"If $f(x) = x^{3} − 2x + 5$, then $f′(x) =$ ?","stemZh":"$若 f(x) = x^{3} − 2x + 5，则 f′(x) = ？$","options":[{"id":"A","text":"$3x^{2} + 5$"},{"id":"B","text":"$3x^{2} − 2x$"},{"id":"C","text":"$3x^{2} − 2$"},{"id":"D","text":"$x^{2} − 2$"}],"answer":"C","explanation":"Differentiate term by $term: (x^{3})′ = 3x^{2}, (−2x)′ = −2$, and the constant 5 contributes 0. Hence $f′(x) = 3x^{2} − 2$.","explanationZh":"$逐项求导：(x^{3})′ = 3x^{2}，(−2x)′ = −2，常数项 5$ 的导数为 $0，所以 f′(x) = 3x^{2} − 2。$","tags":["easy","sum-rule","polynomial"],"status":"published"},{"id":"math-derivatives-007","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":2,"stem":"If $f(x) = 2^{x}$, then $f′(x) =$ ?","stemZh":"$若 f(x) = 2^{x}，则 f′(x) = ？$","options":[{"id":"A","text":"$2^{x}$"},{"id":"B","text":"$2^{x}·\\ln 2$"},{"id":"C","text":"$x·2^{x−1}$"},{"id":"D","text":"$2^{x}/\\ln 2$"}],"answer":"B","explanation":"For a general $exponential, (a^{x})′ = a^{x}·\\ln a$, so $(2^{x})′ = 2^{x}·\\ln 2$. The power rule $x·2^{x−1}$ would only apply to $x^{2}$, not to $2^{x}$.","explanationZh":"一般指数函数的求导公式为 $(a^{x})′ = a^{x}·\\ln$ a，故 $(2^{x})′ = 2^{x}·\\ln 2。幂函数公式只适用于 x^{2}$ 这类形式，不能用于 $2^{x}。$","tags":["easy","exponential-derivative","derivative-formula"],"status":"published"},{"id":"math-derivatives-008","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":2,"stem":"If f is differentiable and $f′(2) = 3$, what does the number 3 represent?","stemZh":"若 f 可导且 $f′(2) = 3，则数 3$ 表示什么？","options":[{"id":"A","text":"The value of f at $x = 2$"},{"id":"B","text":"The y-intercept of the tangent line at $x = 2$"},{"id":"C","text":"The slope of the normal line to $y = f(x)$ at $x = 2$"},{"id":"D","text":"The slope of the tangent line to $y = f(x)$ at $x = 2$"}],"answer":"D","explanation":"$Geometrically, f′(x_{0})$ is the slope of the tangent line to $y = f(x)$ at the point $(x_{0}, f(x_{0}))$. It is not the function value $f(2)$, and the normal would have slope −1/3.","explanationZh":"$几何意义：f′(x_{0})$ 表示曲线 $y = f(x)$ 在点 $(x_{0}, f(x_{0}))$ 处切线的斜率。它不是函数值 $f(2)，法线斜率应为 −1/3。$","tags":["easy","geometric-meaning","tangent"],"status":"published"},{"id":"math-derivatives-009","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":2,"stem":"$(\\sin x)′ =$ ?","stemZh":"$(\\sin x)′ = ？$","options":[{"id":"A","text":"$\\cos x$"},{"id":"B","text":"$−\\cos x$"},{"id":"C","text":"$\\sin x$"},{"id":"D","text":"$−\\sin x$"}],"answer":"A","explanation":"The basic formula is $(\\sin x)′ = \\cos$ x. The minus sign appears in $(\\cos x)′ = −\\sin x$, not here.","explanationZh":"基本公式为 $(\\sin x)′ = \\cos$ x。负号出现在 $(\\cos x)′ = −\\sin x$ 中，此处没有负号。","tags":["easy","trigonometric-derivative","derivative-formula"],"status":"published"},{"id":"math-derivatives-010","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":2,"stem":"$(\\cos x)′ =$ ?","stemZh":"$(\\cos x)′ = ？$","options":[{"id":"A","text":"$\\cos x$"},{"id":"B","text":"$\\sin x$"},{"id":"C","text":"$−\\cos x$"},{"id":"D","text":"$−\\sin x$"}],"answer":"D","explanation":"The derivative of cosine is negative $sine: (\\cos x)′ = −\\sin$ x. Forgetting the minus sign is the most common slip.","explanationZh":"$余弦的导数是负的正弦：(\\cos x)′ = −\\sin$ x。最容易出错的地方是漏掉负号。","tags":["easy","trigonometric-derivative","derivative-formula"],"status":"published"},{"id":"math-derivatives-011","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":3,"stem":"If $f(x) = x^{2}·e^x$, then $f′(x) =$ ?","stemZh":"$若 f(x) = x^{2}·e^x，则 f′(x) = ？$","options":[{"id":"A","text":"$e^x(x^{2} + 2x)$"},{"id":"B","text":"$2x·e^x$"},{"id":"C","text":"$x^{2}·e^x$"},{"id":"D","text":"$e^x(2x − x^{2})$"}],"answer":"A","explanation":"Use the product $rule: f′ = (x^{2})′·e^x + x^{2}·(e^x)′ = 2x·e^x + x^{2}·e^x = e^x(x^{2} + 2x)$. Options B and C each keep only one of the two terms.","explanationZh":"$用乘法法则：f′ = (x^{2})′·e^x + x^{2}·(e^x)′ = 2x·e^x + x^{2}·e^x = e^x(x^{2} + 2x)。B、C$ 两项都只算了一半。","tags":["medium","product-rule","derivative-formula"],"status":"published"},{"id":"math-derivatives-012","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":3,"stem":"If $f(x) = x/(x + 1)$, then $f′(x) =$ ?","stemZh":"$若 f(x) = x/(x + 1)，则 f′(x) = ？$","options":[{"id":"A","text":"$1/(x + 1)$"},{"id":"B","text":"$−1/(x + 1)^{2}$"},{"id":"C","text":"$1/(x + 1)^{2}$"},{"id":"D","text":"$x/(x + 1)^{2}$"}],"answer":"C","explanation":"By the quotient $rule, f′ = [(x + 1)·1 − x·1]/(x + 1)^{2} = 1/(x + 1)^{2}$. The numerator simplifies to 1, not to −1.","explanationZh":"$由除法法则：f′ = [(x + 1)·1 − x·1]/(x + 1)^{2} = 1/(x + 1)^{2}。分子化简后为 1，不是 −1。$","tags":["medium","quotient-rule","derivative-formula"],"status":"published"},{"id":"math-derivatives-013","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":3,"stem":"If $f(x) = (2x + 1)^{3}$, then $f′(x) =$ ?","stemZh":"$若 f(x) = (2x + 1)^{3}，则 f′(x) = ？$","options":[{"id":"A","text":"$3(2x + 1)^{2}$"},{"id":"B","text":"$(2x + 1)^{2}$"},{"id":"C","text":"$6(2x + 1)$"},{"id":"D","text":"$6(2x + 1)^{2}$"}],"answer":"D","explanation":"Chain rule: bring down the exponent 3, keep the inner function, then multiply by the derivative of the inside, 2. So $f′ = 3(2x + 1)^{2}·2 = 6(2x + 1)^{2}$.","explanationZh":"复合函数求导：先把指数 3 提到前面，保留内层，再乘内层的导数 $2，即 f′ = 3(2x + 1)^{2}·2 = 6(2x + 1)^{2}。漏乘 2$ 是常见错误。","tags":["medium","chain-rule","derivative-formula"],"status":"published"},{"id":"math-derivatives-014","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":3,"stem":"The derivative of $f(x) = \\sin 2x$ is","stemZh":"函数 $f(x) = \\sin 2x$ 的导数是","options":[{"id":"A","text":"$\\cos 2x$"},{"id":"B","text":"$2 \\cos 2x$"},{"id":"C","text":"$−2 \\cos 2x$"},{"id":"D","text":"$2 \\sin 2x$"}],"answer":"B","explanation":"Chain $rule: (\\sin 2x)′ = \\cos(2x)·(2x)′ = 2 \\cos 2x$. The minus sign in option C belongs to the derivative of $\\cos$, not $\\sin$.","explanationZh":"$复合函数求导：(\\sin 2x)′ = \\cos(2x)·(2x)′ = 2 \\cos 2x。C$ 项的负号属于 $\\cos$ 的导数，不属于 $\\sin。$","tags":["medium","chain-rule","trigonometric-derivative"],"status":"published"},{"id":"math-derivatives-015","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":3,"stem":"The equation of the tangent line to $y = x^{2}$ at the point $(1, 1)$ is","stemZh":"曲线 $y = x^{2}$ 在点 $(1, 1)$ 处的切线方程是","options":[{"id":"A","text":"$y = 2x − 1$"},{"id":"B","text":"$y = 2x + 1$"},{"id":"C","text":"$y = x$"},{"id":"D","text":"$y = 2x$"}],"answer":"A","explanation":"$f′(x) = 2x$, so the tangent slope is $f′(1) = 2$. Using point-slope $form: y − 1 = 2(x − 1)$, i.e. $y = 2x − 1$.","explanationZh":"$f′(x) = 2x，切线斜率 k = f′(1) = 2。由点斜式 y − 1 = 2(x − 1)，得 y = 2x − 1。$","tags":["medium","tangent-line","geometric-meaning"],"status":"published"},{"id":"math-derivatives-016","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":3,"stem":"On which interval is $f(x) = x^{3} − 3x$ decreasing?","stemZh":"函数 $f(x) = x^{3} − 3x$ 在哪个区间上单调递减？","options":[{"id":"A","text":"$(−∞, −1)$"},{"id":"B","text":"$(1, +∞)$"},{"id":"C","text":"$(−1, 1)$"},{"id":"D","text":"$(−∞, −1) ∪ (1, +∞)$"}],"answer":"C","explanation":"$f′(x) = 3x^{2} − 3 = 3(x − 1)(x + 1)$. Then $f′(x) < 0$ exactly when $−1 < x < 1$, so f decreases on $(−1, 1)$ and increases on the two outer intervals.","explanationZh":"$f′(x) = 3x^{2} − 3 = 3(x − 1)(x + 1)。令 f′(x) < 0 得 −1 < x < 1，故减区间为 (−1, 1)，两侧为增区间。$","tags":["medium","monotonicity","derivative-sign"],"status":"published"},{"id":"math-derivatives-017","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":3,"stem":"The minimum value of $f(x) = x^{2} − 4x + 3$ on the interval $[0, 3]$ is","stemZh":"函数 $f(x) = x^{2} − 4x + 3$ 在区间 $[0, 3]$ 上的最小值是","options":[{"id":"A","text":"0"},{"id":"B","text":"3"},{"id":"C","text":"−4"},{"id":"D","text":"−1"}],"answer":"D","explanation":"$f′(x) = 2x − 4$ vanishes at $x = 2$, which lies inside $[0, 3]$. Compare $f(0) = 3, f(2) = −1$ and $f(3) = 0$; the smallest of these is −1.","explanationZh":"$f′(x) = 2x − 4，得驻点 x = 2$ 在区间 $[0, 3]$ 内。比较 $f(0) = 3、f(2) = −1、f(3) = 0，最小值为 −1。$","tags":["medium","extreme-value","closed-interval"],"status":"published"},{"id":"math-derivatives-018","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":3,"stem":"At which value of x does $f(x) = x^{3} − 3x^{2}$ have a local minimum?","stemZh":"函数 $f(x) = x^{3} − 3x^{2} 在 x$ 取何值时取得极小值？","options":[{"id":"A","text":"$x = 0$"},{"id":"B","text":"$x = 2$"},{"id":"C","text":"$x = 1$"},{"id":"D","text":"$x = −2$"}],"answer":"B","explanation":"$f′(x) = 3x^{2} − 6x = 3x(x − 2)$. Its sign is + on $(−∞, 0), −$ on $(0, 2)$ and + on $(2, +∞)$. The sign changes from − to + at $x = 2$, giving a local minimum; at $x = 0$ it changes from + to −, giving a local maximum.","explanationZh":"$f′(x) = 3x^{2} − 6x = 3x(x − 2)。f′(x) 在 (−∞, 0) 为正，(0, 2) 为负，(2, +∞)$ 为正。在 $x = 2$ 处由负变正，取极小值；在 $x = 0$ 处由正变负，取极大值。","tags":["medium","local-extremum","derivative-sign"],"status":"published"},{"id":"math-derivatives-019","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":3,"stem":"The average rate of change of $f(x) = x^{2}$ over the interval $[1, 3]$ is","stemZh":"函数 $f(x) = x^{2}$ 在区间 $[1, 3]$ 上的平均变化率是","options":[{"id":"A","text":"2"},{"id":"B","text":"3"},{"id":"C","text":"4"},{"id":"D","text":"8"}],"answer":"C","explanation":"Average rate of change $= [f(3) − f(1)]/(3 − 1) = (9 − 1)/2 = 4$. It is the slope of the secant line, not the derivative at an endpoint.","explanationZh":"平均变化率 $= [f(3) − f(1)]/(3 − 1) = (9 − 1)/2 = 4，即割线斜率，不是端点处的导数。$","tags":["medium","rate-of-change","application"],"status":"published"},{"id":"math-derivatives-020","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":3,"stem":"If $f(x) = \\ln(2x)$ with $x > 0$, then $f′(x) =$ ?","stemZh":"$若 f(x) = \\ln(2x)（x > 0），则 f′(x) = ？$","options":[{"id":"A","text":"1/x"},{"id":"B","text":"$1/(2x)$"},{"id":"C","text":"2/x"},{"id":"D","text":"$(\\ln 2)/x$"}],"answer":"A","explanation":"Chain $rule: f′ = [1/(2x)]·(2x)′ = [1/(2x)]·2 = 1/x$. Equivalently $\\ln(2x) = \\ln 2 + \\ln x$, whose derivative is 0 + 1/x.","explanationZh":"$复合函数求导：f′ = [1/(2x)]·(2x)′ = [1/(2x)]·2 = 1/x。也可化为 \\ln(2x) = \\ln 2 + \\ln$ x，求导得 0 + 1/x。","tags":["medium","chain-rule","logarithm-derivative"],"status":"published"},{"id":"math-derivatives-021","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":4,"stem":"The tangent line to the curve $y = x^{2} − 3x + 2$ at the point with $x = 2$ has equation","stemZh":"曲线 $y = x^{2} − 3x + 2$ 在横坐标 $x = 2$ 的点处的切线方程是","options":[{"id":"A","text":"$y = x − 2$"},{"id":"B","text":"$y = x + 2$"},{"id":"C","text":"$y = 2x − 2$"},{"id":"D","text":"$y =$ −x + 2"}],"answer":"A","explanation":"$f(2) = 4 − 6 + 2 = 0$, so the point is $(2, 0)$. Since $f′(x) = 2x − 3$, the slope is $f′(2) = 1$. The tangent is $y − 0 = 1(x − 2)$, i.e. $y = x − 2$.","explanationZh":"$f(2) = 4 − 6 + 2 = 0，切点为 (2, 0)；f′(x) = 2x − 3，斜率 k = f′(2) = 1。故切线为 y − 0 = 1·(x − 2)，即 y = x − 2。$","tags":["hard","tangent-line","geometric-meaning"],"status":"published"},{"id":"math-derivatives-022","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":4,"stem":"Let $f(x) = x^{3}$. Which statement about f at $x = 0$ is correct?","stemZh":"$设 f(x) = x^{3}，关于 f 在 x = 0$ 处的结论正确的是","options":[{"id":"A","text":"f has a local minimum at $x = 0$"},{"id":"B","text":"f has a local maximum at $x = 0$"},{"id":"C","text":"$f′(0) = 0$, but f has no local extremum at $x = 0$"},{"id":"D","text":"f is decreasing on $(−∞, +∞)$"}],"answer":"C","explanation":"$f′(x) = 3x^{2}$ gives $f′(0) = 0$, but $3x^{2} ≥ 0$ on both sides of 0, so the derivative never changes sign and f is increasing throughout. Hence $x = 0$ is a stationary point that is not an extremum.","explanationZh":"$f′(x) = 3x^{2}，有 f′(0) = 0；但 3x^{2} ≥ 0 在 0$ 的两侧恒成立，导数不变号，函数始终递增，故 $x = 0$ 是驻点但不是极值点。","tags":["hard","local-extremum","derivative-sign"],"status":"published"},{"id":"math-derivatives-023","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":4,"stem":"On which interval is $f(x) = e^x − x$ decreasing?","stemZh":"函数 $f(x) = e^x − x$ 在哪个区间上单调递减？","options":[{"id":"A","text":"$(0, +∞)$"},{"id":"B","text":"$(−∞, 0)$"},{"id":"C","text":"$(−∞, +∞)$"},{"id":"D","text":"$(1, +∞)$"}],"answer":"B","explanation":"$f′(x) = e^x − 1$. Since $e^x < 1$ exactly when $x < 0$, we get $f′(x) < 0$ on $(−∞, 0)$ and $f′(x) > 0$ on $(0, +∞)$. So f decreases on $(−∞, 0)$ and increases afterwards.","explanationZh":"$f′(x) = e^x − 1。当且仅当 x < 0 时 e^x < 1，故 f′(x) < 0 在 (−∞, 0)$ 上成立，在 $(0, +∞) 上 f′(x) > 0，即减区间为 (−∞, 0)。$","tags":["hard","monotonicity","exponential-derivative"],"status":"published"},{"id":"math-derivatives-024","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":4,"stem":"For $f(x) = x^{3} − 3x$, what is the slope of the normal line to the curve at $x = 2$?","stemZh":"对于 $f(x) = x^{3} − 3x，曲线在 x = 2$ 处的法线斜率是","options":[{"id":"A","text":"−1/9"},{"id":"B","text":"9"},{"id":"C","text":"1/9"},{"id":"D","text":"−9"}],"answer":"A","explanation":"$f′(2) = 3·2^{2} − 3 = 9$ is the tangent slope. The normal is perpendicular to the tangent, so its slope is the negative reciprocal: −1/9.","explanationZh":"$f′(2) = 3\\times2^{2} − 3 = 9$ 是切线斜率；法线与切线垂直，其斜率为负倒数，即 −1/9。","tags":["hard","normal-line","geometric-meaning"],"status":"published"},{"id":"math-derivatives-025","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":4,"stem":"If $f(x) = \\cos 2x$, then $f′(x) =$ ?","stemZh":"$若 f(x) = \\cos 2x，则 f′(x) = ？$","options":[{"id":"A","text":"$−\\sin 2x$"},{"id":"B","text":"$2 \\sin 2x$"},{"id":"C","text":"$2 \\cos 2x$"},{"id":"D","text":"$−2 \\sin 2x$"}],"answer":"D","explanation":"Chain rule with the cosine $formula: (\\cos 2x)′ = −\\sin(2x)·(2x)′ = −2 \\sin 2x$. Both the minus sign and the factor 2 are required.","explanationZh":"$复合函数求导：(\\cos 2x)′ = −\\sin(2x)·(2x)′ = −2 \\sin 2x。负号和系数 2$ 缺一不可。","tags":["hard","chain-rule","trigonometric-derivative"],"status":"published"},{"id":"math-derivatives-026","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":4,"stem":"A particle moves along a line with position $s(t) = t^{3} − 6t^{2} + 9t (t ≥ 0)$. At which time(s) is its instantaneous velocity zero?","stemZh":"质点沿直线运动，位置 $s(t) = t^{3} − 6t^{2} + 9t（t ≥ 0）。瞬时速度为 0$ 的时刻是","options":[{"id":"A","text":"$t = 0$ only"},{"id":"B","text":"$t = 1$ and $t = 3$"},{"id":"C","text":"$t = 3$ only"},{"id":"D","text":"$t = 0$ and $t = 3$"}],"answer":"B","explanation":"Velocity is $v(t) = s′(t) = 3t^{2} − 12t + 9 = 3(t − 1)(t − 3)$, which is zero at $t = 1$ and $t = 3$. Note $s(0) = 0$ is the initial position, not a zero of the velocity.","explanationZh":"速度 $v(t) = s′(t) = 3t^{2} − 12t + 9 = 3(t − 1)(t − 3)，令其为 0 得 t = 1 或 t = 3。注意 s(0) = 0$ 是初始位置，不是速度为 0 的时刻。","tags":["hard","rate-of-change","application"],"status":"published"},{"id":"math-derivatives-027","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":5,"stem":"If $f(x) = x^{2}·\\ln x$ with $x > 0$, then $f′(x) =$ ?","stemZh":"$若 f(x) = x^{2}·\\ln$ x（x $> 0），则 f′(x) = ？$","options":[{"id":"A","text":"$2x·\\ln x$"},{"id":"B","text":"$2x·\\ln x + x$"},{"id":"C","text":"$2x·\\ln x + 2$"},{"id":"D","text":"$x·\\ln x + x^{2}$"}],"answer":"B","explanation":"Product $rule: f′ = (x^{2})′·\\ln x + x^{2}·(\\ln x)′ = 2x·\\ln x + x^{2}·(1/x) = 2x·\\ln x +$ x. Option A drops the second term; option C wrongly treats $(\\ln x)′$ as $1/x^{2}$.","explanationZh":"$乘法法则：f′ = (x^{2})′·\\ln x + x^{2}·(\\ln x)′ = 2x·\\ln x + x^{2}·(1/x) = 2x·\\ln x +$ x。A 项漏掉第二项，C 项把 $(\\ln x)′$ 错当成 $1/x^{2}。$","tags":["hard","product-rule","logarithm-derivative"],"status":"published"},{"id":"math-derivatives-028","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":5,"stem":"If $f(x) = e^x/x$, then $f′(x) =$ ?","stemZh":"$若 f(x) = e^x/x，则 f′(x) = ？$","options":[{"id":"A","text":"$e^x/x$"},{"id":"B","text":"$e^x(1 − x)/x^{2}$"},{"id":"C","text":"$e^x/x^{2}$"},{"id":"D","text":"$e^x(x − 1)/x^{2}$"}],"answer":"D","explanation":"Quotient $rule: f′ = [(e^x)′·x − e^x·(x)′]/x^{2} = (e^x·x − e^x)/x^{2} = e^x(x − 1)/x^{2}$. Option B reverses the order of the terms in the numerator, which flips the sign.","explanationZh":"$除法法则：f′ = [(e^x)′·x − e^x·(x)′]/x^{2} = (e^x·x − e^x)/x^{2} = e^x(x − 1)/x^{2}。B$ 项把分子两项顺序颠倒，导致符号相反。","tags":["hard","quotient-rule","exponential-derivative"],"status":"published"},{"id":"math-derivatives-029","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":5,"stem":"An open box is made from a 12 cm by 12 cm square sheet by cutting equal squares of side x cm from the four corners and folding up the sides. Its volume is $V(x) = x(12 − 2x)^{2}$. Which value of x maximises V?","stemZh":"用一块 12 cm $\\times 12$ cm 的正方形铁皮，四角各剪去边长为 x cm 的小正方形后折成无盖盒子，容积 $V(x) = x(12 − 2x)^{2}。使 V$ 最大的 x 是","options":[{"id":"A","text":"3"},{"id":"B","text":"4"},{"id":"C","text":"6"},{"id":"D","text":"2"}],"answer":"D","explanation":"$V′(x) = (12 − 2x)^{2} + x·2(12 − 2x)(−2) = (12 − 2x)(12 − 6x)$, so $V′ = 0$ at $x = 2$ or $x = 6$. For $0 < x < 6$ the domain excludes $x = 6 (there V = 0)$, and $V′ > 0$ on $(0, 2), V′ < 0$ on $(2, 6)$. Hence the maximum occurs at $x = 2$.","explanationZh":"$V′(x) = (12 − 2x)^{2} + x·2(12 − 2x)(−2) = (12 − 2x)(12 − 6x)，令 V′ = 0 得 x = 2 或 x = 6。定义域为 0 < x < 6，x = 6 时 V = 0$ 舍去；且 $V′ 在 (0, 2) 为正、(2, 6)$ 为负，故 $x = 2$ 时容积最大。","tags":["hard","optimisation","application"],"status":"published"},{"id":"math-derivatives-030","subject":"math","moduleId":"functions","unitId":"calculus","topicId":"derivatives","type":"single-choice","difficulty":5,"stem":"The maximum value of $f(x) = 2x^{3} − 3x^{2} − 12x + 1$ on the interval $[−2, 3]$ is","stemZh":"函数 $f(x) = 2x^{3} − 3x^{2} − 12x + 1$ 在区间 $[−2, 3]$ 上的最大值是","options":[{"id":"A","text":"−3"},{"id":"B","text":"−19"},{"id":"C","text":"8"},{"id":"D","text":"−8"}],"answer":"C","explanation":"$f′(x) = 6x^{2} − 6x − 12 = 6(x − 2)(x + 1)$, so the critical points are $x = −1$ and $x = 2$, both inside $[−2, 3]$. Compare $values: f(−2) = −3, f(−1) = 8, f(2) = −19, f(3) = −8$. The largest is 8.","explanationZh":"$f′(x) = 6x^{2} − 6x − 12 = 6(x − 2)(x + 1)，驻点 x = −1 与 x = 2$ 均在 $[−2, 3] 内。比较：f(−2) = −3，f(−1) = 8，f(2) = −19，f(3) = −8，最大值为 8。$","tags":["hard","extreme-value","closed-interval"],"status":"published"},{"id":"math-sequence-basics-001","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":1,"stem":"A sequence is given by $a_{n} = 3n − 2 (n ∈ ℕ⁺)$. What are its first three terms?","stemZh":"已知数列 $a_{n} = 3n − 2（n ∈ ℕ⁺），则它的前三项为？$","options":[{"id":"A","text":"1, 3, 5"},{"id":"B","text":"0, 3, 6"},{"id":"C","text":"1, 4, 7"},{"id":"D","text":"2, 5, 8"}],"answer":"C","explanation":"$a_{1} = 3·1 − 2 = 1, a_{2} = 3·2 − 2 = 4, a_{3} = 3·3 − 2 = 7$. The index of a sequence starts at $n = 1$, so the list beginning with 0 is wrong.","explanationZh":"$a_{1} = 1，a_{2} = 4，a_{3} = 7。数列的项数从 n = 1$ 开始，因此以 0 开头的选项是错误的。","tags":["easy","general-term","indexing"],"status":"published"},{"id":"math-sequence-basics-002","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":1,"stem":"A sequence ${a_{n}}$ can be regarded as a function $f(n) = a_{n}$. Its domain is","stemZh":"数列 ${a_{n}}$ 可以看成函数 $f(n) = a_{n}，它的定义域是$","options":[{"id":"A","text":"$ℕ⁺ (or {1, 2,\\text{ …, }n}\\text{ for }a\\text{ finite }sequence)$"},{"id":"B","text":"$ℝ$"},{"id":"C","text":"$ℤ$"},{"id":"D","text":"$[0, +∞)$"}],"answer":"A","explanation":"A sequence is a function whose domain is the set of positive integers $ℕ⁺$, or the finite set ${1, 2,\\text{ …, }n}$ when the sequence has only n terms.","explanationZh":"数列是定义域为正整数集 $ℕ⁺（有限数列时为 {1, 2, …, n}）的函数。$","tags":["easy","definition","sequence-as-function"],"status":"published"},{"id":"math-sequence-basics-003","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":1,"stem":"How many terms does the sequence 1, 3, 5, 7, …, 99 contain?","stemZh":"数列 1, 3, 5, 7, …, 99 共有多少项？","options":[{"id":"A","text":"49"},{"id":"B","text":"51"},{"id":"C","text":"100"},{"id":"D","text":"50"}],"answer":"D","explanation":"The terms are the odd numbers from 1 to 99. Writing $2k − 1 = 99$ gives $k = 50$, so there are 50 terms.","explanationZh":"这些项是从 1 到 99 的所有奇数。由 $2k − 1 = 99 得 k = 50，因此共有 50$ 项。","tags":["easy","counting-terms","finite-sequence"],"status":"published"},{"id":"math-sequence-basics-004","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":2,"stem":"If $a_{n} = n^{2} − n$, then $a_{4} =$","stemZh":"$若 a_{n} = n^{2} −$ n，则 $a_{4} =$","options":[{"id":"A","text":"6"},{"id":"B","text":"12"},{"id":"C","text":"16"},{"id":"D","text":"20"}],"answer":"B","explanation":"$a_{4} = 4^{2} − 4 = 16 − 4 = 12. (16\\text{ is only }n^{2},\\text{ forgetting to subtract }n.)$","explanationZh":"$a_{4} = 4^{2} − 4 = 16 − 4 = 12。（只算 n^{2} 得 16，是漏减 n$ 的错误。）","tags":["easy","general-term","evaluation"],"status":"published"},{"id":"math-sequence-basics-005","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":2,"stem":"A sequence satisfies $a_{1} = 1$ and $a_{n+1} = a_{n} + 2$. Its first four terms are","stemZh":"数列满足 $a_{1} = 1，a_{n+1} = a_{n} + 2，则它的前四项为$","options":[{"id":"A","text":"1, 2, 3, 4"},{"id":"B","text":"1, 3, 5, 7"},{"id":"C","text":"1, 3, 6, 10"},{"id":"D","text":"2, 4, 6, 8"}],"answer":"B","explanation":"$a_{2} = 1 + 2 = 3, a_{3} = 3 + 2 = 5, a_{4} = 5 + 2 = 7$, so the terms are 1, 3, 5, 7.","explanationZh":"$a_{2} = 3，a_{3} = 5，a_{4} = 7，所以前四项是 1, 3, 5, 7。$","tags":["easy","recurrence","listing-terms"],"status":"published"},{"id":"math-sequence-basics-006","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":2,"stem":"Which of the following is an infinite sequence?","stemZh":"下列哪一项是无穷数列？","options":[{"id":"A","text":"1, 2, 3, 4, 5"},{"id":"B","text":"The first 100 positive even numbers written in order"},{"id":"C","text":"$a_{n} = 1/n$ for $n = 1, 2, …, 10$"},{"id":"D","text":"$a_{n} = 1/n$ for $n ∈ ℕ⁺$"}],"answer":"D","explanation":"Only the last option has domain $ℕ⁺$ with no last term. The others have a finite number of terms $(5, 100$ and $10 respectively)$.","explanationZh":"只有最后一项的定义域是全体正整数 $ℕ⁺，没有末项。其余各项分别只有 5 项、100 项、10$ 项。","tags":["easy","finite-vs-infinite","definition"],"status":"published"},{"id":"math-sequence-basics-007","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":2,"stem":"A general term for the sequence 2, 5, 10, 17, 26, … is","stemZh":"数列 2, 5, 10, 17, 26, … 的一个通项公式是","options":[{"id":"A","text":"$a_{n} = n^{2} + 1$"},{"id":"B","text":"$a_{n} = n^{2} − 1$"},{"id":"C","text":"$a_{n} = 2n$"},{"id":"D","text":"$a_{n} = n^{2} + n$"}],"answer":"A","explanation":"Each term is one more than a perfect $square: 2 = 1^{2} + 1, 5 = 2^{2} + 1, 10 = 3^{2} + 1, 17 = 4^{2} + 1$, so $a_{n} = n^{2} + 1$.","explanationZh":"每一项都比完全平方数大 $1：2 = 1^{2} + 1，5 = 2^{2} + 1，10 = 3^{2} + 1，故 a_{n} = n^{2} + 1。$","tags":["easy","general-term","pattern-recognition"],"status":"published"},{"id":"math-sequence-basics-008","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":1,"stem":"For $a_{n} = (−1)^{n}$, the terms $a_{1}, a_{2}, a_{3}$ are","stemZh":"$若 a_{n} = (−1)^{n}，则 a_{1}, a_{2}, a_{3}$ 分别为","options":[{"id":"A","text":"1, 1, 1"},{"id":"B","text":"1, −1, 1"},{"id":"C","text":"−1, 1, −1"},{"id":"D","text":"−1, −1, −1"}],"answer":"C","explanation":"$(−1)^{1} = −1, (−1)^{2} = 1, (−1)^{3} = −1$. The option 1, −1, 1 comes from using $(−1)^{n-1}$ instead of $(−1)^{n}$.","explanationZh":"$(−1)^{1} = −1，(−1)^{2} = 1，(−1)^{3} = −1。选项 1, −1, 1$ 是把通项误作 $(−1)^{n-1}$ 的结果。","tags":["easy","oscillating-sequence","sign"],"status":"published"},{"id":"math-sequence-basics-009","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":2,"stem":"The sequence 1, 1/2, 1/3, 1/4, … is","stemZh":"数列 1, 1/2, 1/3, 1/4, … 是","options":[{"id":"A","text":"increasing"},{"id":"B","text":"constant"},{"id":"C","text":"decreasing"},{"id":"D","text":"oscillating"}],"answer":"C","explanation":"$a_{n} = 1/n$ and $a_{n+1} = 1/(n+1) < 1/n = a_{n}$ for every n, so the sequence is strictly decreasing.","explanationZh":"$a_{n} = 1/n，a_{n+1} = 1/(n+1) < 1/n = a_{n}，所以该数列单调递减。$","tags":["easy","monotonicity","classification"],"status":"published"},{"id":"math-sequence-basics-010","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":2,"stem":"If $a_{n} = 2^{n} − 1$, then $a_{5} − a_{3} =$","stemZh":"$若 a_{n} = 2^{n} − 1，则 a_{5} − a_{3} =$","options":[{"id":"A","text":"6"},{"id":"B","text":"24"},{"id":"C","text":"28"},{"id":"D","text":"30"}],"answer":"B","explanation":"$a_{5} = 2^{5} − 1 = 31$ and $a_{3} = 2^{3} − 1 = 7$, so $a_{5} − a_{3} = 31 − 7 = 24$.","explanationZh":"$a_{5} = 2^{5} − 1 = 31，a_{3} = 2^{3} − 1 = 7，故 a_{5} − a_{3} = 31 − 7 = 24。$","tags":["easy","general-term","evaluation"],"status":"published"},{"id":"math-sequence-basics-011","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"Let $S_{n}$ be the sum of the first n terms of a sequence and suppose $S_{n} = n^{2} + 1$. For $n ≥ 2, a_{n} =$","stemZh":"$设 S_{n}$ 为数列前 n 项和，且 $S_{n} = n^{2} + 1，则当 n ≥ 2 时 a_{n} =$","options":[{"id":"A","text":"2n + 1"},{"id":"B","text":"$n^{2} − 1$"},{"id":"C","text":"2n + 2"},{"id":"D","text":"2n − 1"}],"answer":"D","explanation":"For $n ≥ 2, a_{n} = S_{n} − S_{n}-_{1} = (n^{2} + 1) − ((n − 1)^{2} + 1) = 2n − 1$. Note that $a_{1} = S_{1} = 2$ does not fit this formula, which is why the question restricts to $n ≥ 2$.","explanationZh":"$当 n ≥ 2 时，a_{n} = S_{n} − S_{n}-_{1} = (n^{2} + 1) − ((n − 1)^{2} + 1) = 2n − 1。注意 a_{1} = S_{1} = 2$ 不满足该式，所以题目限定 $n ≥ 2。$","tags":["medium","sum-to-term","a_{n}-equals-S_{n}-minus-S_{n}-₁"],"status":"published"},{"id":"math-sequence-basics-012","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"$S_{n} = 3n^{2} − n$ is the sum of the first n terms of a sequence. Then $a_{5} =$","stemZh":"数列前 n 项和 $S_{n} = 3n^{2} −$ n，则 $a_{5} =$","options":[{"id":"A","text":"26"},{"id":"B","text":"24"},{"id":"C","text":"28"},{"id":"D","text":"30"}],"answer":"A","explanation":"$a_{5} = S_{5} − S_{4} = (3·25 − 5) − (3·16 − 4) = 70 − 44 = 26. (Equivalently a_{n} = 6n − 4\\text{ for }n ≥ 2.)$","explanationZh":"$a_{5} = S_{5} − S_{4} = (75 − 5) − (48 − 4) = 70 − 44 = 26。（也可先得 n ≥ 2 时 a_{n} = 6n − 4。）$","tags":["medium","sum-to-term","computation"],"status":"published"},{"id":"math-sequence-basics-013","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"A sequence satisfies $a_{1} = 1$ and $a_{n+1} = 2a_{n} + 1$. Then $a_{4} =$","stemZh":"数列满足 $a_{1} = 1，a_{n+1} = 2a_{n} + 1，则 a_{4} =$","options":[{"id":"A","text":"7"},{"id":"B","text":"15"},{"id":"C","text":"16"},{"id":"D","text":"31"}],"answer":"B","explanation":"$a_{2} = 2·1 + 1 = 3, a_{3} = 2·3 + 1 = 7, a_{4} = 2·7 + 1 = 15$.","explanationZh":"$a_{2} = 3，a_{3} = 7，a_{4} = 15。$","tags":["medium","recurrence","iteration"],"status":"published"},{"id":"math-sequence-basics-014","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"Compute $1/(1·2) + 1/(2·3) + … + 1/(9·10)$.","stemZh":"计算 $1/(1·2) + 1/(2·3) + … + 1/(9·10)。$","options":[{"id":"A","text":"1"},{"id":"B","text":"8/9"},{"id":"C","text":"9/10"},{"id":"D","text":"1/10"}],"answer":"C","explanation":"Since $1/(n(n+1)) = 1/n − 1/(n+1)$, the sum $telescopes: (1 − 1/2) + (1/2 − 1/3) + … + (1/9 − 1/10) = 1 − 1/10 = 9/10$.","explanationZh":"因为 $1/(n(n+1)) = 1/n − 1/(n+1)，原式 = (1 − 1/2) + (1/2 − 1/3) + … + (1/9 − 1/10) = 1 − 1/10 = 9/10。$","tags":["medium","telescoping","summation"],"status":"published"},{"id":"math-sequence-basics-015","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"The sequence with general term $a_{n} = n/(n+1)$ is","stemZh":"通项为 $a_{n} = n/(n+1)$ 的数列是","options":[{"id":"A","text":"increasing"},{"id":"B","text":"decreasing"},{"id":"C","text":"constant"},{"id":"D","text":"oscillating"}],"answer":"A","explanation":"$a_{n} = 1 − 1/(n+1)$. As $n grows, 1/(n+1) decreases$, so $a_{n}$ increases $(1/2, 2/3, 3/4, …)$.","explanationZh":"$a_{n} = 1 − 1/(n+1)，随着 n 增大，1/(n+1)$ 减小，所以 $a_{n} 递增（1/2, 2/3, 3/4$, …）。","tags":["medium","monotonicity","fraction"],"status":"published"},{"id":"math-sequence-basics-016","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"A sequence satisfies $a_{1} = 2$ and $a_{n+1} = a_{n}/(a_{n} + 1)$. Then $a_{3} =$","stemZh":"数列满足 $a_{1} = 2，a_{n+1} = a_{n}/(a_{n} + 1)，则 a_{3} =$","options":[{"id":"A","text":"2/3"},{"id":"B","text":"3/5"},{"id":"C","text":"1/2"},{"id":"D","text":"2/5"}],"answer":"D","explanation":"$a_{2} = 2/(2 + 1) = 2/3$, then $a_{3} = (2/3)/(2/3 + 1) = (2/3)/(5/3) = 2/5$.","explanationZh":"$a_{2} = 2/3，a_{3} = (2/3)/(5/3) = 2/5。$","tags":["medium","recurrence","fractions"],"status":"published"},{"id":"math-sequence-basics-017","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"$S_{n} = 2^{n} − 1$ is the sum of the first n terms of a sequence. Then $a_{3} =$","stemZh":"数列前 n 项和 $S_{n} = 2^{n} − 1，则 a_{3} =$","options":[{"id":"A","text":"3"},{"id":"B","text":"4"},{"id":"C","text":"7"},{"id":"D","text":"8"}],"answer":"B","explanation":"$a_{3} = S_{3} − S_{2} = (2^{3} − 1) − (2^{2} − 1) = 7 − 3 = 4. (7\\text{ is }S_{3},\\text{ not }a_{3}.)$","explanationZh":"$a_{3} = S_{3} − S_{2} = 7 − 3 = 4。（7 是 S_{3}$ 而不是 $a_{3}。）$","tags":["medium","sum-to-term","computation"],"status":"published"},{"id":"math-sequence-basics-018","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"Which recurrence relation produces the sequence 1, 2, 4, 8, …?","stemZh":"下列哪个递推关系可以得到数列 1, 2, 4, 8, …？","options":[{"id":"A","text":"$a_{1} = 1, a_{n+1} = a_{n} + 2$"},{"id":"B","text":"$a_{1} = 2, a_{n+1} = 2a_{n}$"},{"id":"C","text":"$a_{1} = 1, a_{n+1} = 2a_{n}$"},{"id":"D","text":"$a_{1} = 1, a_{n+1} = a_{n}^{2}$"}],"answer":"C","explanation":"Starting from $a_{1} = 1$ and doubling each time gives 1, 2, 4, 8, …. With $a_{1} = 2$ we would get 2, 4, 8, 16, … instead.","explanationZh":"$从 a_{1} = 1$ 开始每次乘以 2 得 1, 2, 4, 8, …；若 $a_{1} = 2$ 则得到 2, 4, 8, 16, …。","tags":["medium","recurrence","initial-term"],"status":"published"},{"id":"math-sequence-basics-019","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"Compute $1/(1·3) + 1/(3·5) + … + 1/(9·11)$.","stemZh":"计算 $1/(1·3) + 1/(3·5) + … + 1/(9·11)。$","options":[{"id":"A","text":"1/11"},{"id":"B","text":"10/11"},{"id":"C","text":"1/2"},{"id":"D","text":"5/11"}],"answer":"D","explanation":"Because $1/((2k−1)(2k+1)) = (1/2)(1/(2k−1) − 1/(2k+1))$, the sum is $(1/2)(1 − 1/11) = 5/11$. Forgetting the factor 1/2 gives 10/11.","explanationZh":"因为 $1/((2k−1)(2k+1)) = (1/2)(1/(2k−1) − 1/(2k+1))，原式 = (1/2)(1 − 1/11) = 5/11。漏掉系数 1/2$ 就会得到 10/11。","tags":["medium","telescoping","summation"],"status":"published"},{"id":"math-sequence-basics-020","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"The sequence with general term $a_{n} = n^{2} − 5n + 6$ has a term equal to 0 when $n =$","stemZh":"通项为 $a_{n} = n^{2} − 5n + 6$ 的数列中，等于 0 的项对应的 n 为","options":[{"id":"A","text":"2 or 3"},{"id":"B","text":"1 or 6"},{"id":"C","text":"2"},{"id":"D","text":"3"}],"answer":"A","explanation":"$n^{2} − 5n + 6 = (n − 2)(n − 3) = 0$ gives $n = 2$ or $n = 3$, so both $a_{2}$ and $a_{3}$ are $0. (1$ and 6 are the factors of the constant term, not the roots.)","explanationZh":"$n^{2} − 5n + 6 = (n − 2)(n − 3) = 0，得 n = 2 或 n = 3，即 a_{2} = a_{3} = 0。（1 和 6$ 是常数项的因数，不是方程的根。）","tags":["medium","general-term","quadratic"],"status":"published"},{"id":"math-sequence-basics-021","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":4,"stem":"Let $S_{n} = n^{2} + 2n$ be the sum of the first n terms of a sequence. Which statement is correct?","stemZh":"设数列前 n 项和 $S_{n} = n^{2} + 2n，下列说法正确的是？$","options":[{"id":"A","text":"$a_{n} = 2n + 1$ only for $n ≥ 2; a_{1}$ does not fit the formula"},{"id":"B","text":"$a_{n} = 2n + 1$ for every $n ∈ ℕ⁺$"},{"id":"C","text":"$a_{n} = 2n + 2$ for every $n ∈ ℕ⁺$"},{"id":"D","text":"$a_{n} = n^{2} + 2n$ for every $n ∈ ℕ⁺$"}],"answer":"B","explanation":"For $n ≥ 2, a_{n} = S_{n} − S_{n}-_{1} = (n^{2} + 2n) − ((n − 1)^{2} + 2(n − 1)) = 2n + 1$, and $a_{1} = S_{1} = 3 = 2·1 + 1$ also fits. Because $S_{n}$ has no constant term, the formula holds for all n here.","explanationZh":"$n ≥ 2 时 a_{n} = S_{n} − S_{n}-_{1} = 2n + 1，且 a_{1} = S_{1} = 3 = 2·1 + 1$ 也符合。由于 $S_{n}$ 没有常数项，该式对所有 n 成立。","tags":["hard","sum-to-term","n-equals-1-case"],"status":"published"},{"id":"math-sequence-basics-022","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":4,"stem":"A sequence satisfies $a_{1} = 1$ and $a_{n+1} = a_{n} + 2n$. Then $a_{5} =$","stemZh":"数列满足 $a_{1} = 1，a_{n+1} = a_{n} + 2n，则 a_{5} =$","options":[{"id":"A","text":"11"},{"id":"B","text":"19"},{"id":"C","text":"21"},{"id":"D","text":"25"}],"answer":"C","explanation":"$a_{2} = 1 + 2·1 = 3, a_{3} = 3 + 2·2 = 7, a_{4} = 7 + 2·3 = 13, a_{5} = 13 + 2·4 = 21$. The added amount depends on n, so the sequence is not arithmetic.","explanationZh":"$a_{2} = 3，a_{3} = 7，a_{4} = 13，a_{5} = 21。每次加上的量与 n$ 有关，所以该数列不是等差数列。","tags":["hard","recurrence","non-arithmetic"],"status":"published"},{"id":"math-sequence-basics-023","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":4,"stem":"Compute 1/2 + 1/6 + 1/12 + … + 1/110.","stemZh":"计算 1/2 + 1/6 + 1/12 + … + 1/110。","options":[{"id":"A","text":"10/11"},{"id":"B","text":"9/10"},{"id":"C","text":"11/12"},{"id":"D","text":"1/11"}],"answer":"A","explanation":"$1\\cdot 2$, $2\\cdot 3$, …, $10\\cdot 11$, so the sum telescopes to $1 − 1/11 = 10/11$. ($9/10$ would stop at $1/90$.)","explanationZh":"分母依次为 $1·2, 2·3, …, 10·11，故原式 = 1 − 1/11 = 10/11。（若只加到 1/90，结果为 9/10。）$","tags":["hard","telescoping","summation"],"status":"published"},{"id":"math-sequence-basics-024","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":4,"stem":"Let $a_{n} = 1/(n(n+1))$ and let $S_{n}$ be the sum of the first n terms. The least n with $S_{n} > 10/11$ is","stemZh":"$设 a_{n} = 1/(n(n+1))，S_{n}$ 为前 n 项和，则使 $S_{n} > 10/11$ 成立的最小 n 为","options":[{"id":"A","text":"9"},{"id":"B","text":"10"},{"id":"C","text":"12"},{"id":"D","text":"11"}],"answer":"D","explanation":"$S_{n} = 1 − 1/(n+1)$. Then $S_{n} > 10/11$ means $1/(n+1) < 1/11$, i.e. $n + 1 > 11$, so $n ≥ 11$. Note $S_{10} = 10/11 exactly$, so $n = 10$ is not enough for a strict inequality.","explanationZh":"$S_{n} = 1 − 1/(n+1)。由 1 − 1/(n+1) > 10/11 得 1/(n+1) < 1/11，即 n + 1 > 11，所以 n ≥ 11。注意 S_{10}$ 恰好等于 $10/11，故 n = 10$ 不满足严格大于。","tags":["hard","telescoping","inequality"],"status":"published"},{"id":"math-sequence-basics-025","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":4,"stem":"A sequence satisfies $a_{1} = 3$ and $a_{n+1} − a_{n} = 2^{n}$. Then $a_{4} =$","stemZh":"数列满足 $a_{1} = 3，a_{n+1} − a_{n} = 2^{n}，则 a_{4} =$","options":[{"id":"A","text":"15"},{"id":"B","text":"17"},{"id":"C","text":"19"},{"id":"D","text":"11"}],"answer":"B","explanation":"$a_{2} = 3 + 2^{1} = 5, a_{3} = 5 + 2^{2} = 9, a_{4} = 9 + 2^{3} = 17$.","explanationZh":"$a_{2} = 3 + 2 = 5，a_{3} = 5 + 4 = 9，a_{4} = 9 + 8 = 17。$","tags":["hard","recurrence","indexing"],"status":"published"},{"id":"math-sequence-basics-026","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":5,"stem":"For the sequence $a_{n} = (n − 3)/(n + 1)$, which statement is correct?","stemZh":"对于数列 $a_{n} = (n − 3)/(n + 1)，下列说法正确的是？$","options":[{"id":"A","text":"It is strictly decreasing for all $n ∈ ℕ⁺$"},{"id":"B","text":"It decreases while $n ≤ 3$ and increases afterwards"},{"id":"C","text":"It is strictly increasing for all $n ∈ ℕ⁺$"},{"id":"D","text":"It increases first and then decreases"}],"answer":"C","explanation":"$a_{n} = (n + 1 − 4)/(n + 1) = 1 − 4/(n + 1)$. As $n increases, 4/(n + 1) decreases$, so $a_{n}$ increases for every $n: a_{1} = −1, a_{2} = −1/3, a_{3} = 0, a_{4} = 1/5$, …. The sign change at $n = 3$ does not change the monotonicity.","explanationZh":"$a_{n} = 1 − 4/(n + 1)。n$ 增大时 $4/(n + 1)$ 减小，故 $a_{n}$ 对一切 $n 递增：a_{1} = −1，a_{2} = −1/3，a_{3} = 0，a_{4} = 1/5$, …。n $= 3$ 处的变号并不改变单调性。","tags":["hard","monotonicity","algebraic-rewrite"],"status":"published"},{"id":"math-sequence-basics-027","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":5,"stem":"$S_{n} = n^{2} + n + 1$ is the sum of the first n terms of a sequence. Then $a_{2} =$","stemZh":"数列前 n 项和 $S_{n} = n^{2} + n + 1，则 a_{2} =$","options":[{"id":"A","text":"4"},{"id":"B","text":"3"},{"id":"C","text":"5"},{"id":"D","text":"2"}],"answer":"A","explanation":"$a_{2} = S_{2} − S_{1} = (4 + 2 + 1) − (1 + 1 + 1) = 7 − 3 = 4$. Because $S_{n}$ has a non-zero constant term, the sequence is not arithmetic and $a_{n} ≠ S_{n} − S_{n}-_{1}$ at $n = 1$; here $a_{1} = 3$.","explanationZh":"$a_{2} = S_{2} − S_{1} = 7 − 3 = 4。由于 S_{n}$ 含有非零常数项，该数列不是等差数列，且 $n = 1$ 时不能用 $a_{n} = S_{n} − S_{n}-_{1}，此时 a_{1} = S_{1} = 3。$","tags":["hard","sum-to-term","constant-term-trap"],"status":"published"},{"id":"math-sequence-basics-028","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":5,"stem":"Compute $(2 + 4 + 6 + … + 100) − (1 + 3 + 5 + … + 99)$.","stemZh":"计算 $(2 + 4 + 6 + … + 100) − (1 + 3 + 5 + … + 99)。$","options":[{"id":"A","text":"100"},{"id":"B","text":"0"},{"id":"C","text":"25"},{"id":"D","text":"50"}],"answer":"D","explanation":"Pair the $terms: (2 − 1) + (4 − 3) + … + (100 − 99)$. There are 50 pairs and each pair equals 1, so the difference is 50.","explanationZh":"$配对相减：(2 − 1) + (4 − 3) + … + (100 − 99)，共 50$ 对，每对等于 1，故差为 50。","tags":["hard","pairing","summation"],"status":"published"},{"id":"math-sequence-basics-029","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":5,"stem":"A sequence satisfies $a_{1} = 2$ and $a_{n+1} = a_{n} + 3$. What is the least n for which $S_{n} > 100$?","stemZh":"数列满足 $a_{1} = 2，a_{n+1} = a_{n} + 3，则使 S_{n} > 100$ 成立的最小 n 为","options":[{"id":"A","text":"8"},{"id":"B","text":"9"},{"id":"C","text":"10"},{"id":"D","text":"7"}],"answer":"B","explanation":"$a_{n} = 3n − 1$ and $S_{n} = n(2 + 3n − 1)/2 = n(3n + 1)/2. S_{8} = 8·25/2 = 100$ and $S_{9} = 9·28/2 = 126$, so the least n with $S_{n}$ strictly greater than 100 is 9.","explanationZh":"$a_{n} = 3n − 1，S_{n} = n(3n + 1)/2。S_{8} = 100，S_{9} = 126，因此满足 S_{n} > 100$ 的最小 n 是 9。","tags":["hard","strict-inequality","summation"],"status":"published"},{"id":"math-sequence-basics-030","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":5,"stem":"$S_{n} = 3^{n} − 2$ is the sum of the first n terms of a sequence. For $n ≥ 2, a_{n} =$","stemZh":"数列前 n 项和 $S_{n} = 3^{n} − 2，则当 n ≥ 2 时 a_{n} =$","options":[{"id":"A","text":"$2·3^{n}$"},{"id":"B","text":"$3^{n-1}$"},{"id":"C","text":"$2·3^{n-1}$"},{"id":"D","text":"$3^{n} − 2$"}],"answer":"C","explanation":"For $n ≥ 2, a_{n} = S_{n} − S_{n}-_{1} = (3^{n} − 2) − (3^{n-1} − 2) = 3^{n} − 3^{n-1} = 2·3^{n-1}$. Note $a_{1} = S_{1} = 1$, which does not fit this formula — that is why $n = 1$ is excluded.","explanationZh":"$n ≥ 2 时，a_{n} = S_{n} − S_{n}-_{1} = (3^{n} − 2) − (3^{n-1} − 2) = 3^{n} − 3^{n-1} = 2·3^{n-1}。注意 a_{1} = S_{1} = 1$ 不满足该式，所以需排除 $n = 1。$","tags":["hard","sum-to-term","exponential"],"status":"published"},{"id":"math-arithmetic-sequence-001","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":1,"stem":"In the arithmetic sequence 3, 7, 11, …, the common difference d is","stemZh":"等差数列 3, 7, 11, … 的公差 d 为","options":[{"id":"A","text":"3"},{"id":"B","text":"4"},{"id":"C","text":"7"},{"id":"D","text":"11"}],"answer":"B","explanation":"$d = 7 − 3 = 11 − 7 = 4$, the constant difference between consecutive terms.","explanationZh":"$d = 7 − 3 = 11 − 7 = 4，即相邻两项的差。$","tags":["easy","common-difference","definition"],"status":"published"},{"id":"math-arithmetic-sequence-002","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":1,"stem":"The 10th term of the arithmetic sequence with $a_{1} = 2$ and $d = 3$ is","stemZh":"首项 $a_{1} = 2、公差 d = 3$ 的等差数列的第 10 项为","options":[{"id":"A","text":"30"},{"id":"B","text":"32"},{"id":"C","text":"27"},{"id":"D","text":"29"}],"answer":"D","explanation":"$a_{10} = a_{1} + 9d = 2 + 9·3 = 29. (Using a_{1} + 10d\\text{ would give }32.)$","explanationZh":"$a_{10} = a_{1} + 9d = 2 + 27 = 29。（若误用 a_{1} + 10d$ 会得到 32。）","tags":["easy","general-term","indexing"],"status":"published"},{"id":"math-arithmetic-sequence-003","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":1,"stem":"Which of the following is an arithmetic sequence?","stemZh":"下列哪一项是等差数列？","options":[{"id":"A","text":"1, 4, 7, 10"},{"id":"B","text":"1, 2, 4, 8"},{"id":"C","text":"1, 4, 9, 16"},{"id":"D","text":"1, 1/2, 1/3, 1/4"}],"answer":"A","explanation":"Only 1, 4, 7, 10 has a constant difference $(d = 3)$. The others are geometric, squares and reciprocals.","explanationZh":"只有 1, 4, 7, 10 相邻两项的差恒定（d $= 3），其余分别是等比数列、平方数列和倒数数列。$","tags":["easy","recognition","definition"],"status":"published"},{"id":"math-arithmetic-sequence-004","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":2,"stem":"An arithmetic sequence has $a_{1} = 1, a_{5} = 9$. Its sum $S_{5}$ is","stemZh":"等差数列中 $a_{1} = 1，a_{5} = 9，则 S_{5} 为$","options":[{"id":"A","text":"20"},{"id":"B","text":"45"},{"id":"C","text":"25"},{"id":"D","text":"50"}],"answer":"C","explanation":"$S_{5} = 5(a_{1} + a_{5})/2 = 5(1 + 9)/2 = 25$.","explanationZh":"$S_{5} = 5(a_{1} + a_{5})/2 = 5 \\times 10/2 = 25。$","tags":["easy","sum-formula","computation"],"status":"published"},{"id":"math-arithmetic-sequence-005","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":2,"stem":"If 2, x, 8 form an arithmetic sequence, then $x =$","stemZh":"若 2, x, 8 成等差数列，则 $x =$","options":[{"id":"A","text":"4"},{"id":"B","text":"6"},{"id":"C","text":"5"},{"id":"D","text":"10"}],"answer":"C","explanation":"For three consecutive $terms, 2x = 2 + 8 = 10$, so $x = 5. (4\\text{ would come from }a\\text{ geometric sequence, where }x^{2} = 16.)$","explanationZh":"三数成等差，则 $2x = 2 + 8 = 10，x = 5。（4$ 是等比数列 $x^{2} = 16$ 的结果。）","tags":["easy","arithmetic-mean","computation"],"status":"published"},{"id":"math-arithmetic-sequence-006","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":1,"stem":"For the arithmetic sequence with general term $a_{n} = 3n − 1$, the common difference is","stemZh":"通项为 $a_{n} = 3n − 1$ 的等差数列，其公差为","options":[{"id":"A","text":"3"},{"id":"B","text":"−1"},{"id":"C","text":"2"},{"id":"D","text":"3n"}],"answer":"A","explanation":"$d = a_{n+1} − a_{n} = (3n + 2) − (3n − 1) = 3$, which is the coefficient of n.","explanationZh":"$d = a_{n+1} − a_{n} = 3，也就是 n$ 的系数。","tags":["easy","common-difference","general-term"],"status":"published"},{"id":"math-arithmetic-sequence-007","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":2,"stem":"Compute 1 + 2 + 3 + … + 100.","stemZh":"计算 1 + 2 + 3 + … + 100。","options":[{"id":"A","text":"5000"},{"id":"B","text":"4950"},{"id":"C","text":"5500"},{"id":"D","text":"5050"}],"answer":"D","explanation":"$S_{100} = 100(1 + 100)/2 = 5050$.","explanationZh":"$S_{100} = 100(1 + 100)/2 = 5050。$","tags":["easy","sum-formula","computation"],"status":"published"},{"id":"math-arithmetic-sequence-008","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":1,"stem":"In an arithmetic sequence $a_{3} = 7$ and $a_{5} = 11$. Then $d =$","stemZh":"等差数列中 $a_{3} = 7，a_{5} = 11，则 d =$","options":[{"id":"A","text":"4"},{"id":"B","text":"2"},{"id":"C","text":"1"},{"id":"D","text":"3"}],"answer":"B","explanation":"$d = (a_{5} − a_{3})/(5 − 3) = (11 − 7)/2 = 2$. There are two steps from $a_{3}$ to $a_{5}$, so divide by 2, not by 1.","explanationZh":"$d = (a_{5} − a_{3})/(5 − 3) = 4/2 = 2。从 a_{3} 到 a_{5}$ 相隔 2 个公差，要除以 2。","tags":["easy","common-difference","indexing"],"status":"published"},{"id":"math-arithmetic-sequence-009","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":2,"stem":"An arithmetic sequence has $d = 0$ and $a_{1} = 5$. Then $a_{10} =$","stemZh":"等差数列的公差 $d = 0，首项 a_{1} = 5，则 a_{10} =$","options":[{"id":"A","text":"0"},{"id":"B","text":"50"},{"id":"C","text":"15"},{"id":"D","text":"5"}],"answer":"D","explanation":"With $d = 0$ every term equals $a_{1} = 5$, so $a_{10} = 5. A$ constant sequence is a special arithmetic sequence.","explanationZh":"$d = 0$ 时每一项都等于 $a_{1} = 5，故 a_{10} = 5。常数列是公差为 0$ 的等差数列。","tags":["easy","constant-sequence","edge-case"],"status":"published"},{"id":"math-arithmetic-sequence-010","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":2,"stem":"An arithmetic sequence has $a_{1} = 10$ and $d = −2$. Its first four terms are","stemZh":"等差数列 $a_{1} = 10，d = −2，则它的前四项为$","options":[{"id":"A","text":"10, 8, 6, 4"},{"id":"B","text":"10, 12, 14, 16"},{"id":"C","text":"10, 9, 8, 7"},{"id":"D","text":"8, 6, 4, 2"}],"answer":"A","explanation":"Each term is 2 less than the previous one: 10, 8, 6, 4. A negative d means the sequence decreases.","explanationZh":"每一项比前一项少 $2：10, 8, 6, 4。d < 0$ 时数列递减。","tags":["easy","listing-terms","negative-d"],"status":"published"},{"id":"math-arithmetic-sequence-011","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"An arithmetic sequence has $a_{1} = 2$ and $d = 4$. If $S_{n} = 200$, then $n =$","stemZh":"等差数列 $a_{1} = 2，d = 4，若 S_{n} = 200，则 n =$","options":[{"id":"A","text":"8"},{"id":"B","text":"9"},{"id":"C","text":"10"},{"id":"D","text":"12"}],"answer":"C","explanation":"$S_{n} = na_{1} + n(n−1)d/2 = 2n + 2n(n − 1) = 2n^{2}$. Setting $2n^{2} = 200$ gives $n^{2} = 100$, so $n = 10$.","explanationZh":"$S_{n} = na_{1} + n(n−1)d/2 = 2n + 2n(n − 1) = 2n^{2}。由 2n^{2} = 200 得 n^{2} = 100，故 n = 10。$","tags":["medium","sum-formula","solve-for-n"],"status":"published"},{"id":"math-arithmetic-sequence-012","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"The sum of the first n terms of an arithmetic sequence is $S_{n} = 2n^{2} −$ n. Its first term and common difference are","stemZh":"等差数列前 n 项和 $S_{n} = 2n^{2} −$ n，则它的首项和公差为","options":[{"id":"A","text":"$a_{1} = 1, d = 4$"},{"id":"B","text":"$a_{1} = 2, d = 4$"},{"id":"C","text":"$a_{1} = 1, d = 2$"},{"id":"D","text":"$a_{1} = 2, d = 2$"}],"answer":"A","explanation":"$a_{1} = S_{1} = 2 − 1 = 1; a_{2} = S_{2} − S_{1} = 6 − 1 = 5$, so $d = a_{2} − a_{1} = 4. Check: S_{n} = n·1 + n(n−1)·4/2 = 2n^{2} −$ n.","explanationZh":"$a_{1} = S_{1} = 1，a_{2} = S_{2} − S_{1} = 6 − 1 = 5，故 d = 4。验证：S_{n} = n + 2n(n − 1) = 2n^{2} −$ n。","tags":["medium","sum-as-quadratic","find-a_{1}-and-d"],"status":"published"},{"id":"math-arithmetic-sequence-013","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"In an arithmetic sequence $a_{2} + a_{8} = 20$. Then $a_{5} =$","stemZh":"等差数列中 $a_{2} + a_{8} = 20，则 a_{5} =$","options":[{"id":"A","text":"20"},{"id":"B","text":"5"},{"id":"C","text":"40"},{"id":"D","text":"10"}],"answer":"D","explanation":"Since $2 + 8 = 5 + 5$, the equidistant-term property gives $a_{2} + a_{8} = 2a_{5}$, so $a_{5} = 10$.","explanationZh":"因为 $2 + 8 = 5 + 5，由等距离项性质 a_{2} + a_{8} = 2a_{5}，得 a_{5} = 10。$","tags":["medium","equidistant-terms","property"],"status":"published"},{"id":"math-arithmetic-sequence-014","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"An arithmetic sequence has $a_{1} = 20$ and $d = −3$. The largest n with $a_{n} > 0$ is","stemZh":"等差数列 $a_{1} = 20，d = −3，则使 a_{n} > 0$ 成立的最大 n 为","options":[{"id":"A","text":"6"},{"id":"B","text":"7"},{"id":"C","text":"8"},{"id":"D","text":"9"}],"answer":"B","explanation":"$a_{n} = 20 − 3(n − 1) = 23 − 3n$. Then $23 − 3n > 0$ gives $n < 23/3 ≈ 7.67$, so the largest integer n is $7 (a_{7} = 2, a_{8} = −1)$.","explanationZh":"$a_{n} = 23 − 3n，由 23 − 3n > 0 得 n < 23/3 ≈ 7.67，故最大整数 n = 7（a_{7} = 2，a_{8} = −1）。$","tags":["medium","general-term","inequality"],"status":"published"},{"id":"math-arithmetic-sequence-015","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"The sum of the first n terms of an arithmetic sequence is $S_{n} = n^{2} + 3n$. Then $a_{4} =$","stemZh":"等差数列前 n 项和 $S_{n} = n^{2} + 3n，则 a_{4} =$","options":[{"id":"A","text":"8"},{"id":"B","text":"10"},{"id":"C","text":"12"},{"id":"D","text":"14"}],"answer":"B","explanation":"$a_{4} = S_{4} − S_{3} = (16 + 12) − (9 + 9) = 28 − 18 = 10$. Equivalently $a_{n} = 2n + 2$ for all n here.","explanationZh":"$a_{4} = S_{4} − S_{3} = 28 − 18 = 10。（也可先得 a_{n} = 2n + 2。）$","tags":["medium","sum-to-term","computation"],"status":"published"},{"id":"math-arithmetic-sequence-016","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"In an arithmetic sequence $a_{5} = 12$ and $a_{10} = 27$. Then $a_{15} =$","stemZh":"等差数列中 $a_{5} = 12，a_{10} = 27，则 a_{15} =$","options":[{"id":"A","text":"37"},{"id":"B","text":"40"},{"id":"C","text":"42"},{"id":"D","text":"45"}],"answer":"C","explanation":"$d = (27 − 12)/(10 − 5) = 3$, so $a_{15} = a_{10} + 5d = 27 + 15 = 42. (Equivalently a_{5}, a_{10}, a_{15}\\text{ are themselves in arithmetic }progression.)$","explanationZh":"$d = (27 − 12)/5 = 3，故 a_{15} = a_{10} + 5d = 27 + 15 = 42。（a_{5}、a_{10}、a_{15}$ 也成等差数列。）","tags":["medium","general-term","property"],"status":"published"},{"id":"math-arithmetic-sequence-017","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"Which expression gives $S_{n}$ for an arithmetic sequence with first term $a_{1}$ and common difference d?","stemZh":"首项为 $a_{1}、公差为 d$ 的等差数列，其前 n 项和 $S_{n}$ 的表达式是","options":[{"id":"A","text":"$na_{1} + n(n−1)d/2$"},{"id":"B","text":"$na_{1} + n(n+1)d/2$"},{"id":"C","text":"$n(a_{1} + a_{n})d/2$"},{"id":"D","text":"$n(a_{1} + d)/2$"}],"answer":"A","explanation":"$S_{n} = n(a_{1} + a_{n})/2 = n(a_{1} + a_{1} + (n − 1)d)/2 = na_{1} + n(n−1)d/2$.","explanationZh":"$S_{n} = n(a_{1} + a_{n})/2 = n(2a_{1} + (n − 1)d)/2 = na_{1} + n(n−1)d/2。$","tags":["medium","sum-formula","formula-recall"],"status":"published"},{"id":"math-arithmetic-sequence-018","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"Three consecutive terms of an arithmetic sequence satisfy $a_{1} + a_{2} + a_{3} = 12$ and $a_{1}a_{2}a_{3} = 48$. Then $a_{2} =$","stemZh":"等差数列连续三项满足 $a_{1} + a_{2} + a_{3} = 12，a_{1}a_{2}a_{3} = 48，则 a_{2} =$","options":[{"id":"A","text":"3"},{"id":"B","text":"6"},{"id":"C","text":"12"},{"id":"D","text":"4"}],"answer":"D","explanation":"By the arithmetic mean property $a_{1} + a_{3} = 2a_{2}$, so $3a_{2} = 12$ and $a_{2} = 4$. The product condition is consistent $(a_{1} = 2, a_{3} = 6)$ but is not needed.","explanationZh":"由等差中项性质 $a_{1} + a_{3} = 2a_{2}，故 3a_{2} = 12，a_{2} = 4。积的条件与之相符（a_{1} = 2，a_{3} = 6），但并非必需。$","tags":["medium","arithmetic-mean","property"],"status":"published"},{"id":"math-arithmetic-sequence-019","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"An arithmetic sequence has $a_{1} = −5$ and $d = 2$. Then $S_{10} =$","stemZh":"等差数列 $a_{1} = −5，d = 2，则 S_{10} =$","options":[{"id":"A","text":"30"},{"id":"B","text":"35"},{"id":"C","text":"40"},{"id":"D","text":"50"}],"answer":"C","explanation":"$S_{10} = 10a_{1} + 10·9·d/2 = −50 + 90 = 40$.","explanationZh":"$S_{10} = 10 \\times (−5) + 45 \\times 2 = −50 + 90 = 40。$","tags":["medium","sum-formula","negative-a_{1}"],"status":"published"},{"id":"math-arithmetic-sequence-020","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"An arithmetic sequence satisfies $S_{5} = 25$ and $S_{10} = 100$. Then $S_{15} =$","stemZh":"等差数列满足 $S_{5} = 25，S_{10} = 100，则 S_{15} =$","options":[{"id":"A","text":"200"},{"id":"B","text":"225"},{"id":"C","text":"250"},{"id":"D","text":"175"}],"answer":"B","explanation":"$S_{5}, S_{10} − S_{5}, S_{15} − S_{10}$ are also in arithmetic progression: 25, 75, 125. Hence $S_{15} = 100 + 125 = 225$.","explanationZh":"$S_{5}、S_{10} − S_{5}、S_{15} − S_{10} 也成等差数列：25, 75, 125，故 S_{15} = 100 + 125 = 225。$","tags":["medium","block-sums","property"],"status":"published"},{"id":"math-arithmetic-sequence-021","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":4,"stem":"An arithmetic sequence has $S_{n} = −n^{2} + 20n$. The maximum value of $S_{n}$ is","stemZh":"等差数列 $S_{n} = −n^{2} + 20n，则 S_{n}$ 的最大值为","options":[{"id":"A","text":"20"},{"id":"B","text":"90"},{"id":"C","text":"110"},{"id":"D","text":"100"}],"answer":"D","explanation":"$S_{n} = −(n − 10)^{2} + 100$, so the maximum is 100 at $n = 10$. Equivalently $a_{n} = −2n + 21$ stays positive up to $n = 10$ and turns negative at $n = 11$.","explanationZh":"$S_{n} = −(n − 10)^{2} + 100，故最大值为 100（n = 10）。也可由 a_{n} = −2n + 21$ 知前 10 项为正、第 11 项起为负。","tags":["hard","maximum-of-S_{n}","quadratic"],"status":"published"},{"id":"math-arithmetic-sequence-022","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":4,"stem":"An arithmetic sequence has $a_{1} = 21$ and $d = −3$. The value(s) of n at which $S_{n}$ is maximal:","stemZh":"等差数列 $a_{1} = 21，d = −3，则 S_{n}$ 取最大值时的 n 为","options":[{"id":"A","text":"$n = 7$ only"},{"id":"B","text":"$n = 7$ or $n = 8$"},{"id":"C","text":"$n = 8$ only"},{"id":"D","text":"$n = 9$"}],"answer":"B","explanation":"$a_{n} = 24 − 3n$, so $a_{7} = 3 > 0, a_{8} = 0$ and $a_{9} = −3 < 0$. Since $a_{8} = 0$ adds $nothing, S_{7} = S_{8} = 84$ is the maximum, attained at both $n = 7$ and $n = 8$.","explanationZh":"$a_{n} = 24 − 3n，a_{7} = 3 > 0，a_{8} = 0，a_{9} = −3 < 0。由于第 8$ 项为 $0，S_{7} = S_{8} = 84$ 同为最大值，故 $n = 7 或 8。$","tags":["hard","maximum-of-S_{n}","zero-term-trap"],"status":"published"},{"id":"math-arithmetic-sequence-023","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":5,"stem":"In a non-constant arithmetic sequence $S_{7} = S_{12}$. Which statement must be true?","stemZh":"在非恒等的等差数列中 $S_{7} = S_{12}，则下列一定正确的是$","options":[{"id":"A","text":"$S_{19} = 0$"},{"id":"B","text":"$S_{18} = 0$"},{"id":"C","text":"$a_{9} = 0$"},{"id":"D","text":"$a_{11} = 0$"}],"answer":"A","explanation":"Write $S_{n} = An^{2} +$ Bn with $A ≠ 0. S_{7} = S_{12}$ gives $19A + B = 0$, hence $S_{19} = 19(19A + B) = 0$. The other options give −18A, −2A and 2A, which are not zero.","explanationZh":"$设 S_{n} = An^{2} +$ Bn（A $≠ 0）。由 S_{7} = S_{12} 得 19A + B = 0，所以 S_{19} = 19(19A + B) = 0。其余各项分别为 −18A、−2A、2A，均不为 0。$","tags":["hard","sum-as-quadratic","symmetry"],"status":"published"},{"id":"math-arithmetic-sequence-024","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":4,"stem":"In an arithmetic sequence $a_{1} + a_{4} + a_{7} = 39$ and $a_{2} + a_{5} + a_{8} = 33$. Then $a_{3} + a_{6} + a_{9} =$","stemZh":"等差数列中 $a_{1} + a_{4} + a_{7} = 39，a_{2} + a_{5} + a_{8} = 33，则 a_{3} + a_{6} + a_{9} =$","options":[{"id":"A","text":"21"},{"id":"B","text":"24"},{"id":"C","text":"27"},{"id":"D","text":"30"}],"answer":"C","explanation":"Let $T_{1} = a_{1} + a_{4} + a_{7}, T_{2} = a_{2} + a_{5} + a_{8}, T_{3} = a_{3} + a_{6} + a_{9}$. Each is obtained from the previous by adding 3d, so $T_{1}, T_{2}, T_{3}$ are arithmetic: 39, 33, 27.","explanationZh":"$记 T_{1} = a_{1} + a_{4} + a_{7}，T_{2} = a_{2} + a_{5} + a_{8}，T_{3} = a_{3} + a_{6} + a_{9}，后一个比前一个多 3d，故 T_{1}、T_{2}、T_{3} 成等差数列：39, 33, 27。$","tags":["hard","block-sums","property"],"status":"published"},{"id":"math-arithmetic-sequence-025","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":5,"stem":"An arithmetic sequence has $a_{1} > 0, d < 0$ and $S_{5} = S_{11}$. For which n is $S_{n}$ maximal?","stemZh":"等差数列 $a_{1} > 0，d < 0 且 S_{5} = S_{11}，则 S_{n}$ 取最大值时 $n =$","options":[{"id":"A","text":"7"},{"id":"B","text":"9"},{"id":"C","text":"10"},{"id":"D","text":"8"}],"answer":"D","explanation":"With $S_{n} = An^{2} + Bn, S_{5} = S_{11}$ gives $16A + B = 0$ and $A = d/2 < 0$. Then $a_{n} = 2An + B − A = A(2n − 17)$, which is positive for $n ≤ 8$ and negative from $n = 9$. Hence the maximum is at $n = 8$.","explanationZh":"$设 S_{n} = An^{2} +$ Bn，由 $S_{5} = S_{11} 得 16A + B = 0，且 A = d/2 < 0。a_{n} = A(2n − 17)，当 n ≤ 8$ 时为正，n $≥ 9$ 时为负，故最大值在 $n = 8$ 处取得。","tags":["hard","maximum-of-S_{n}","quadratic"],"status":"published"},{"id":"math-arithmetic-sequence-026","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":5,"stem":"Two arithmetic sequences ${a_{n}}$ and ${b_{n}}$ have sums $S_{n}$ and $T_{n}$ with $S_{n}/T_{n} = (2n + 1)/(3n + 2)$. Then $a_{5}/b_{5} =$","stemZh":"两个等差数列 ${a_{n}}、{b_{n}}$ 的前 n 项和分别为 $S_{n}、T_{n}，且 S_{n}/T_{n} = (2n + 1)/(3n + 2)，则 a_{5}/b_{5} =$","options":[{"id":"A","text":"19/29"},{"id":"B","text":"11/17"},{"id":"C","text":"9/14"},{"id":"D","text":"3/5"}],"answer":"A","explanation":"$a_{5}/b_{5} = 2a_{5}/2b_{5} = (a_{1} + a_{9})/(b_{1} + b_{9}) = (2S_{9}/9)/(2T_{9}/9) = S_{9}/T_{9} = (2·9 + 1)/(3·9 + 2) = 19/29$. Using $n = 5$ instead of $n = 9$ gives the distractor 11/17.","explanationZh":"$a_{5}/b_{5} = (a_{1} + a_{9})/(b_{1} + b_{9}) = S_{9}/T_{9} = (2\\times9 + 1)/(3\\times9 + 2) = 19/29。若误取 n = 5$ 则得 11/17。","tags":["hard","sum-ratio","middle-term"],"status":"published"},{"id":"math-arithmetic-sequence-027","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":4,"stem":"In an arithmetic sequence $a_{1} + … + a_{10} = 100$ and $a_{11} + … + a_{20} = 300$. Then $d =$","stemZh":"等差数列中 $a_{1} + … + a_{10} = 100，a_{11} + … + a_{20} = 300，则 d =$","options":[{"id":"A","text":"1"},{"id":"B","text":"2"},{"id":"C","text":"3"},{"id":"D","text":"4"}],"answer":"B","explanation":"$S_{10} = 10a_{1} + 45d = 100$ and $S_{20} − S_{10} = 10a_{1} + 145d = 300$. Subtracting gives $100d = 200$, so $d = 2$.","explanationZh":"$S_{10} = 10a_{1} + 45d = 100，S_{20} − S_{10} = 10a_{1} + 145d = 300，两式相减得 100d = 200，d = 2。$","tags":["hard","block-sums","solve-for-d"],"status":"published"},{"id":"math-arithmetic-sequence-028","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":5,"stem":"An arithmetic sequence has $a_{1} = 13$ and $S_{3} = S_{11}$. The value of n at which $S_{n}$ is maximal is","stemZh":"等差数列 $a_{1} = 13 且 S_{3} = S_{11}，则 S_{n}$ 取最大值时 $n =$","options":[{"id":"A","text":"6"},{"id":"B","text":"8"},{"id":"C","text":"7"},{"id":"D","text":"9"}],"answer":"C","explanation":"$3a_{1} + 3d = 11a_{1} + 55d$ gives $8a_{1} + 52d = 0$, so with $a_{1} = 13$ we get $d = −2$. Then $a_{n} = 13 − 2(n − 1) = 15 − 2n: a_{7} = 1 > 0$ and $a_{8} = −1 < 0$, so the maximum is at $n = 7$.","explanationZh":"$由 3a_{1} + 3d = 11a_{1} + 55d 得 8a_{1} + 52d = 0，代入 a_{1} = 13 得 d = −2。a_{n} = 15 − 2n，a_{7} = 1 > 0，a_{8} = −1 < 0，故最大值在 n = 7。$","tags":["hard","maximum-of-S_{n}","two-conditions"],"status":"published"},{"id":"math-arithmetic-sequence-029","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":4,"stem":"An arithmetic sequence satisfies $S_{10} = 100$ and $S_{20} = 400$. Then $S_{30} =$","stemZh":"等差数列满足 $S_{10} = 100，S_{20} = 400，则 S_{30} =$","options":[{"id":"A","text":"900"},{"id":"B","text":"800"},{"id":"C","text":"700"},{"id":"D","text":"1000"}],"answer":"A","explanation":"$S_{10}, S_{20} − S_{10}, S_{30} − S_{20}$ are in arithmetic progression: 100, 300, 500, so $S_{30} = 400 + 500 = 900$.","explanationZh":"$S_{10}、S_{20} − S_{10}、S_{30} − S_{20} 成等差数列：100, 300, 500，故 S_{30} = 400 + 500 = 900。$","tags":["hard","block-sums","property"],"status":"published"},{"id":"math-arithmetic-sequence-030","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":5,"stem":"In an arithmetic sequence $a_{m} = n$ and $a_{n} = m$ with $m ≠$ n. Then $a_{m+n} =$","stemZh":"等差数列中 $a_{m} = n，a_{n} =$ m（m $≠$ n），则 $a_{m+n} =$","options":[{"id":"A","text":"m + n"},{"id":"B","text":"m − n"},{"id":"C","text":"1"},{"id":"D","text":"0"}],"answer":"D","explanation":"$d = (a_{n} − a_{m})/(n − m) = (m − n)/(n − m) = −1$, and $a_{1} = a_{m} − (m − 1)d = n + m − 1$. Thus $a_{m+n} = a_{1} + (m + n − 1)d = (m + n − 1) − (m + n − 1) = 0$.","explanationZh":"$d = (m − n)/(n − m) = −1，a_{1} = n + m − 1，故 a_{m+n} = a_{1} + (m + n − 1)d = 0。$","tags":["hard","general-term","symbolic"],"status":"published"},{"id":"math-geometric-sequence-001","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":1,"stem":"In the geometric sequence 2, 6, 18, …, the common ratio q is","stemZh":"等比数列 2, 6, 18, … 的公比 q 为","options":[{"id":"A","text":"3"},{"id":"B","text":"2"},{"id":"C","text":"6"},{"id":"D","text":"1/3"}],"answer":"A","explanation":"$q = 6/2 = 18/6 = 3$, the constant ratio of consecutive terms.","explanationZh":"$q = 6/2 = 18/6 = 3，即相邻两项的比。$","tags":["easy","common-ratio","definition"],"status":"published"},{"id":"math-geometric-sequence-002","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":1,"stem":"The 5th term of the geometric sequence with $a_{1} = 2$ and $q = 3$ is","stemZh":"首项 $a_{1} = 2、公比 q = 3$ 的等比数列第 5 项为","options":[{"id":"A","text":"54"},{"id":"B","text":"81"},{"id":"C","text":"162"},{"id":"D","text":"243"}],"answer":"C","explanation":"$a_{5} = a_{1}q^{4} = 2·3^{4} = 2·81 = 162$.","explanationZh":"$a_{5} = a_{1}q^{4} = 2 \\times 81 = 162。$","tags":["easy","general-term","indexing"],"status":"published"},{"id":"math-geometric-sequence-003","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":1,"stem":"Which of the following is a geometric sequence?","stemZh":"下列哪一项是等比数列？","options":[{"id":"A","text":"1, 3, 5, 7"},{"id":"B","text":"2, 4, 8, 16"},{"id":"C","text":"1, 4, 9, 16"},{"id":"D","text":"1, 2, 3, 5"}],"answer":"B","explanation":"Only 2, 4, 8, 16 has a constant ratio $(q = 2)$. The first is arithmetic, the third is squares, the fourth has no fixed ratio.","explanationZh":"只有 2, 4, 8, 16 相邻两项的比恒定（q $= 2）。第一个是等差数列，第三个是平方数列，第四个比值不固定。$","tags":["easy","recognition","definition"],"status":"published"},{"id":"math-geometric-sequence-004","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":2,"stem":"For a geometric sequence the common ratio q must satisfy","stemZh":"等比数列的公比 q 必须满足","options":[{"id":"A","text":"$q > 0$"},{"id":"B","text":"$q ≥ 1$"},{"id":"C","text":"$q ∈ ℤ$"},{"id":"D","text":"$q ≠ 0$"}],"answer":"D","explanation":"The only requirement is $q ≠ 0 (and\\text{ every term must be }non-zero); q$ may be negative or a fraction.","explanationZh":"唯一要求是 $q ≠ 0（且各项均不为 0）；q$ 可以是负数或分数。","tags":["easy","common-ratio","edge-case"],"status":"published"},{"id":"math-geometric-sequence-005","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":2,"stem":"A geometric sequence has $a_{1} = 1$ and $q = 1$. Then $S_{10} =$","stemZh":"等比数列 $a_{1} = 1，q = 1，则 S_{10} =$","options":[{"id":"A","text":"1"},{"id":"B","text":"0"},{"id":"C","text":"$2^{10}$"},{"id":"D","text":"10"}],"answer":"D","explanation":"When $q = 1$ the sum formula $a_{1}(1 − q^{n})/(1 − q)$ is not valid; instead every term is 1 and $S_{10} = 10$.","explanationZh":"$q = 1$ 时不能使用 $a_{1}(1 − q^{n})/(1 − q)，此时每项都是 1，故 S_{10} = 10。$","tags":["easy","q-equals-1","edge-case"],"status":"published"},{"id":"math-geometric-sequence-006","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":2,"stem":"If 2, x, 8 form a geometric sequence with $x > 0$, then $x =$","stemZh":"若 2, x, 8 成等比数列且 $x > 0，则 x =$","options":[{"id":"A","text":"$\\pm4$"},{"id":"B","text":"4"},{"id":"C","text":"−4"},{"id":"D","text":"5"}],"answer":"B","explanation":"$x^{2} = 2·8 = 16$ gives $x = \\pm4$, and the condition $x > 0$ selects $x = 4. (Without$ the condition both signs are possible.)","explanationZh":"$x^{2} = 2 \\times 8 = 16，得 x = \\pm4，再由 x > 0 取 x = 4。（没有该条件时两解都成立。）$","tags":["easy","geometric-mean","sign-care"],"status":"published"},{"id":"math-geometric-sequence-007","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":1,"stem":"In a geometric sequence $a_{2} = 6$ and $a_{3} = 18$. Then $a_{1} =$","stemZh":"等比数列中 $a_{2} = 6，a_{3} = 18，则 a_{1} =$","options":[{"id":"A","text":"2"},{"id":"B","text":"3"},{"id":"C","text":"6"},{"id":"D","text":"1"}],"answer":"A","explanation":"$q = a_{3}/a_{2} = 3$, so $a_{1} = a_{2}/q = 6/3 = 2$.","explanationZh":"$q = 18/6 = 3，故 a_{1} = a_{2}/q = 2。$","tags":["easy","common-ratio","backwards"],"status":"published"},{"id":"math-geometric-sequence-008","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":2,"stem":"A geometric sequence has general term $a_{n} = 3·2^{n-1}$. Then $a_{1} =$","stemZh":"等比数列通项 $a_{n} = 3·2^{n-1}，则 a_{1} =$","options":[{"id":"A","text":"6"},{"id":"B","text":"2"},{"id":"C","text":"3"},{"id":"D","text":"1"}],"answer":"C","explanation":"$a_{1} = 3·2^{0} = 3$. The first term is the coefficient in front of the power, not the base 2.","explanationZh":"$a_{1} = 3·2^{0} = 3。首项是幂前面的系数，而不是底数 2。$","tags":["easy","general-term","indexing"],"status":"published"},{"id":"math-geometric-sequence-009","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":2,"stem":"For a geometric sequence with $a_{1} = 1$ and $q = 2, S_{5} =$","stemZh":"等比数列 $a_{1} = 1，q = 2，则 S_{5} =$","options":[{"id":"A","text":"16"},{"id":"B","text":"31"},{"id":"C","text":"32"},{"id":"D","text":"63"}],"answer":"B","explanation":"$S_{5} = a_{1}(1 − q^{5})/(1 − q) = (1 − 32)/(−1) = 31$, which also equals 1 + 2 + 4 + 8 + 16.","explanationZh":"$S_{5} = (1 − 2^{5})/(1 − 2) = 31，即 1 + 2 + 4 + 8 + 16 = 31。$","tags":["easy","sum-formula","computation"],"status":"published"},{"id":"math-geometric-sequence-010","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":2,"stem":"A geometric sequence has $a_{1} = 3$ and $q = −1$. Its first four terms are","stemZh":"等比数列 $a_{1} = 3，q = −1，则它的前四项为$","options":[{"id":"A","text":"3, 3, 3, 3"},{"id":"B","text":"−3, 3, −3, 3"},{"id":"C","text":"3, −3, 3, −3"},{"id":"D","text":"3, −1, 1/3, −1/9"}],"answer":"C","explanation":"Each term is the previous one multiplied by −1, so the signs alternate while the absolute value stays 3: 3, −3, 3, −3.","explanationZh":"每次乘以 −1，符号交替而绝对值不变：3, −3, 3, −3。","tags":["easy","negative-ratio","listing-terms"],"status":"published"},{"id":"math-geometric-sequence-011","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"A geometric sequence has $a_{1} = 1$ and $q = 3$. If $a_{n} = 81$, then $n =$","stemZh":"等比数列 $a_{1} = 1，q = 3，若 a_{n} = 81，则 n =$","options":[{"id":"A","text":"3"},{"id":"B","text":"4"},{"id":"C","text":"6"},{"id":"D","text":"5"}],"answer":"D","explanation":"$a_{n} = 3^{n-1} = 81 = 3^{4}$, so $n − 1 = 4$ and $n = 5$.","explanationZh":"$3^{n-1} = 81 = 3^{4}，故 n − 1 = 4，n = 5。$","tags":["medium","solve-for-n","general-term"],"status":"published"},{"id":"math-geometric-sequence-012","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"A geometric sequence with all terms positive has $a_{3} = 4$ and $a_{5} = 16$. Then $a_{4} =$","stemZh":"各项均为正的等比数列中 $a_{3} = 4，a_{5} = 16，则 a_{4} =$","options":[{"id":"A","text":"$\\pm8$"},{"id":"B","text":"8"},{"id":"C","text":"−8"},{"id":"D","text":"10"}],"answer":"B","explanation":"$a_{4}^{2} = a_{3}a_{5} = 64$, so $a_{4} = \\pm8$; since all terms are $positive, a_{4} = 8$.","explanationZh":"$a_{4}^{2} = a_{3}a_{5} = 64，得 a_{4} = \\pm8；由各项为正知 a_{4} = 8。$","tags":["medium","geometric-mean","sign-care"],"status":"published"},{"id":"math-geometric-sequence-013","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"The sum of the first n terms of a geometric sequence is $S_{n} = 3(2^{n} − 1)$. Then $a_{1} =$","stemZh":"等比数列前 n 项和 $S_{n} = 3(2^{n} − 1)，则 a_{1} =$","options":[{"id":"A","text":"3"},{"id":"B","text":"6"},{"id":"C","text":"2"},{"id":"D","text":"1"}],"answer":"A","explanation":"$a_{1} = S_{1} = 3(2 − 1) = 3. (Indeed a_{1} = 3\\text{ and }q = 2\\text{ give }S_{n} = 3(2^{n} − 1)/(2 − 1).)$","explanationZh":"$a_{1} = S_{1} = 3(2 − 1) = 3。（a_{1} = 3、q = 2$ 时确有 $S_{n} = 3(2^{n} − 1)。）$","tags":["medium","sum-formula","find-a_{1}"],"status":"published"},{"id":"math-geometric-sequence-014","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"A geometric sequence has $a_{1} = 2$ and $q = 3$. Then $S_{4} =$","stemZh":"等比数列 $a_{1} = 2，q = 3，则 S_{4} =$","options":[{"id":"A","text":"40"},{"id":"B","text":"54"},{"id":"C","text":"80"},{"id":"D","text":"162"}],"answer":"C","explanation":"$S_{4} = 2(3^{4} − 1)/(3 − 1) = 2·80/2 = 80$, which agrees with 2 + 6 + 18 + 54.","explanationZh":"$S_{4} = 2(3^{4} − 1)/(3 − 1) = 80，与 2 + 6 + 18 + 54$ 一致。","tags":["medium","sum-formula","computation"],"status":"published"},{"id":"math-geometric-sequence-015","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"Compute 1 + 1/2 + 1/4 + 1/8 + 1/16.","stemZh":"计算 1 + 1/2 + 1/4 + 1/8 + 1/16。","options":[{"id":"A","text":"31/16"},{"id":"B","text":"31/32"},{"id":"C","text":"15/8"},{"id":"D","text":"63/32"}],"answer":"A","explanation":"With $a_{1} = 1$ and $q = 1/2: S_{5} = (1 − (1/2)^{5})/(1 − 1/2) = 2(1 − 1/32) = 31/16$.","explanationZh":"$a_{1} = 1，q = 1/2，S_{5} = (1 − (1/2)^{5})/(1 − 1/2) = 2 \\times 31/32 = 31/16。$","tags":["medium","sum-formula","fractions"],"status":"published"},{"id":"math-geometric-sequence-016","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"In a geometric sequence $a_{1}a_{9} = 16$. Then $a_{5} =$","stemZh":"等比数列中 $a_{1}a_{9} = 16，则 a_{5} =$","options":[{"id":"A","text":"4"},{"id":"B","text":"−4"},{"id":"C","text":"8"},{"id":"D","text":"$\\pm4$"}],"answer":"D","explanation":"Since $1 + 9 = 5 + 5, a_{1}a_{9} = a_{5}^{2} = 16$, so $a_{5} = \\pm4$. Without extra sign information both signs are possible (for example $q = 1$ gives $a_{5} = 4$, while alternating-sign sequences can give $−4)$.","explanationZh":"$由 1 + 9 = 5 + 5 得 a_{1}a_{9} = a_{5}^{2} = 16，故 a_{5} = \\pm4。没有附加条件时两种符号都可能。$","tags":["medium","equidistant-terms","sign-care"],"status":"published"},{"id":"math-geometric-sequence-017","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"A geometric sequence has $a_{1} = 1$ and $q = 2$. If $S_{n} = 127$, then $n =$","stemZh":"等比数列 $a_{1} = 1，q = 2，若 S_{n} = 127，则 n =$","options":[{"id":"A","text":"6"},{"id":"B","text":"8"},{"id":"C","text":"7"},{"id":"D","text":"9"}],"answer":"C","explanation":"$S_{n} = (2^{n} − 1)/(2 − 1) = 2^{n} − 1 = 127$, so $2^{n} = 128$ and $n = 7$.","explanationZh":"$2^{n} − 1 = 127，得 2^{n} = 128，故 n = 7。$","tags":["medium","solve-for-n","sum-formula"],"status":"published"},{"id":"math-geometric-sequence-018","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"The geometric sequence with $a_{1} = 1$ and $q = −2$ is","stemZh":"$a_{1} = 1、q = −2$ 的等比数列是","options":[{"id":"A","text":"increasing"},{"id":"B","text":"oscillating"},{"id":"C","text":"decreasing"},{"id":"D","text":"constant"}],"answer":"B","explanation":"The terms 1, −2, 4, −8, … alternate in sign while growing in absolute value, so the sequence is oscillating — it is neither increasing nor decreasing.","explanationZh":"各项为 1, −2, 4, −8, …，符号交替而绝对值递增，因此是摆动数列，既非递增也非递减。","tags":["medium","negative-ratio","monotonicity"],"status":"published"},{"id":"math-geometric-sequence-019","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"For a geometric sequence with $q ≠ 1$, the sum of the first n terms is","stemZh":"公比 $q ≠ 1$ 的等比数列，其前 n 项和为","options":[{"id":"A","text":"$a_{1}(1 − q^{n})/(1 + q)$"},{"id":"B","text":"$a_{1}(1 − q^{n})/(1 − q)$"},{"id":"C","text":"$na_{1}$"},{"id":"D","text":"$a_{1}q^{n}/(1 − q)$"}],"answer":"B","explanation":"$S_{n} = a_{1}(1 − q^{n})/(1 − q)$ for $q ≠ 1 (equivalently a_{1}(q^{n} − 1)/(q − 1)); na_{1}$ only applies when $q = 1$.","explanationZh":"$q ≠ 1 时 S_{n} = a_{1}(1 − q^{n})/(1 − q)（也可写成 a_{1}(q^{n} − 1)/(q − 1)）；na_{1}$ 只适用于 $q = 1。$","tags":["medium","sum-formula","formula-recall"],"status":"published"},{"id":"math-geometric-sequence-020","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"A geometric sequence has $a_{1} = 3$ and $q = 1/2$. Then $S_{3} =$","stemZh":"等比数列 $a_{1} = 3，q = 1/2，则 S_{3} =$","options":[{"id":"A","text":"21/8"},{"id":"B","text":"7/2"},{"id":"C","text":"15/4"},{"id":"D","text":"21/4"}],"answer":"D","explanation":"$S_{3} = 3 + 3/2 + 3/4 = 21/4$. Using the $formula: 3(1 − 1/8)/(1 − 1/2) = 3·(7/8)·2 = 21/4$.","explanationZh":"$S_{3} = 3 + 3/2 + 3/4 = 21/4；用公式得 3(1 − 1/8)/(1 − 1/2) = 21/4。$","tags":["medium","sum-formula","fractions"],"status":"published"},{"id":"math-geometric-sequence-021","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":4,"stem":"In a geometric sequence with positive terms $a_{2}a_{4} + 2a_{3}a_{5} + a_{4}a_{6} = 25$. Then $a_{3} + a_{5} =$","stemZh":"各项为正的等比数列中 $a_{2}a_{4} + 2a_{3}a_{5} + a_{4}a_{6} = 25，则 a_{3} + a_{5} =$","options":[{"id":"A","text":"25"},{"id":"B","text":"$\\pm5$"},{"id":"C","text":"5"},{"id":"D","text":"10"}],"answer":"C","explanation":"Since $a_{2}a_{4} = a_{3}^{2}$ and $a_{4}a_{6} = a_{5}^{2}$, the left side is $a_{3}^{2} + 2a_{3}a_{5} + a_{5}^{2} = (a_{3} + a_{5})^{2} = 25$. With positive terms $a_{3} + a_{5} = 5$.","explanationZh":"$由 a_{2}a_{4} = a_{3}^{2}，a_{4}a_{6} = a_{5}^{2}，左边 = a_{3}^{2} + 2a_{3}a_{5} + a_{5}^{2} = (a_{3} + a_{5})^{2} = 25；各项为正，故 a_{3} + a_{5} = 5。$","tags":["hard","equidistant-terms","algebraic-identity"],"status":"published"},{"id":"math-geometric-sequence-022","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":4,"stem":"A geometric sequence satisfies $S_{3} = 7$ and $S_{6} = 63$. Then $q =$","stemZh":"等比数列满足 $S_{3} = 7，S_{6} = 63，则 q =$","options":[{"id":"A","text":"2"},{"id":"B","text":"3"},{"id":"C","text":"1/2"},{"id":"D","text":"−2"}],"answer":"A","explanation":"$S_{6}/S_{3} = (1 − q^{6})/(1 − q^{3}) = 1 + q^{3} = 9$, so $q^{3} = 8$ and $q = 2. (Then a_{1} = 1,\\text{ giving }S_{3} = 7\\text{ and }S_{6} = 63.)$","explanationZh":"$S_{6}/S_{3} = (1 − q^{6})/(1 − q^{3}) = 1 + q^{3} = 9，故 q^{3} = 8，q = 2。（此时 a_{1} = 1，验算 S_{3} = 7，S_{6} = 63。）$","tags":["hard","sum-ratio","solve-for-q"],"status":"published"},{"id":"math-geometric-sequence-023","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":4,"stem":"If $S_{n} = 3^{n} + k$ is the sum of the first n terms of a geometric sequence, then $k =$","stemZh":"$若 S_{n} = 3^{n} + k$ 是某等比数列的前 n 项和，则 $k =$","options":[{"id":"A","text":"0"},{"id":"B","text":"1"},{"id":"C","text":"3"},{"id":"D","text":"−1"}],"answer":"D","explanation":"A geometric sum has the form $A(q^{n} − 1)$, so $S_{0} = A(q^{0} − 1) = 0$; here $3^{0} + k = 1 + k = 0$, hence $k = −1$. Indeed $3^{n} − 1 = 2(3^{n} − 1)/(3 − 1)$ is the sum for $a_{1} = 2, q = 3$.","explanationZh":"等比前 n 项和形如 $A(q^{n} − 1)，故 S_{0} = 0，即 3^{0} + k = 0，k = −1。此时 3^{n} − 1$ 正是 $a_{1} = 2、q = 3$ 的前 n 项和。","tags":["hard","sum-form","constant-term-trap"],"status":"published"},{"id":"math-geometric-sequence-024","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":4,"stem":"For a geometric sequence with $a_{1} = 1$ and $q = 2$, the three block sums $S_{n}, S_{2n} − S_{n}, S_{3n} − S_{2n}$","stemZh":"等比数列 $a_{1} = 1，q = 2，则 S_{n}、S_{2n} − S_{n}、S_{3n} − S_{2n}$ 这三个分段和","options":[{"id":"A","text":"form an arithmetic sequence"},{"id":"B","text":"form a geometric sequence with ratio $2^{n}$"},{"id":"C","text":"are all equal"},{"id":"D","text":"form a geometric sequence with ratio 2"}],"answer":"B","explanation":"$S_{n} = 2^{n} − 1, S_{2n} − S_{n} = 2^{n}(2^{n} − 1)$ and $S_{3n} − S_{2n} = 2^{2n}(2^{n} − 1)$; each block is $q^{n} = 2^{n}$ times the previous one, so they form a geometric sequence with ratio $2^{n} (not 2)$.","explanationZh":"$S_{n} = 2^{n} − 1，S_{2n} − S_{n} = 2^{n}(2^{n} − 1)，S_{3n} − S_{2n} = 2^{2n}(2^{n} − 1)，每段是前一段的 q^{n} = 2^{n}$ 倍，故公比为 $2^{n}$ 而非 2。","tags":["hard","block-sums","property"],"status":"published"},{"id":"math-geometric-sequence-025","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":5,"stem":"A geometric sequence has $a_{1} = 1$ and $q = 3$. The least n with $S_{n} > 100$ is","stemZh":"等比数列 $a_{1} = 1，q = 3，则使 S_{n} > 100$ 成立的最小 n 为","options":[{"id":"A","text":"3"},{"id":"B","text":"4"},{"id":"C","text":"5"},{"id":"D","text":"6"}],"answer":"C","explanation":"$S_{n} = (3^{n} − 1)/2 > 100$ means $3^{n} > 201$. Since $3^{4} = 81$ and $3^{5} = 243$, the least such n is $5 (S_{5} = 121)$.","explanationZh":"$S_{n} = (3^{n} − 1)/2 > 100，即 3^{n} > 201。3^{4} = 81，3^{5} = 243，故最小 n = 5（S_{5} = 121）。$","tags":["hard","logarithm","inequality"],"status":"published"},{"id":"math-geometric-sequence-026","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":5,"stem":"A geometric sequence has $a_{1} = 1$ and $q = −2$. Then $S_{6} =$","stemZh":"等比数列 $a_{1} = 1，q = −2，则 S_{6} =$","options":[{"id":"A","text":"21"},{"id":"B","text":"−63"},{"id":"C","text":"63"},{"id":"D","text":"−21"}],"answer":"D","explanation":"$S_{6} = a_{1}(1 − q^{6})/(1 − q) = (1 − 64)/(1 + 2) = −63/3 = −21$. Direct $check: 1 − 2 + 4 − 8 + 16 − 32 = −21. (Using 1 − q = 1\\text{ instead of }3\\text{ gives }−63.)$","explanationZh":"$S_{6} = (1 − (−2)^{6})/(1 − (−2)) = (1 − 64)/3 = −21。直接相加：1 − 2 + 4 − 8 + 16 − 32 = −21。（若把分母误作 1$ 则得 −63。）","tags":["hard","negative-ratio","sign-error"],"status":"published"},{"id":"math-geometric-sequence-027","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":5,"stem":"In a geometric sequence with all terms $positive, a_{1}a_{2}a_{3} = 8$ and $a_{7}a_{8}a_{9} = 512$. Then $a_{4}a_{5}a_{6} =$","stemZh":"各项为正的等比数列中 $a_{1}a_{2}a_{3} = 8，a_{7}a_{8}a_{9} = 512，则 a_{4}a_{5}a_{6} =$","options":[{"id":"A","text":"64"},{"id":"B","text":"32"},{"id":"C","text":"128"},{"id":"D","text":"216"}],"answer":"A","explanation":"$a_{1}a_{2}a_{3} = a_{2}^{3} = 8$ gives $a_{2} = 2$, and $a_{7}a_{8}a_{9} = a_{8}^{3} = 512$ gives $a_{8} = 8$. The terms $a_{2}, a_{5}, a_{8}$ are themselves geometric, so $a_{5}^{2} = a_{2}a_{8} = 16$ and $a_{5} = 4$. Hence $a_{4}a_{5}a_{6} = a_{5}^{3} = 64$.","explanationZh":"$a_{1}a_{2}a_{3} = a_{2}^{3} = 8，得 a_{2} = 2；a_{7}a_{8}a_{9} = a_{8}^{3} = 512，得 a_{8} = 8。a_{2}、a_{5}、a_{8} 成等比，a_{5}^{2} = a_{2}a_{8} = 16，a_{5} = 4，故 a_{4}a_{5}a_{6} = a_{5}^{3} = 64。$","tags":["hard","product-property","equidistant-terms"],"status":"published"},{"id":"math-geometric-sequence-028","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":4,"stem":"For a geometric sequence with $a_{1} > 0$, which condition makes it decreasing?","stemZh":"首项 $a_{1} > 0$ 的等比数列，在什么条件下单调递减？","options":[{"id":"A","text":"$q > 1$"},{"id":"B","text":"$0 < q < 1$"},{"id":"C","text":"$q < 0$"},{"id":"D","text":"$q = 1$"}],"answer":"B","explanation":"With $a_{1} > 0: q > 1$ gives an increasing $sequence, 0 < q < 1$ gives a decreasing $one, q = 1$ gives a constant sequence, and $q < 0$ makes the terms alternate in sign (not monotone).","explanationZh":"$a_{1} > 0$ 时：q $> 1 递增，0 < q < 1$ 递减，q $= 1$ 为常数列，q $< 0$ 时符号交替（不是单调）。","tags":["hard","monotonicity","sign-of-q"],"status":"published"},{"id":"math-geometric-sequence-029","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":5,"stem":"A geometric sequence has $q = 2$ and $a_{1} + a_{3} + a_{5} = 21$. Then $a_{2} + a_{4} + a_{6} =$","stemZh":"等比数列 $q = 2，且 a_{1} + a_{3} + a_{5} = 21，则 a_{2} + a_{4} + a_{6} =$","options":[{"id":"A","text":"21"},{"id":"B","text":"63"},{"id":"C","text":"42"},{"id":"D","text":"84"}],"answer":"C","explanation":"$a_{2} + a_{4} + a_{6} = q(a_{1} + a_{3} + a_{5}) = 2·21 = 42$, since each even-indexed term is q times the preceding odd-indexed term.","explanationZh":"$a_{2} + a_{4} + a_{6} = q(a_{1} + a_{3} + a_{5}) = 2 \\times 21 = 42，因为每个偶数项等于前一个奇数项乘以$ q。","tags":["hard","sub-sequence","property"],"status":"published"},{"id":"math-geometric-sequence-030","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":5,"stem":"How many terms of the geometric sequence 1, 2, 4, … must be taken for the sum first to exceed 1000?","stemZh":"等比数列 1, 2, 4, … 至少取多少项时，其和首次超过 1000？","options":[{"id":"A","text":"10"},{"id":"B","text":"9"},{"id":"C","text":"11"},{"id":"D","text":"12"}],"answer":"A","explanation":"$S_{n} = 2^{n} − 1$. Since $S_{9} = 511 < 1000$ and $S_{10} = 1023 > 1000$, ten terms are needed; equivalently $n = \\log_{_{2}}1024 = 10$.","explanationZh":"$S_{n} = 2^{n} − 1。S_{9} = 511 < 1000，S_{10} = 1023 > 1000，故需取 10$ 项（即 $n = \\log_{_{2}}1024 = 10）。$","tags":["hard","logarithm","inequality"],"status":"published"},{"id":"math-line-circle-001","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":1,"stem":"Find the slope of the line passing through $A(1, 2)$ and $B(3, 6)$.","stemZh":"求过 $A(1, 2)、B(3, 6)$ 两点的直线的斜率。","options":[{"id":"A","text":"1/2"},{"id":"B","text":"2"},{"id":"C","text":"−2"},{"id":"D","text":"4"}],"answer":"B","explanation":"Using $k = (y_{2} − y_{1})/(x_{2} − x_{1})$, we get $k = (6 − 2)/(3 − 1) = 4/2 = 2$. Option A is the result of swapping numerator and denominator.","explanationZh":"由斜率公式 $k = (y_{2} − y_{1})/(x_{2} − x_{1})，得 k = (6 − 2)/(3 − 1) = 4/2 = 2。A$ 项是把分子分母颠倒的结果。","tags":["easy","slope","two-point-form"],"status":"published"},{"id":"math-line-circle-002","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":1,"stem":"Find the slope of the line passing through $P(2, 5)$ and $Q(2, −1)$.","stemZh":"求过 $P(2, 5)、Q(2, −1)$ 两点的直线的斜率。","options":[{"id":"A","text":"0"},{"id":"B","text":"4"},{"id":"C","text":"The slope does not exist"},{"id":"D","text":"−3"}],"answer":"C","explanation":"Both points have $x = 2$, so the line is vertical $(x = 2). A$ vertical line has no $slope: k = (y_{2} − y_{1})/(x_{2} − x_{1})$ would require division by 0. Its angle of inclination is 90°, but the slope is undefined.","explanationZh":"两点横坐标都是 $2，直线为 x = 2，垂直于 x$ 轴。此时 $x_{2} − x_{1} = 0，斜率不存在（倾斜角为 90°，但斜率无定义）。$","tags":["easy","slope","vertical-line"],"status":"published"},{"id":"math-line-circle-003","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":1,"stem":"Find the distance between the points $A(1, 2)$ and $B(4, 6)$.","stemZh":"$求 A(1, 2) 与 B(4, 6)$ 两点间的距离。","options":[{"id":"A","text":"5"},{"id":"B","text":"7"},{"id":"C","text":"$\\sqrt{13}$"},{"id":"D","text":"25"}],"answer":"A","explanation":"$|AB| = \\sqrt{(4 − 1)^{2} + (6 − 2)^{2}} = \\sqrt{9 + 16} = \\sqrt{25} = 5$. Option D is the value inside the square root before taking the root.","explanationZh":"$|AB| = \\sqrt{(4 − 1)^{2} + (6 − 2)^{2}} = \\sqrt{9 + 16} = \\sqrt{25} = 5。D$ 项是忘记开平方的结果。","tags":["easy","distance-formula"],"status":"published"},{"id":"math-line-circle-004","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":2,"stem":"Find the midpoint of the segment joining $A(−2, 3)$ and $B(4, −1)$.","stemZh":"$求 A(−2, 3) 与 B(4, −1)$ 连线的中点坐标。","options":[{"id":"A","text":"$(2, 2)$"},{"id":"B","text":"$(−6, −3)$"},{"id":"C","text":"$(1, −1)$"},{"id":"D","text":"$(1, 1)$"}],"answer":"D","explanation":"Midpoint $= ((x_{1} + x_{2})/2, (y_{1} + y_{2})/2) = ((−2 + 4)/2, (3 + (−1))/2) = (1, 1)$. Option A adds without halving; option B multiplies instead of dividing.","explanationZh":"中点坐标 $= ((x_{1} + x_{2})/2, (y_{1} + y_{2})/2) = ((−2 + 4)/2, (3 − 1)/2) = (1, 1)。A$ 项忘记除以 2，B 项错误地相乘。","tags":["easy","midpoint"],"status":"published"},{"id":"math-line-circle-005","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":1,"stem":"What is the angle of inclination of the line $y =$ x?","stemZh":"直线 $y = x$ 的倾斜角是多少？","options":[{"id":"A","text":"30°"},{"id":"B","text":"45°"},{"id":"C","text":"60°"},{"id":"D","text":"90°"}],"answer":"B","explanation":"The slope is $k = 1$ and $k = \\tan θ$ with $0° ≤ θ < 180°$, so $\\tan θ = 1$ gives $θ = 45°$.","explanationZh":"斜率 $k = 1，又 k = \\tan θ（0° ≤ θ < 180°），由 \\tan θ = 1 得 θ = 45°。$","tags":["easy","angle-of-inclination"],"status":"published"},{"id":"math-line-circle-006","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":1,"stem":"Write the equation of the line with slope 2 and y-intercept 3.","stemZh":"写出斜率为 2、在 y 轴上截距为 3 的直线方程。","options":[{"id":"A","text":"$y = 3x + 2$"},{"id":"B","text":"$y = 2x − 3$"},{"id":"C","text":"$y = 2x + 3$"},{"id":"D","text":"$2x + 3y = 1$"}],"answer":"C","explanation":"In slope-intercept form $y =$ kx + b, k is the slope and b is the y-intercept, so $y = 2x + 3$. Option A swaps the two numbers.","explanationZh":"斜截式 $y =$ kx + b 中 k 为斜率、b 为 y 轴截距，故方程为 $y = 2x + 3。A$ 项把两个数交换了位置。","tags":["easy","slope-intercept-form","intercept"],"status":"published"},{"id":"math-line-circle-007","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":2,"stem":"Write the standard equation of the circle with centre $(1, −2)$ and radius 3.","stemZh":"写出圆心为 $(1, −2)、半径为 3$ 的圆的标准方程。","options":[{"id":"A","text":"$(x + 1)^{2} + (y − 2)^{2} = 9$"},{"id":"B","text":"$(x − 1)^{2} + (y + 2)^{2} = 3$"},{"id":"C","text":"$(x − 1)^{2} + (y − 2)^{2} = 9$"},{"id":"D","text":"$(x − 1)^{2} + (y + 2)^{2} = 9$"}],"answer":"D","explanation":"The standard form is $(x − a)^{2} + (y − b)^{2} = r^{2}$ with centre $(a, b)$. Here $(x − 1)^{2} + (y − (−2))^{2} = 3^{2}$, i.e. $(x − 1)^{2} + (y + 2)^{2} = 9$. Note the right-hand side is $r^{2} = 9$, not $r = 3$.","explanationZh":"标准方程为 $(x − a)^{2} + (y − b)^{2} = r^{2}，其中 (a, b)$ 为圆心。代入得 $(x − 1)^{2} + (y + 2)^{2} = 3^{2} = 9，注意右边是 r^{2}$ 而不是 r。","tags":["easy","circle-equation","standard-equation"],"status":"published"},{"id":"math-line-circle-008","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":1,"stem":"Where does the point $P(3, 4)$ lie relative to the circle $x^{2} + y^{2} = 25$?","stemZh":"$点 P(3, 4)$ 与圆 $x^{2} + y^{2} = 25$ 的位置关系是？","options":[{"id":"A","text":"On the circle"},{"id":"B","text":"Inside the circle"},{"id":"C","text":"Outside the circle"},{"id":"D","text":"At the centre"}],"answer":"A","explanation":"Substitute the point into the left $side: 3^{2} + 4^{2} = 9 + 16 = 25$, which equals $r^{2}$. Hence P lies exactly on the circle.","explanationZh":"$把点代入左边：3^{2} + 4^{2} = 9 + 16 = 25 = r^{2}，所以点 P$ 在圆上。","tags":["easy","point-and-circle"],"status":"published"},{"id":"math-line-circle-009","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":2,"stem":"Which of the following lines is parallel to $y = 3x + 1$?","stemZh":"下列哪条直线与 $y = 3x + 1$ 平行？","options":[{"id":"A","text":"$y = −3x + 1$"},{"id":"B","text":"$y = (1/3)x + 1$"},{"id":"C","text":"$y = 3x − 5$"},{"id":"D","text":"$x + 3y = 1$"}],"answer":"C","explanation":"Two non-vertical lines are parallel when their slopes are equal. The given line has slope 3, and $y = 3x − 5$ also has slope $3 (with a$ different intercept, so the lines are distinct). Option B has the negative reciprocal slope and option A the negative slope.","explanationZh":"两直线平行（不重合）的条件是斜率相等。已知直线斜率为 $3，y = 3x − 5$ 斜率也是 3 且截距不同，故平行。B 项是负倒数斜率，A 项是相反数。","tags":["easy","parallel-lines","slope"],"status":"published"},{"id":"math-line-circle-010","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":2,"stem":"Which of the following lines is perpendicular to $y = 2x + 1$?","stemZh":"下列哪条直线与 $y = 2x + 1$ 垂直？","options":[{"id":"A","text":"$y = 2x − 1$"},{"id":"B","text":"$y = −(1/2)x + 4$"},{"id":"C","text":"$y = (1/2)x + 4$"},{"id":"D","text":"$y = −2x + 4$"}],"answer":"B","explanation":"For two lines with slopes $k_{1}$ and $k_{2}$, perpendicular means $k_{1}k_{2} = −1$. Here $k_{1} = 2$, so $k_{2} = −1/2$, giving $y = −(1/2)x + 4$. Option D uses the negative slope instead of the negative reciprocal.","explanationZh":"两直线垂直的条件是 $k_{1}k_{2} = −1。已知 k_{1} = 2，故 k_{2} = −1/2，即 y = −(1/2)x + 4。D$ 项错误地取了相反数而不是负倒数。","tags":["easy","perpendicular","slope"],"status":"published"},{"id":"math-line-circle-011","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Find the distance from the point $P(1, 2)$ to the line $3x + 4y − 5 = 0$.","stemZh":"求点 $P(1, 2)$ 到直线 $3x + 4y − 5 = 0$ 的距离。","options":[{"id":"A","text":"6/5"},{"id":"B","text":"6"},{"id":"C","text":"1"},{"id":"D","text":"5/6"}],"answer":"A","explanation":"$d = |Ax_0 + By_0 + C| / \\sqrt{A^{2} + B^{2}} = |3(1) + 4(2) − 5| / \\sqrt{9 + 16} = |3 + 8 − 5|/5 = 6/5$. Option B forgets to divide by $\\sqrt{A^{2} + B^{2}} = 5$, and option D divides the wrong way round.","explanationZh":"$d = |Ax_0 + By_0 + C| / \\sqrt{A^{2} + B^{2}} = |3 + 8 − 5| / \\sqrt{9 + 16} = 6/5。B$ 项忘记除以 $\\sqrt{A^{2} + B^{2}} = 5，D$ 项分子分母颠倒。","tags":["medium","point-to-line-distance"],"status":"published"},{"id":"math-line-circle-012","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Find the point of intersection of the lines $2x + y = 5$ and $x − y = 1$.","stemZh":"求直线 $2x + y = 5 与 x − y = 1$ 的交点坐标。","options":[{"id":"A","text":"$(1, 2)$"},{"id":"B","text":"$(2, −1)$"},{"id":"C","text":"$(3, 1)$"},{"id":"D","text":"$(2, 1)$"}],"answer":"D","explanation":"Adding the two equations gives $3x = 6$, so $x = 2$; substituting into $x − y = 1$ gives $y = 1$. The intersection point is $(2, 1)$, which satisfies both equations.","explanationZh":"两式相加得 $3x = 6，x = 2；代入 x − y = 1 得 y = 1。交点为 (2, 1)，代入两方程均成立。$","tags":["medium","intersection"],"status":"published"},{"id":"math-line-circle-013","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Find the equation of the line through $(−1, 3)$ that is parallel to $2x − y + 4 = 0$.","stemZh":"求过点 $(−1, 3)$ 且与直线 $2x − y + 4 = 0$ 平行的直线方程。","options":[{"id":"A","text":"$2x − y − 5 = 0$"},{"id":"B","text":"$2x − y + 5 = 0$"},{"id":"C","text":"$x + 2y − 5 = 0$"},{"id":"D","text":"$2x + y + 5 = 0$"}],"answer":"B","explanation":"The line $2x − y + 4 = 0$ has slope 2, so the required line has slope 2 as $well: y − 3 = 2(x + 1)$, i.e. $y = 2x + 5$, or $2x − y + 5 = 0. Check: 2(−1) − 3 + 5 = 0$.","explanationZh":"直线 $2x − y + 4 = 0$ 的斜率为 $2，所求直线斜率也为 2：y − 3 = 2(x + 1)，即 2x − y + 5 = 0。验证：2(−1) − 3 + 5 = 0。$","tags":["medium","parallel-lines","point-slope-form"],"status":"published"},{"id":"math-line-circle-014","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Find the centre and radius of the circle $x^{2} + y^{2} − 4x + 6y − 3 = 0$.","stemZh":"求圆 $x^{2} + y^{2} − 4x + 6y − 3 = 0$ 的圆心和半径。","options":[{"id":"A","text":"centre $(−2, 3)$, radius 4"},{"id":"B","text":"centre $(2, −3)$, radius 16"},{"id":"C","text":"centre $(2, −3)$, radius 4"},{"id":"D","text":"centre $(2, 3)$, radius 4"}],"answer":"C","explanation":"Complete the $square: (x^{2} − 4x + 4) + (y^{2} + 6y + 9) = 3 + 4 + 9$, so $(x − 2)^{2} + (y + 3)^{2} = 16$. The centre is $(2, −3)$ and $r = \\sqrt{16} = 4$. Option B mistakes $r^{2} = 16$ for the radius.","explanationZh":"$配方：(x^{2} − 4x + 4) + (y^{2} + 6y + 9) = 3 + 4 + 9，即 (x − 2)^{2} + (y + 3)^{2} = 16。圆心为 (2, −3)，半径 r = \\sqrt{16} = 4。B$ 项把 $r^{2} = 16$ 当成了半径。","tags":["medium","completing-the-square","general-form"],"status":"published"},{"id":"math-line-circle-015","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"For the line $3x + 4y + c = 0$ to be tangent to the circle $x^{2} + y^{2} = 25$, what must $|c|$ be?","stemZh":"若直线 $3x + 4y + c = 0$ 与圆 $x^{2} + y^{2} = 25$ 相切，则 $|c|$ 等于多少？","options":[{"id":"A","text":"5"},{"id":"B","text":"−25"},{"id":"C","text":"125"},{"id":"D","text":"25"}],"answer":"D","explanation":"A line is tangent to a circle when the distance from the centre to the line equals the $radius: |c|/\\sqrt{3^{2} + 4^{2}} = 5$, so $|c|/5 = 5$ and $|c| = 25 (c = 25$ or $c = −25)$. Option A is the radius itself; option B cannot be an absolute value.","explanationZh":"$相切条件是圆心到直线的距离等于半径：|c|/\\sqrt{3^{2} + 4^{2}} = 5，即 |c|/5 = 5，得 |c| = 25（c = \\pm25）。A$ 项是半径本身；B 项绝对值不可能为负。","tags":["medium","tangent-condition","point-to-line-distance"],"status":"published"},{"id":"math-line-circle-016","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Under which condition does $x^{2} + y^{2} +$ Dx + Ey $+ F = 0$ represent a (real) circle?","stemZh":"方程 $x^{2} + y^{2} +$ Dx + Ey $+ F = 0$ 表示圆的充要条件是？","options":[{"id":"A","text":"$D^{2} + E^{2} − 4F > 0$"},{"id":"B","text":"$D^{2} + E^{2} − 4F < 0$"},{"id":"C","text":"$D^{2} + E^{2} − 4F = 0$"},{"id":"D","text":"$D + E − F > 0$"}],"answer":"A","explanation":"Completing the square gives $(x + D/2)^{2} + (y + E/2)^{2} = (D^{2} + E^{2} − 4F)/4$. This is a real circle exactly when the right side is positive, i.e. $D^{2} + E^{2} − 4F > 0$. If it equals 0 the equation is a single point, and if it is negative no real points satisfy it.","explanationZh":"配方得 $(x + D/2)^{2} + (y + E/2)^{2} = (D^{2} + E^{2} − 4F)/4。当 D^{2} + E^{2} − 4F > 0$ 时表示实圆；等于 0 时表示一个点；小于 0 时不表示任何实图形。","tags":["medium","general-form","circle-equation"],"status":"published"},{"id":"math-line-circle-017","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Find the equation of the line through the intercepts $(0, 4)$ and $(2, 0)$.","stemZh":"求过 $(0, 4) 和 (2, 0)$ 两点的直线方程。","options":[{"id":"A","text":"$x + 2y = 4$"},{"id":"B","text":"$2x + y = 4$"},{"id":"C","text":"$2x − y = 4$"},{"id":"D","text":"$x − 2y = 4$"}],"answer":"B","explanation":"The slope is $(0 − 4)/(2 − 0) = −2$, so $y = −2x + 4$, i.e. $2x + y = 4$. Equivalently use the intercept form $x/2 + y/4 = 1. Check: (0, 4)$ gives $4 = 4$ and $(2, 0)$ gives $4 = 4$.","explanationZh":"斜率 $k = (0 − 4)/(2 − 0) = −2，故 y = −2x + 4，即 2x + y = 4。也可用截距式 x/2 + y/4 = 1。代入两点均成立。$","tags":["medium","intercept-form","two-point-form"],"status":"published"},{"id":"math-line-circle-018","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"A line has slope −1. What is its angle of inclination?","stemZh":"一条直线的斜率为 −1，则它的倾斜角为多少？","options":[{"id":"A","text":"45°"},{"id":"B","text":"−45°"},{"id":"C","text":"135°"},{"id":"D","text":"90°"}],"answer":"C","explanation":"$k = \\tan θ$ with $0° ≤ θ < 180°$. Since $\\tan θ = −1 < 0, θ$ is $obtuse: θ = 180° − 45° = 135°. A$ negative angle such as −45° is not allowed for an angle of inclination.","explanationZh":"$由 k = \\tan θ 且 0° ≤ θ < 180°，\\tan θ = −1 < 0$ 说明 $θ 为钝角：θ = 180° − 45° = 135°。倾斜角的取值范围不允许 −45°。$","tags":["medium","angle-of-inclination"],"status":"published"},{"id":"math-line-circle-019","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Find the distance between the parallel lines $3x + 4y + 1 = 0$ and $3x + 4y − 9 = 0$.","stemZh":"求两条平行直线 $3x + 4y + 1 = 0 与 3x + 4y − 9 = 0$ 之间的距离。","options":[{"id":"A","text":"1"},{"id":"B","text":"5"},{"id":"C","text":"4"},{"id":"D","text":"2"}],"answer":"D","explanation":"For parallel lines Ax + By $+ C_1 = 0$ and Ax + By $+ C_2 = 0, d = |C_1 − C_2| / \\sqrt{A^{2} + B^{2}} = |1 − (−9)| / \\sqrt{9 + 16} = 10/5 = 2$. Equivalently take any point on one line and use the point-to-line distance formula.","explanationZh":"两平行线 Ax + By $+ C_1 = 0 与$ Ax + By $+ C_2 = 0$ 的距离为 $d = |C_1 − C_2| / \\sqrt{A^{2} + B^{2}} = |1 − (−9)|/5 = 10/5 = 2。也可在其中一条上取一点代入点到直线距离公式。$","tags":["medium","parallel-lines","point-to-line-distance"],"status":"published"},{"id":"math-line-circle-020","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Where does the point $P(1, 1)$ lie relative to the circle $(x − 2)^{2} + (y − 3)^{2} = 4$?","stemZh":"$点 P(1, 1)$ 与圆 $(x − 2)^{2} + (y − 3)^{2} = 4$ 的位置关系是？","options":[{"id":"A","text":"Outside the circle"},{"id":"B","text":"On the circle"},{"id":"C","text":"Inside the circle"},{"id":"D","text":"At the centre"}],"answer":"A","explanation":"The centre is $C(2, 3)$ and $r = 2. |PC|^{2} = (1 − 2)^{2} + (1 − 3)^{2} = 1 + 4 = 5$, so $|PC| = \\sqrt{5} > 2 =$ r. Hence P lies outside the circle.","explanationZh":"圆心 $C(2, 3)，半径 r = 2。|PC|^{2} = (1 − 2)^{2} + (1 − 3)^{2} = 5，故 |PC| = \\sqrt{5} > 2，点在圆外。$","tags":["medium","point-and-circle"],"status":"published"},{"id":"math-line-circle-021","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":4,"stem":"Which of the following equations does NOT represent a circle?","stemZh":"下列方程中不表示圆的是？","options":[{"id":"A","text":"$x^{2} + y^{2} − 2x + 4y + 1 = 0$"},{"id":"B","text":"$x^{2} + y^{2} − 2x + 4y + 5 = 0$"},{"id":"C","text":"$x^{2} + y^{2} − 2x + 4y − 1 = 0$"},{"id":"D","text":"$x^{2} + y^{2} − 2x + 4y + 4 = 0$"}],"answer":"B","explanation":"All four have $D = −2, E = 4$, so $r^{2} = (D^{2} + E^{2} − 4F)/4 = (4 + 16 − 4F)/4 = 5 −$ F. For $B, F = 5$ gives $r^{2} = 0: (x − 1)^{2} + (y + 2)^{2} = 0$, which is the single point $(1, −2)$, not a circle. The others give $r^{2} = 4, 6$ and 1 respectively.","explanationZh":"四式的 $D = −2、E = 4，故 r^{2} = (D^{2} + E^{2} − 4F)/4 = 5 −$ F。B $项 F = 5 时 r^{2} = 0，即 (x − 1)^{2} + (y + 2)^{2} = 0，只表示点 (1, −2)，不是圆。其余各项 r^{2}$ 分别为 4、6、1。","tags":["hard","general-form","circle-equation"],"status":"published"},{"id":"math-line-circle-022","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":4,"stem":"Find the distance from $P(1, −1)$ to the line passing through $A(2, 1)$ and $B(−1, 3)$.","stemZh":"求点 $P(1, −1)$ 到过 $A(2, 1)、B(−1, 3)$ 两点的直线的距离。","options":[{"id":"A","text":"$8\\sqrt{13}/13$"},{"id":"B","text":"$4\\sqrt{13}/13$"},{"id":"C","text":"8/13"},{"id":"D","text":"$2\\sqrt{13}/13$"}],"answer":"A","explanation":"Slope of $AB: k = (3 − 1)/(−1 − 2) = −2/3$. Using point-slope form at $A: y − 1 = (−2/3)(x − 2) → 2x + 3y − 7 = 0$. Then $d = |2(1) + 3(−1) − 7| / \\sqrt{4 + 9} = 8/\\sqrt{13} = 8\\sqrt{13}/13$.","explanationZh":"AB 的斜率 $k = (3 − 1)/(−1 − 2) = −2/3，由点斜式得 y − 1 = (−2/3)(x − 2)，即 2x + 3y − 7 = 0。故 d = |2 − 3 − 7| / \\sqrt{4 + 9} = 8/\\sqrt{13} = 8\\sqrt{13}/13。$","tags":["hard","point-to-line-distance","two-point-form"],"status":"published"},{"id":"math-line-circle-023","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":4,"stem":"For what value of m does $x^{2} + y^{2} + 4x − 2y + m = 0$ represent a circle of radius 3?","stemZh":"当 m 为何值时，方程 $x^{2} + y^{2} + 4x − 2y + m = 0$ 表示半径为 3 的圆？","options":[{"id":"A","text":"−5"},{"id":"B","text":"4"},{"id":"C","text":"−4"},{"id":"D","text":"0"}],"answer":"C","explanation":"Complete the $square: (x + 2)^{2} + (y − 1)^{2} = 5 −$ m. Setting $r^{2} = 9$ gives $5 − m = 9$, so $m = −4$. The sign is the trap: a negative constant term makes the right-hand side larger.","explanationZh":"配方得 $(x + 2)^{2} + (y − 1)^{2} = 5 −$ m。令 $r^{2} = 9，即 5 − m = 9，得 m = −4。注意符号：常数项为负时右边反而更大。$","tags":["hard","completing-the-square","general-form"],"status":"published"},{"id":"math-line-circle-024","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":4,"stem":"Find the equation of the tangent to the circle $x^{2} + y^{2} = 13$ at the point $(2, 3)$.","stemZh":"求圆 $x^{2} + y^{2} = 13$ 在点 $(2, 3)$ 处的切线方程。","options":[{"id":"A","text":"$2x + 3y = 13$"},{"id":"B","text":"$3x + 2y = 13$"},{"id":"C","text":"$2x − 3y = 13$"},{"id":"D","text":"$2x + 3y = −13$"}],"answer":"A","explanation":"First check that $(2, 3)$ is on the $circle: 2^{2} + 3^{2} = 13$. For $x^{2} + y^{2} = r^{2}$ the tangent at $(x_{0}, y_{0})$ is $x_{0}x + y_{0}y = r^{2}$, giving $2x + 3y = 13$. The radius to $(2, 3)$ has slope 3/2, and the tangent's slope −2/3 is its negative reciprocal.","explanationZh":"先验证 $(2, 3) 在圆上：2^{2} + 3^{2} = 13。圆 x^{2} + y^{2} = r^{2} 在 (x_{0}, y_{0})$ 处的切线为 $x_{0}x + y_{0}y = r^{2}，即 2x + 3y = 13。半径斜率为 3/2，切线斜率 −2/3$ 恰为其负倒数。","tags":["hard","tangent-condition","circle-equation"],"status":"published"},{"id":"math-line-circle-025","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":5,"stem":"How many tangent lines can be drawn from the point $P(3, 4)$ to the circle $x^{2} + y^{2} = 4$?","stemZh":"过点 $P(3, 4)$ 可作圆 $x^{2} + y^{2} = 4$ 的几条切线？","options":[{"id":"A","text":"0"},{"id":"B","text":"1"},{"id":"C","text":"2"},{"id":"D","text":"Infinitely many"}],"answer":"C","explanation":"The centre is $O(0, 0)$ and $r = 2. |OP| = \\sqrt{9 + 16} = 5 > 2$, so P lies outside the circle, and exactly two tangents can be drawn from it. (One tangent if P is on the circle, none if P is inside.)","explanationZh":"圆心 $O(0, 0)，半径 r = 2。|OP| = \\sqrt{9 + 16} = 5 > 2，点在圆外，故可作两条切线。（点在圆上只能作一条，点在圆内则没有切线。）$","tags":["hard","tangent-condition","point-and-circle"],"status":"published"},{"id":"math-line-circle-026","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":5,"stem":"The lines kx $+ y + 1 = 0$ and x + ky $+ 2 = 0$ are perpendicular. Find k.","stemZh":"若直线 kx $+ y + 1 = 0 与 x +$ ky $+ 2 = 0$ 垂直，求 k。","options":[{"id":"A","text":"−1"},{"id":"B","text":"1"},{"id":"C","text":"2"},{"id":"D","text":"0"}],"answer":"D","explanation":"For general-form lines $A_{1}x + B_{1}y + C_{1} = 0$ and $A_{2}x + B_{2}y + C_{2} = 0$, perpendicular means $A_{1}A_{2} + B_{1}B_{2} = 0$. Here $k·1 + 1·k = 2k = 0$, so $k = 0$. Indeed the lines become $y = −1 (horizontal)$ and $x = −2 (vertical)$. Using $k_{1}k_{2} = −1$ fails here because one line is vertical and has no slope.","explanationZh":"一般式下两直线垂直的条件为 $A_{1}A_{2} + B_{1}B_{2} = 0，即 k·1 + 1·k = 2k = 0，得 k = 0。此时两直线为 y = −1（水平）与 x = −2（竖直）。若用 k_{1}k_{2} = −1$ 会失效，因为其中一条直线斜率不存在。","tags":["hard","perpendicular","general-form"],"status":"published"},{"id":"math-line-circle-027","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":5,"stem":"Find the point on the line $2x + y − 4 = 0$ that is closest to the origin.","stemZh":"求直线 $2x + y − 4 = 0$ 上到原点距离最小的点。","options":[{"id":"A","text":"$(4/5, 8/5)$"},{"id":"B","text":"$(8/5, 4/5)$"},{"id":"C","text":"$(2, 0)$"},{"id":"D","text":"$(0, 4)$"}],"answer":"B","explanation":"The closest point is the foot of the perpendicular from O. The given line has slope −2, so the perpendicular through O has slope $1/2: y = x/2$, i.e. $x = 2y$. Substituting into $2x + y = 4$ gives $5y = 4$, so $y = 4/5$ and $x = 8/5$. The point is $(8/5, 4/5)$.","explanationZh":"最近点是原点到该直线的垂足。已知直线斜率为 $−2，过原点的垂线斜率为 1/2，即 y = x/2（x = 2y）。代入 2x + y = 4 得 5y = 4，y = 4/5，x = 8/5，即 (8/5, 4/5)。$","tags":["hard","perpendicular","foot-of-perpendicular"],"status":"published"},{"id":"math-line-circle-028","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":5,"stem":"A circle passes through $A(1, 1), B(3, 1)$ and $C(1, 3)$. Find its centre.","stemZh":"一圆过 $A(1, 1)、B(3, 1)、C(1, 3)$ 三点，求圆心坐标。","options":[{"id":"A","text":"$(2, 1)$"},{"id":"B","text":"$(1, 2)$"},{"id":"C","text":"$(3, 3)$"},{"id":"D","text":"$(2, 2)$"}],"answer":"D","explanation":"The centre lies on the perpendicular bisector of every chord. Chord AB is horizontal from $(1, 1)$ to $(3, 1)$, so its perpendicular bisector is $x = 2$. Chord AC is vertical from $(1, 1)$ to $(1, 3)$, so its perpendicular bisector is $y = 2$. The centre is $(2, 2)$, with $r = \\sqrt{2}$.","explanationZh":"圆心在任意弦的垂直平分线上。弦 AB 水平，其中垂线为 $x = 2；弦$ AC 竖直，其中垂线为 $y = 2。故圆心为 (2, 2)，半径 r = \\sqrt{2}。$","tags":["hard","circle-equation","perpendicular"],"status":"published"},{"id":"math-line-circle-029","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":4,"stem":"Find the area enclosed by the circle $x^{2} + y^{2} − 6x − 8y = 0$.","stemZh":"求圆 $x^{2} + y^{2} − 6x − 8y = 0$ 的面积。","options":[{"id":"A","text":"$5π$"},{"id":"B","text":"$10π$"},{"id":"C","text":"$25π$"},{"id":"D","text":"$100π$"}],"answer":"C","explanation":"Completing the $square: (x − 3)^{2} + (y − 4)^{2} = 9 + 16 = 25$, so $r = 5 (note F = 0\\text{ here, so }r^{2} = (D^{2} + E^{2})/4)$. The area is $πr^{2} = 25π$.","explanationZh":"配方得 $(x − 3)^{2} + (y − 4)^{2} = 9 + 16 = 25，故 r = 5（此处 F = 0，r^{2} = (D^{2} + E^{2})/4）。面积 S = πr^{2} = 25π。$","tags":["hard","completing-the-square","general-form"],"status":"published"},{"id":"math-line-circle-030","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":5,"stem":"The circle $(x − 1)^{2} + (y − 2)^{2} = 4$ has two tangents parallel to $3x + 4y = 0$. One of them is $3x + 4y − 1 = 0$. Find the other.","stemZh":"$圆 (x − 1)^{2} + (y − 2)^{2} = 4$ 有两条与 $3x + 4y = 0$ 平行的切线，其中一条为 $3x + 4y − 1 = 0，求另一条。$","options":[{"id":"A","text":"$3x + 4y − 21 = 0$"},{"id":"B","text":"$3x + 4y + 21 = 0$"},{"id":"C","text":"$3x + 4y − 11 = 0$"},{"id":"D","text":"$3x + 4y + 1 = 0$"}],"answer":"A","explanation":"Write a parallel tangent as $3x + 4y − c = 0$. Tangency requires the distance from the centre $(1, 2)$ to equal $r = 2: |3(1) + 4(2) − c|/5 = 2 → |11 − c| = 10 → c = 1$ or $c = 21$. So the other tangent is $3x + 4y − 21 = 0$.","explanationZh":"设平行切线为 $3x + 4y − c = 0。相切要求圆心 (1, 2)$ 到直线的距离等于 $r = 2：|3 + 8 − c|/5 = 2，即 |11 − c| = 10，解得 c = 1 或 c = 21。故另一条切线为 3x + 4y − 21 = 0。$","tags":["hard","tangent-condition","parallel-lines"],"status":"published"},{"id":"math-conics-001","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":1,"stem":"For the ellipse $x^{2}/25 + y^{2}/16 = 1$, which axis contains the major axis?","stemZh":"椭圆 $x^{2}/25 + y^{2}/16 = 1$ 的长轴在哪个轴上？","options":[{"id":"A","text":"The x-axis"},{"id":"B","text":"The y-axis"},{"id":"C","text":"The line $y = x$"},{"id":"D","text":"Neither axis"}],"answer":"A","explanation":"In $x^{2}/a^{2} + y^{2}/b^{2} = 1$ with $a > b > 0$, the larger denominator is $a^{2}$. Here $25 > 16$, so $a^{2} = 25$ sits under $x^{2}$, and the major axis lies on the x-axis with length $2a = 10$.","explanationZh":"标准方程 $x^{2}/a^{2} + y^{2}/b^{2} = 1（a > b > 0）中，较大的分母是 a^{2}。此处 25 > 16，a^{2} = 25 在 x^{2}$ 下方，故长轴在 x 轴上，长轴长 $2a = 10。$","tags":["easy","ellipse","major-axis"],"status":"published"},{"id":"math-conics-002","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":1,"stem":"For the ellipse $x^{2}/25 + y^{2}/16 = 1$, find the focal half-distance c.","stemZh":"椭圆 $x^{2}/25 + y^{2}/16 = 1$ 的半焦距 c 为多少？","options":[{"id":"A","text":"9"},{"id":"B","text":"3"},{"id":"C","text":"41"},{"id":"D","text":"$\\sqrt{41}$"}],"answer":"B","explanation":"For an $ellipse, c^{2} = a^{2} − b^{2} = 25 − 16 = 9$, so $c = 3 (the$ foci are at $(\\pm3, 0))$. Option A is $c^{2}$, and $\\sqrt{41}$ would come from wrongly adding as in a hyperbola.","explanationZh":"椭圆中 $c^{2} = a^{2} − b^{2} = 25 − 16 = 9，故 c = 3（焦点为 (\\pm3, 0)）。A$ 项是 $c^{2}；\\sqrt{41}$ 是误用双曲线的 $c^{2} = a^{2} + b^{2}$ 得出的。","tags":["easy","ellipse","focal-distance"],"status":"published"},{"id":"math-conics-003","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":2,"stem":"Find the eccentricity of the ellipse $x^{2}/25 + y^{2}/16 = 1$.","stemZh":"求椭圆 $x^{2}/25 + y^{2}/16 = 1$ 的离心率。","options":[{"id":"A","text":"3/5"},{"id":"B","text":"4/5"},{"id":"C","text":"5/3"},{"id":"D","text":"3/4"}],"answer":"A","explanation":"$a = 5, b = 4$ and $c^{2} = a^{2} − b^{2} = 9$, so $c = 3$. Hence $e = c/a = 3/5$. Option B is b/a and option C is a/c, neither of which is the eccentricity.","explanationZh":"$a = 5，b = 4，c^{2} = a^{2} − b^{2} = 9，c = 3，故 e = c/a = 3/5。B$ 项是 b/a，C 项是 a/c，都不是离心率。","tags":["easy","ellipse","eccentricity"],"status":"published"},{"id":"math-conics-004","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":2,"stem":"For the hyperbola $x^{2}/9 − y^{2}/16 = 1$, find the focal half-distance c.","stemZh":"双曲线 $x^{2}/9 − y^{2}/16 = 1$ 的半焦距 c 为多少？","options":[{"id":"A","text":"$\\sqrt{7}$"},{"id":"B","text":"7"},{"id":"C","text":"5"},{"id":"D","text":"25"}],"answer":"C","explanation":"For $a hyperbola, c^{2} = a^{2} + b^{2} = 9 + 16 = 25$, so $c = 5$ and the foci are $(\\pm5, 0)$. Note the relation differs from the ellipse: here $a^{2}$ and $b^{2}$ are added, not subtracted.","explanationZh":"双曲线中 $c^{2} = a^{2} + b^{2} = 9 + 16 = 25，故 c = 5，焦点为 (\\pm5, 0)。注意与椭圆不同：这里是 a^{2} + b^{2}$ 而不是相减。","tags":["easy","hyperbola","focal-distance"],"status":"published"},{"id":"math-conics-005","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":1,"stem":"Find the equations of the asymptotes of the hyperbola $x^{2}/9 − y^{2}/16 = 1$.","stemZh":"求双曲线 $x^{2}/9 − y^{2}/16 = 1$ 的渐近线方程。","options":[{"id":"A","text":"$y = \\pm(3/4)x$"},{"id":"B","text":"$y = \\pm(9/16)x$"},{"id":"C","text":"$y = \\pm(16/9)x$"},{"id":"D","text":"$y = \\pm(4/3)x$"}],"answer":"D","explanation":"For $x^{2}/a^{2} − y^{2}/b^{2} = 1$ the asymptotes are $y = \\pm(b/a)x$. Here $a = 3$ and $b = 4$, so $y = \\pm(4/3)x$. Option A inverts the ratio b/a into a/b.","explanationZh":"双曲线 $x^{2}/a^{2} − y^{2}/b^{2} = 1$ 的渐近线为 $y = \\pm(b/a)x。此处 a = 3、b = 4，故为 y = \\pm(4/3)x。A$ 项把 b/a 错成了 a/b。","tags":["easy","hyperbola","asymptote"],"status":"published"},{"id":"math-conics-006","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":1,"stem":"Find the focus of the parabola $y^{2} = 8x$.","stemZh":"求抛物线 $y^{2} = 8x$ 的焦点坐标。","options":[{"id":"A","text":"$(4, 0)$"},{"id":"B","text":"$(2, 0)$"},{"id":"C","text":"$(0, 2)$"},{"id":"D","text":"$(−2, 0)$"}],"answer":"B","explanation":"Comparing $y^{2} = 8x$ with $y^{2} = 2px$ gives $2p = 8$, so $p = 4$. The focus is $(p/2, 0) = (2, 0)$ and the directrix is $x = −2$. Option A uses p itself instead of p/2.","explanationZh":"$与 y^{2} = 2px$ 比较得 $2p = 8，p = 4。焦点为 (p/2, 0) = (2, 0)，准线为 x = −2。A$ 项误把 p 当成了焦点横坐标（应为 p/2）。","tags":["easy","parabola","focus"],"status":"published"},{"id":"math-conics-007","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":2,"stem":"Find the directrix of the parabola $y^{2} = 8x$.","stemZh":"求抛物线 $y^{2} = 8x$ 的准线方程。","options":[{"id":"A","text":"$x = 2$"},{"id":"B","text":"$x = −4$"},{"id":"C","text":"$x = −2$"},{"id":"D","text":"$y = −2$"}],"answer":"C","explanation":"$y^{2} = 2px$ with $2p = 8$ gives $p = 4$, so the directrix is $x = −p/2 = −2$. The focus $(2, 0)$ and the directrix $x = −2$ are symmetric about the vertex (the origin).","explanationZh":"$由 y^{2} = 2px 得 2p = 8，p = 4，故准线为 x = −p/2 = −2。焦点 (2, 0)$ 与准线 $x = −2$ 关于顶点（原点）对称。","tags":["easy","parabola","directrix"],"status":"published"},{"id":"math-conics-008","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":1,"stem":"Which of the following equations represents a hyperbola?","stemZh":"下列方程中表示双曲线的是？","options":[{"id":"A","text":"$x^{2}/4 − y^{2}/9 = 1$"},{"id":"B","text":"$x^{2}/4 + y^{2}/9 = 1$"},{"id":"C","text":"$y^{2} = 4x$"},{"id":"D","text":"$x^{2} + y^{2} = 4$"}],"answer":"A","explanation":"A minus sign between the two squared terms gives a hyperbola. Option B (plus sign) is an ellipse, option C is a parabola (only one variable is $squared)$, and option D is a circle.","explanationZh":"两个平方项之间为减号时表示双曲线。B 项为加号，是椭圆；C 项只有一个变量平方，是抛物线；D 项是圆。","tags":["easy","hyperbola","standard-equation"],"status":"published"},{"id":"math-conics-009","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":2,"stem":"For the ellipse $x^{2}/25 + y^{2}/16 = 1$, what is the length of the major axis?","stemZh":"椭圆 $x^{2}/25 + y^{2}/16 = 1$ 的长轴长为多少？","options":[{"id":"A","text":"5"},{"id":"B","text":"8"},{"id":"C","text":"16"},{"id":"D","text":"10"}],"answer":"D","explanation":"$a^{2} = 25$, so $a = 5$ and the major axis has length $2a = 10$. The minor axis has length $2b = 8$, which is option B — a common confusion.","explanationZh":"$a^{2} = 25，a = 5，长轴长为 2a = 10。短轴长为 2b = 8（即 B$ 项），这是常见的混淆点。","tags":["easy","ellipse","major-axis"],"status":"published"},{"id":"math-conics-010","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":1,"stem":"In which direction does the parabola $y^{2} = 2px (p > 0)$ open?","stemZh":"抛物线 $y^{2} = 2px（p > 0）的开口方向是？$","options":[{"id":"A","text":"Upward"},{"id":"B","text":"To the right"},{"id":"C","text":"To the left"},{"id":"D","text":"Downward"}],"answer":"B","explanation":"Since $y^{2} ≥ 0$ and $p > 0$, we need $x ≥ 0$, so the curve opens towards the positive x-direction, i.e. to the right. Its axis of symmetry is the x-axis and its focus is $(p/2, 0)$.","explanationZh":"$由 y^{2} ≥ 0 且 p > 0$ 可知 $x ≥ 0，故开口朝 x$ 轴正方向，即向右。对称轴为 x 轴，焦点为 $(p/2, 0)。$","tags":["easy","parabola","axis-of-symmetry"],"status":"published"},{"id":"math-conics-011","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"An ellipse centred at the origin has its major axis on the x-axis with $a = 5$ and $c = 3$. Find its standard equation.","stemZh":"中心在原点、长轴在 x 轴上的椭圆，a $= 5，c = 3，求其标准方程。$","options":[{"id":"A","text":"$x^{2}/25 + y^{2}/9 = 1$"},{"id":"B","text":"$x^{2}/25 + y^{2}/34 = 1$"},{"id":"C","text":"$x^{2}/25 + y^{2}/16 = 1$"},{"id":"D","text":"$x^{2}/5 + y^{2}/4 = 1$"}],"answer":"C","explanation":"For an ellipse $c^{2} = a^{2} − b^{2}$, so $b^{2} = a^{2} − c^{2} = 25 − 9 = 16$. The equation is $x^{2}/25 + y^{2}/16 = 1$. Option A puts $c^{2}$ under $y^{2}$, and option D uses a and b instead of $a^{2}$ and $b^{2}$ in the denominators.","explanationZh":"椭圆中 $c^{2} = a^{2} − b^{2}，故 b^{2} = a^{2} − c^{2} = 25 − 9 = 16，方程为 x^{2}/25 + y^{2}/16 = 1。A$ 项把 $c^{2}$ 放到了 $y^{2}$ 下面；D 项分母用了 a、b 而非 $a^{2}、b^{2}。$","tags":["medium","ellipse","standard-equation"],"status":"published"},{"id":"math-conics-012","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"Find the eccentricity of the hyperbola $x^{2}/9 − y^{2}/16 = 1$.","stemZh":"求双曲线 $x^{2}/9 − y^{2}/16 = 1$ 的离心率。","options":[{"id":"A","text":"3/5"},{"id":"B","text":"5/3"},{"id":"C","text":"4/3"},{"id":"D","text":"25/9"}],"answer":"B","explanation":"$c^{2} = a^{2} + b^{2} = 9 + 16 = 25$, so $c = 5$ and $a = 3$, giving $e = c/a = 5/3$. For a hyperbola $e > 1 always$, which immediately rules out option A.","explanationZh":"$c^{2} = a^{2} + b^{2} = 9 + 16 = 25，c = 5，a = 3，故 e = c/a = 5/3。双曲线的离心率恒大于 1，据此可直接排除 A$ 项。","tags":["medium","hyperbola","eccentricity"],"status":"published"},{"id":"math-conics-013","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"A hyperbola centred at the origin has its real axis on the x-axis with $a = 3$ and $c = 5$. Find its standard equation.","stemZh":"中心在原点、实轴在 x 轴上的双曲线，a $= 3，c = 5，求其标准方程。$","options":[{"id":"A","text":"$x^{2}/9 − y^{2}/25 = 1$"},{"id":"B","text":"$x^{2}/9 + y^{2}/16 = 1$"},{"id":"C","text":"$x^{2}/25 − y^{2}/9 = 1$"},{"id":"D","text":"$x^{2}/9 − y^{2}/16 = 1$"}],"answer":"D","explanation":"For a hyperbola $c^{2} = a^{2} + b^{2}$, so $b^{2} = c^{2} − a^{2} = 25 − 9 = 16$. The equation is $x^{2}/9 − y^{2}/16 = 1$. Notice $b^{2}$ comes from subtracting $a^{2}$ from $c^{2}$, the opposite of the ellipse rule.","explanationZh":"双曲线中 $c^{2} = a^{2} + b^{2}，故 b^{2} = c^{2} − a^{2} = 25 − 9 = 16，方程为 x^{2}/9 − y^{2}/16 = 1。注意这里 b^{2} = c^{2} − a^{2}，与椭圆的 b^{2} = a^{2} − c^{2}$ 正好相反。","tags":["medium","hyperbola","standard-equation"],"status":"published"},{"id":"math-conics-014","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"Find the foci of the ellipse $x^{2}/16 + y^{2}/25 = 1$.","stemZh":"求椭圆 $x^{2}/16 + y^{2}/25 = 1$ 的焦点坐标。","options":[{"id":"A","text":"$(\\pm3, 0)$"},{"id":"B","text":"$(0, \\pm5)$"},{"id":"C","text":"$(0, \\pm3)$"},{"id":"D","text":"$(\\pm5, 0)$"}],"answer":"C","explanation":"Here $25 > 16$, so $a^{2} = 25$ is under $y^{2}$ and the major axis lies on the y-axis. Then $c^{2} = a^{2} − b^{2} = 25 − 16 = 9, c = 3$, and the foci are $(0, \\pm3)$. Option A has the right value of c but puts the foci on the wrong axis.","explanationZh":"$因 25 > 16，a^{2} = 25 在 y^{2}$ 下方，长轴在 $y 轴上。c^{2} = a^{2} − b^{2} = 25 − 16 = 9，c = 3，焦点为 (0, \\pm3)。A 项 c$ 值正确但焦点放错了坐标轴。","tags":["medium","ellipse","major-axis"],"status":"published"},{"id":"math-conics-015","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"For the parabola $y^{2} = 12x$, find the distance from the focus to the directrix.","stemZh":"抛物线 $y^{2} = 12x$ 的焦点到准线的距离为多少？","options":[{"id":"A","text":"3"},{"id":"B","text":"12"},{"id":"C","text":"4"},{"id":"D","text":"6"}],"answer":"D","explanation":"$2p = 12$ gives $p = 6$. The focus is $(3, 0)$ and the directrix is $x = −3$, so the distance between them is $p = 6$. Option A is $p/2 (the$ focal distance from the vertex) and option B is 2p.","explanationZh":"$由 2p = 12 得 p = 6。焦点为 (3, 0)，准线为 x = −3，二者距离为 p = 6。A$ 项是 p/2（焦点到顶点的距离），B 项是 2p。","tags":["medium","parabola","directrix"],"status":"published"},{"id":"math-conics-016","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"P is a point on the parabola $y^{2} = 8x$ with x-coordinate 2, and F is its focus. Find $|PF|$.","stemZh":"点 P 在抛物线 $y^{2} = 8x$ 上，横坐标为 2，F 为焦点，求 $|PF|。$","options":[{"id":"A","text":"2"},{"id":"B","text":"4"},{"id":"C","text":"6"},{"id":"D","text":"8"}],"answer":"B","explanation":"By the definition of $a parabola, |PF|$ equals the distance from P to the directrix $x = −2$, which is $2 − (−2) = 4. Directly: y^{2} = 16$ gives $P = (2, 4)$ or $(2, −4)$, and the distance to $F(2, 0)$ is 4.","explanationZh":"$由抛物线定义，|PF|$ 等于 P 到准线 $x = −2$ 的距离，即 $2 − (−2) = 4。直接计算：y^{2} = 16，P 为 (2, 4) 或 (2, −4)，到 F(2, 0)$ 的距离也是 4。","tags":["medium","parabola","focus"],"status":"published"},{"id":"math-conics-017","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"An ellipse has $a = 5$ and $b = 3$. Find the distance between its two foci.","stemZh":"椭圆的 $a = 5，b = 3，求两焦点间的距离。$","options":[{"id":"A","text":"8"},{"id":"B","text":"4"},{"id":"C","text":"16"},{"id":"D","text":"10"}],"answer":"A","explanation":"$c^{2} = a^{2} − b^{2} = 25 − 9 = 16$, so $c = 4$. The distance between the foci (the focal distance) is $2c = 8$, not $c = 4$. Option D is 2a, the length of the major axis.","explanationZh":"$c^{2} = a^{2} − b^{2} = 25 − 9 = 16，c = 4。两焦点间的距离（焦距）为 2c = 8，而不是 c = 4。D 项 2a = 10$ 是长轴长。","tags":["medium","ellipse","focal-distance"],"status":"published"},{"id":"math-conics-018","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"For the hyperbola $x^{2}/4 − y^{2}/9 = 1$, what is the length of the real axis?","stemZh":"双曲线 $x^{2}/4 − y^{2}/9 = 1$ 的实轴长为多少？","options":[{"id":"A","text":"2"},{"id":"B","text":"6"},{"id":"C","text":"9"},{"id":"D","text":"4"}],"answer":"D","explanation":"$a^{2} = 4$ gives $a = 2$, so the real axis has length $2a = 4$. The imaginary axis has length $2b = 6 (option B)$. Do not confuse the axis length with a itself.","explanationZh":"$a^{2} = 4，a = 2，故实轴长为 2a = 4。虚轴长为 2b = 6（B$ 项）。注意区分轴长与半轴长 a。","tags":["medium","hyperbola","real-axis"],"status":"published"},{"id":"math-conics-019","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"Which of the following points lies on the ellipse $x^{2}/25 + y^{2}/16 = 1$?","stemZh":"下列哪个点在椭圆 $x^{2}/25 + y^{2}/16 = 1$ 上？","options":[{"id":"A","text":"$(3, 4)$"},{"id":"B","text":"$(5, 0)$"},{"id":"C","text":"$(4, 3)$"},{"id":"D","text":"$(2, 2)$"}],"answer":"B","explanation":"Substituting $(5, 0)$ gives $25/25 + 0 = 1$, so it lies on the ellipse (it is a vertex). Checking the $others: (3, 4)$ gives $9/25 + 1 = 34/25 ≠ 1, (4, 3)$ gives $16/25 + 9/16 ≠ 1$, and $(2, 2)$ gives $4/25 + 4/16 ≠ 1$.","explanationZh":"代入 $(5, 0)：25/25 + 0 = 1，该点在椭圆上（是长轴端点）。其余：(3, 4) 得 9/25 + 1 = 34/25 ≠ 1；(4, 3) 得 16/25 + 9/16 ≠ 1；(2, 2) 得 4/25 + 4/16 ≠ 1。$","tags":["medium","ellipse","standard-equation"],"status":"published"},{"id":"math-conics-020","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"Which of the following could be the eccentricity of an ellipse?","stemZh":"下列哪个数可能是椭圆的离心率？","options":[{"id":"A","text":"3/5"},{"id":"B","text":"1"},{"id":"C","text":"5/3"},{"id":"D","text":"4/3"}],"answer":"A","explanation":"For an ellipse $e = c/a$ with $0 < c < a$, so $0 < e < 1$. Only 3/5 lies in this $range: e = 1$ corresponds to a parabola and $e > 1 (options C\\text{ and }D)$ to a hyperbola.","explanationZh":"椭圆的 $e = c/a，且 0 < c <$ a，故 $0 < e < 1。只有 3/5$ 在此范围内：e $= 1$ 是抛物线，e $> 1（C、D$ 两项）是双曲线。","tags":["medium","eccentricity","ellipse"],"status":"published"},{"id":"math-conics-021","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":5,"stem":"The equation $x^{2}/m + y^{2}/4 = 1$ represents an ellipse with eccentricity 1/2. Find all possible values of m.","stemZh":"方程 $x^{2}/m + y^{2}/4 = 1$ 表示离心率为 1/2 的椭圆，求 m 的所有可能值。","options":[{"id":"A","text":"3 only"},{"id":"B","text":"16/3 only"},{"id":"C","text":"3 or 16/3"},{"id":"D","text":"4"}],"answer":"C","explanation":"Two cases. If $m > 4$ the major axis is on the $x-axis: e^{2} = (m − 4)/m = 1/4 → 4m − 16 = m → m = 16/3$. If $0 < m < 4$ the major axis is on the $y-axis: e^{2} = (4 − m)/4 = 1/4 → m = 3$. Both are valid, so $m = 3$ or $m = 16/3$.","explanationZh":"分两种情况：若 $m > 4，长轴在 x 轴上，e^{2} = (m − 4)/m = 1/4，得 m = 16/3；若 0 < m < 4，长轴在 y 轴上，e^{2} = (4 − m)/4 = 1/4，得 m = 3。两者都成立，故 m = 3 或 16/3。$","tags":["hard","ellipse","eccentricity"],"status":"published"},{"id":"math-conics-022","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":4,"stem":"The hyperbola $x^{2}/9 − y^{2}/m = 1$ has asymptotes $y = \\pm(2/3)x$. Find m.","stemZh":"双曲线 $x^{2}/9 − y^{2}/m = 1$ 的渐近线为 $y = \\pm(2/3)x，求$ m。","options":[{"id":"A","text":"6"},{"id":"B","text":"4"},{"id":"C","text":"2"},{"id":"D","text":"36"}],"answer":"B","explanation":"Here $a^{2} = 9$ so $a = 3$, and $b^{2} =$ m. The asymptotes are $y = \\pm(b/a)x = \\pm(\\sqrt{m}/3)x$. Setting $\\sqrt{m}/3 = 2/3$ gives $\\sqrt{m} = 2$, so $m = 4$. Option C is the value of b rather than $b^{2}$.","explanationZh":"此处 $a^{2} = 9，a = 3，b^{2} =$ m。渐近线为 $y = \\pm(b/a)x = \\pm(\\sqrt{m}/3)x。令 \\sqrt{m}/3 = 2/3 得 \\sqrt{m} = 2，故 m = 4。C$ 项是 b 而不是 $b^{2}。$","tags":["hard","hyperbola","asymptote"],"status":"published"},{"id":"math-conics-023","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":4,"stem":"Find the equation of the parabola with focus $(0, 2)$ and directrix $y = −2$.","stemZh":"求焦点为 $(0, 2)、准线为 y = −2$ 的抛物线方程。","options":[{"id":"A","text":"$y^{2} = 8x$"},{"id":"B","text":"$x^{2} = −8y$"},{"id":"C","text":"$x^{2} = 4y$"},{"id":"D","text":"$x^{2} = 8y$"}],"answer":"D","explanation":"The vertex is midway between the focus and the directrix, i.e. the origin, and the parabola opens upward along the y-axis. The distance from the focus to the directrix is $p = 4$, and for $x^{2} = 2py$ we get $x^{2} = 8y (check:\\text{ focus }(0, p/2) = (0, 2))$.","explanationZh":"顶点在焦点与准线中间，即原点；开口沿 y 轴向上。焦点到准线的距离 $p = 4，标准方程 x^{2} = 2py 即 x^{2} = 8y（验证：焦点 (0, p/2) = (0, 2)）。$","tags":["hard","parabola","directrix"],"status":"published"},{"id":"math-conics-024","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":5,"stem":"P is a point on the right branch of the hyperbola $x^{2}/9 − y^{2}/16 = 1$. If $|PF_{2}| = 10$, where $F_{2}$ is the right focus, find $|PF_{1}|$ for the left focus $F_{1}$.","stemZh":"点 P 在双曲线 $x^{2}/9 − y^{2}/16 = 1 的右支上，F_{2}$ 为右焦点，若 $|PF_{2}| = 10，求 |PF_{1}|。$","options":[{"id":"A","text":"16"},{"id":"B","text":"4"},{"id":"C","text":"6"},{"id":"D","text":"14"}],"answer":"A","explanation":"By the definition of $a hyperbola, ||PF_{1}| − |PF_{2}|| = 2a = 6$. On the right branch P is closer to $F_{2}$, so $|PF_{1}| − |PF_{2}| = 6$ and $|PF_{1}| = 10 + 6 = 16$. Option B is the value obtained on the left branch, and option C is 2a itself.","explanationZh":"由双曲线定义 $||PF_{1}| − |PF_{2}|| = 2a = 6。P$ 在右支上时离 $F_{2}$ 更近，故 $|PF_{1}| − |PF_{2}| = 6，|PF_{1}| = 10 + 6 = 16。B$ 项是左支上的结果，C 项是 2a 本身。","tags":["hard","hyperbola","focal-distance"],"status":"published"},{"id":"math-conics-025","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":4,"stem":"P is a point on the ellipse $x^{2}/25 + y^{2}/16 = 1$ with foci $F_{1}$ and $F_{2}$. Find the perimeter of triangle $PF_{1}F_{2}$.","stemZh":"点 P 在椭圆 $x^{2}/25 + y^{2}/16 = 1 上，F_{1}、F_{2}$ 为焦点，求 $△PF_{1}F_{2}$ 的周长。","options":[{"id":"A","text":"10"},{"id":"B","text":"6"},{"id":"C","text":"16"},{"id":"D","text":"12"}],"answer":"C","explanation":"By the definition of an $ellipse, |PF_{1}| + |PF_{2}| = 2a = 10$. Also $c^{2} = 25 − 16 = 9$ so $c = 3$ and $|F_{1}F_{2}| = 2c = 6$. The perimeter is $10 + 6 = 16$.","explanationZh":"由椭圆定义 $|PF_{1}| + |PF_{2}| = 2a = 10；又 c^{2} = 25 − 16 = 9，c = 3，|F_{1}F_{2}| = 2c = 6。故周长为 10 + 6 = 16。$","tags":["hard","ellipse","focal-distance"],"status":"published"},{"id":"math-conics-026","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":5,"stem":"Find the equation of the tangent to the parabola $y^{2} = 4x$ at the point $(1, 2)$.","stemZh":"求抛物线 $y^{2} = 4x$ 在点 $(1, 2)$ 处的切线方程。","options":[{"id":"A","text":"$y = x + 1$"},{"id":"B","text":"$y = x − 1$"},{"id":"C","text":"$y =$ −x + 3"},{"id":"D","text":"$y = 2x$"}],"answer":"A","explanation":"First $check: 2^{2} = 4(1)$, so $(1, 2)$ is on the parabola. Substituting $y = x + 1$ into $y^{2} = 4x$ gives $(x + 1)^{2} = 4x → x^{2} − 2x + 1 = 0 → (x − 1)^{2} = 0, a$ double root, so the line touches the curve at $x = 1. (Equivalently$, for $y^{2} = 2px$ the tangent at $(x_{0}, y_{0})$ is $y_{0}y = p(x + x_{0}).)$","explanationZh":"先验证 $2^{2} = 4 \\times 1，(1, 2)$ 在抛物线上。把 $y = x + 1$ 代入 $y^{2} = 4x 得 (x + 1)^{2} = 4x，即 (x − 1)^{2} = 0，有二重根，说明直线与曲线相切于 x = 1。（也可用切线公式 y_{0}y = p(x + x_{0})。）$","tags":["hard","parabola","tangent"],"status":"published"},{"id":"math-conics-027","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":5,"stem":"A hyperbola $x^{2}/a^{2} − y^{2}/b^{2} = 1 (a > 0, b > 0)$ has eccentricity 2. Find the acute angle between its two asymptotes.","stemZh":"双曲线 $x^{2}/a^{2} − y^{2}/b^{2} = 1（a > 0，b > 0）的离心率为 2，求两条渐近线的锐夹角。$","options":[{"id":"A","text":"30°"},{"id":"B","text":"90°"},{"id":"C","text":"120°"},{"id":"D","text":"60°"}],"answer":"D","explanation":"$e = c/a = 2$ gives $c = 2a$, so $b^{2} = c^{2} − a^{2} = 3a^{2}$ and $b/a = \\sqrt{3}$. The asymptotes $y = \\pm\\sqrt{3}x$ make angles of $\\pm60°$ with the x-axis, so the acute angle between them is $60° (the$ supplementary angle is $120°)$.","explanationZh":"$e = c/a = 2，即 c = 2a，故 b^{2} = c^{2} − a^{2} = 3a^{2}，b/a = \\sqrt{3}。渐近线 y = \\pm\\sqrt{3}x 与 x$ 轴夹角为 $\\pm60°，故两渐近线的锐夹角为 60°（另一角为 120°）。$","tags":["hard","hyperbola","asymptote"],"status":"published"},{"id":"math-conics-028","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":4,"stem":"The ellipse $x^{2}/25 + y^{2}/16 = 1$ and the hyperbola $x^{2}/4 − y^{2}/m = 1$ have the same foci. Find m.","stemZh":"椭圆 $x^{2}/25 + y^{2}/16 = 1$ 与双曲线 $x^{2}/4 − y^{2}/m = 1$ 有相同的焦点，求 m。","options":[{"id":"A","text":"4"},{"id":"B","text":"9"},{"id":"C","text":"5"},{"id":"D","text":"13"}],"answer":"C","explanation":"For the $ellipse, c^{2} = 25 − 16 = 9$, so the foci are $(\\pm3, 0)$. For the $hyperbola, c^{2} = a^{2} + b^{2} = 4 + m$, and this must equal 9, so $m = 5$. The key is that the two conics use different relations linking a, b and c.","explanationZh":"椭圆中 $c^{2} = 25 − 16 = 9，焦点为 (\\pm3, 0)。双曲线中 c^{2} = a^{2} + b^{2} = 4 +$ m，令其等于 $9 得 m = 5。关键在于椭圆与双曲线中$ a、b、c 的关系式不同。","tags":["hard","hyperbola","focal-distance"],"status":"published"},{"id":"math-conics-029","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":4,"stem":"For which values of k does the equation $x^{2}/k + y^{2}/4 = 1$ represent a hyperbola?","stemZh":"当 k 取何值时，方程 $x^{2}/k + y^{2}/4 = 1$ 表示双曲线？","options":[{"id":"A","text":"$k > 0$"},{"id":"B","text":"$k < 0$"},{"id":"C","text":"$k > 4$"},{"id":"D","text":"$0 < k < 4$"}],"answer":"B","explanation":"A hyperbola needs the two squared terms to have opposite signs. Since the coefficient of $y^{2}$ is positive $(4 > 0)$, we need $1/k < 0$, i.e. $k < 0$; the equation then reads $y^{2}/4 − x^{2}/(−k) = 1$. For $k > 0$ it is an ellipse.","explanationZh":"$双曲线要求两个平方项系数异号。y^{2} 的系数为正（4 > 0），故需 1/k < 0，即 k < 0，此时方程化为 y^{2}/4 − x^{2}/(−k) = 1。k > 0$ 时为椭圆。","tags":["hard","hyperbola","standard-equation"],"status":"published"},{"id":"math-conics-030","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":5,"stem":"A parabola has its vertex at the origin, the x-axis as its axis of symmetry, opens to the right, and passes through $(2, 4)$. Find the distance from its focus to its directrix.","stemZh":"抛物线顶点在原点，对称轴为 x 轴，开口向右，且过点 $(2, 4)，求其焦点到准线的距离。$","options":[{"id":"A","text":"2"},{"id":"B","text":"8"},{"id":"C","text":"16"},{"id":"D","text":"4"}],"answer":"D","explanation":"The form is $y^{2} = 2px$ with $p > 0$. Substituting $(2, 4): 16 = 2p(2) = 4p$, so $p = 4$. The focus is $(2, 0)$, the directrix is $x = −2$, and the distance between them is $p = 4$. Option B is 2p and option A is p/2.","explanationZh":"设方程为 $y^{2} = 2px（p > 0）。代入 (2, 4)：16 = 2p \\times 2 = 4p，得 p = 4。焦点为 (2, 0)，准线为 x = −2，二者距离为 p = 4。B$ 项是 2p，A 项是 p/2。","tags":["hard","parabola","directrix"],"status":"published"},{"id":"math-vectors-001","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":1,"stem":"Let $\\overrightarrow{a} = (3, −4)$. Find $|\\overrightarrow{a}|$.","stemZh":"已知 $\\overrightarrow{a} = (3, −4)，求 |\\overrightarrow{a}|。$","options":[{"id":"A","text":"5"},{"id":"B","text":"7"},{"id":"C","text":"1"},{"id":"D","text":"$\\sqrt{7}$"}],"answer":"A","explanation":"$|\\overrightarrow{a}| = \\sqrt{3^{2} + (−4)^{2}} = \\sqrt{25} = 5.$","explanationZh":"$|\\overrightarrow{a}| = \\sqrt{3^{2} + (−4)^{2}} = \\sqrt{25} = 5。}$","tags":["easy","magnitude"],"status":"published"},{"id":"math-vectors-002","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":1,"stem":"Let $\\overrightarrow{a} = (1, 2)$ and $\\overrightarrow{b} = (3, −1)$. Find $\\overrightarrow{a} + \\overrightarrow{b}$.","stemZh":"已知 $\\overrightarrow{a} = (1, 2)，\\overrightarrow{b} = (3, −1)，求 \\overrightarrow{a} + \\overrightarrow{b}。$","options":[{"id":"A","text":"$(4, 3)$"},{"id":"B","text":"$(4, 1)$"},{"id":"C","text":"$(2, 1)$"},{"id":"D","text":"$(−2, 3)$"}],"answer":"B","explanation":"Add $coordinates: \\overrightarrow{a} + \\overrightarrow{b} = (1 + 3, 2 + (−1)) = (4, 1)$. Option D is $\\overrightarrow{a} − \\overrightarrow{b}$.","explanationZh":"$坐标相加：\\overrightarrow{a} + \\overrightarrow{b} = (1 + 3, 2 + (−1)) = (4, 1)。D$ 选项是 $\\overrightarrow{a} − \\overrightarrow{b}。$","tags":["easy","addition"],"status":"published"},{"id":"math-vectors-003","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":1,"stem":"Which statement about plane vectors is correct?","stemZh":"下列关于平面向量的说法正确的是？","options":[{"id":"A","text":"$|\\overrightarrow{a}| = |\\overrightarrow{b}|$ implies $\\overrightarrow{a} = \\overrightarrow{b}$"},{"id":"B","text":"$\\overrightarrow{a} ∥ \\overrightarrow{b}$ implies $|\\overrightarrow{a}| = |\\overrightarrow{b}|$"},{"id":"C","text":"Equal vectors have equal magnitude and the same direction"},{"id":"D","text":"Two vectors with the same direction are equal"}],"answer":"C","explanation":"Two vectors are equal exactly when they have the same magnitude AND the same direction. Equal magnitude alone (A) or equal direction alone (D) is not enough; parallel vectors (B) need not have equal length.","explanationZh":"两向量相等当且仅当大小相等且方向相同。仅有大小相等（A）或仅有方向相同（D）都不够；平行向量（B）长度不一定相等。","tags":["easy","concept"],"status":"published"},{"id":"math-vectors-004","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":1,"stem":"Which statement about the zero vector $\\overrightarrow{0}$ is correct?","stemZh":"关于零向量 $\\overrightarrow{0}，下列说法正确的是？$","options":[{"id":"A","text":"The zero vector has magnitude 1"},{"id":"B","text":"The zero vector is perpendicular to every vector"},{"id":"C","text":"$|\\overrightarrow{0}|$ is not defined"},{"id":"D","text":"$|\\overrightarrow{0}| = 0$"}],"answer":"D","explanation":"The zero vector has length 0, so $|\\overrightarrow{0}| = 0$; its direction is regarded as arbitrary.","explanationZh":"零向量的长度为 $0，即 |\\overrightarrow{0}| = 0，其方向规定为任意的。$","tags":["easy","zero-vector"],"status":"published"},{"id":"math-vectors-005","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":2,"stem":"Let $\\overrightarrow{a} = (2, −3)$ and $\\overrightarrow{b} = (−1, 5)$. Find $\\overrightarrow{a} − \\overrightarrow{b}$.","stemZh":"已知 $\\overrightarrow{a} = (2, −3)，\\overrightarrow{b} = (−1, 5)，求 \\overrightarrow{a} − \\overrightarrow{b}。$","options":[{"id":"A","text":"$(3, −8)$"},{"id":"B","text":"$(3, 8)$"},{"id":"C","text":"$(1, −8)$"},{"id":"D","text":"$(−3, −8)$"}],"answer":"A","explanation":"$\\overrightarrow{a} − \\overrightarrow{b} = (2 − (−1), −3 − 5) = (3, −8)$. Watch the double signs.","explanationZh":"$\\overrightarrow{a} − \\overrightarrow{b} = (2 − (−1), −3 − 5) = (3, −8)，注意符号。$","tags":["easy","subtraction"],"status":"published"},{"id":"math-vectors-006","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":2,"stem":"Let $\\overrightarrow{a} = (−2, 4)$. Find $−\\overrightarrow{2a}$.","stemZh":"已知 $\\overrightarrow{a} = (−2, 4)，求 −\\overrightarrow{2a}。$","options":[{"id":"A","text":"$(4, 8)$"},{"id":"B","text":"$(4, −8)$"},{"id":"C","text":"$(−4, 8)$"},{"id":"D","text":"$(−4, −8)$"}],"answer":"B","explanation":"Multiply both coordinates by $−2: −\\overrightarrow{2a} = (−2·(−2), −2·4) = (4, −8)$.","explanationZh":"两个坐标都乘 $−2：−\\overrightarrow{2a} = (−2·(−2), −2·4) = (4, −8)。$","tags":["easy","scalar-multiplication"],"status":"published"},{"id":"math-vectors-007","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":2,"stem":"Let $\\overrightarrow{a} = (2, m)$ and $\\overrightarrow{b} = (4, 6)$. If $\\overrightarrow{a} ∥ \\overrightarrow{b}$, find m.","stemZh":"已知 $\\overrightarrow{a} = (2, m)，\\overrightarrow{b} = (4, 6)，若 \\overrightarrow{a} ∥ \\overrightarrow{b}，求$ m。","options":[{"id":"A","text":"−3"},{"id":"B","text":"12"},{"id":"C","text":"3"},{"id":"D","text":"4"}],"answer":"C","explanation":"For $\\overrightarrow{a} = (x_{1}, y_{1})$ and $\\overrightarrow{b} = (x_{2}, y_{2}), \\overrightarrow{a} ∥ \\overrightarrow{b} ⇔ x_{1}y_{2} − x_{2}y_{1} = 0$, so $2·6 − 4m = 0$ and $m = 3$.","explanationZh":"$\\overrightarrow{a} = (x_{1}, y_{1})，\\overrightarrow{b} = (x_{2}, y_{2})，\\overrightarrow{a} ∥ \\overrightarrow{b} ⇔ x_{1}y_{2} − x_{2}y_{1} = 0，故 2·6 − 4m = 0，m = 3。$","tags":["easy","collinear"],"status":"published"},{"id":"math-vectors-008","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":2,"stem":"Let $\\overrightarrow{a} = (1, 2)$ and $\\overrightarrow{b} = (2, −1)$. Find $\\overrightarrow{a} · \\overrightarrow{b}$.","stemZh":"已知 $\\overrightarrow{a} = (1, 2)，\\overrightarrow{b} = (2, −1)，求 \\overrightarrow{a} · \\overrightarrow{b}。$","options":[{"id":"A","text":"4"},{"id":"B","text":"3"},{"id":"C","text":"1"},{"id":"D","text":"0"}],"answer":"D","explanation":"$\\overrightarrow{a} · \\overrightarrow{b} = 1·2 + 2·(−1) = 2 − 2 = 0$, so $\\overrightarrow{a} ⟂ \\overrightarrow{b}$.","explanationZh":"$\\overrightarrow{a} · \\overrightarrow{b} = 1·2 + 2·(−1) = 0，因此 \\overrightarrow{a} ⟂ \\overrightarrow{b}。$","tags":["easy","dot-product"],"status":"published"},{"id":"math-vectors-009","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":1,"stem":"Given $A(1, 2)$ and $B(4, 6)$, find the coordinate representation of $\\overrightarrow{AB}$.","stemZh":"已知 $A(1, 2)，B(4, 6)，求 \\overrightarrow{AB}$ 的坐标。","options":[{"id":"A","text":"$(3, 4)$"},{"id":"B","text":"$(−3, −4)$"},{"id":"C","text":"$(5, 8)$"},{"id":"D","text":"$(1, 2)$"}],"answer":"A","explanation":"$\\overrightarrow{AB} = (x_B − x_A, y_B − y_A) = (4 − 1, 6 − 2) = (3, 4)$. Option B is $\\overrightarrow{BA}$.","explanationZh":"$\\overrightarrow{AB} = (x_B − x_A, y_B − y_A) = (4 − 1, 6 − 2) = (3, 4)。B$ 选项是 $\\overrightarrow{BA}。$","tags":["easy","coordinates"],"status":"published"},{"id":"math-vectors-010","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":2,"stem":"Let $\\overrightarrow{a} = (3, 4)$. Find the unit vector in the same direction as $\\overrightarrow{a}$.","stemZh":"已知 $\\overrightarrow{a} = (3, 4)，求与 \\overrightarrow{a}$ 同方向的单位向量。","options":[{"id":"A","text":"$(3, 4)$"},{"id":"B","text":"$(3/5, 4/5)$"},{"id":"C","text":"$(4/5, 3/5)$"},{"id":"D","text":"$(1, 0)$"}],"answer":"B","explanation":"$|\\overrightarrow{a}| = 5$, so the unit vector is $\\overrightarrow{a}/|\\overrightarrow{a}| = (3/5, 4/5)$.","explanationZh":"$|\\overrightarrow{a}| = 5，故单位向量为 \\overrightarrow{a}/|\\overrightarrow{a}| = (3/5, 4/5)。$","tags":["easy","unit-vector"],"status":"published"},{"id":"math-vectors-011","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Let $|\\overrightarrow{a}| = 2, |\\overrightarrow{b}| = 1$ and $\\overrightarrow{a} · \\overrightarrow{b} = 1$. Find the angle $θ$ between $\\overrightarrow{a}$ and $\\overrightarrow{b}$.","stemZh":"已知 $|\\overrightarrow{a}| = 2，|\\overrightarrow{b}| = 1，\\overrightarrow{a} · \\overrightarrow{b} = 1，求 \\overrightarrow{a} 与 \\overrightarrow{b}$ 的夹角 $θ。$","options":[{"id":"A","text":"30°"},{"id":"B","text":"45°"},{"id":"C","text":"60°"},{"id":"D","text":"120°"}],"answer":"C","explanation":"$\\cos θ = (\\overrightarrow{a} · \\overrightarrow{b})/(|\\overrightarrow{a}||\\overrightarrow{b}|) = 1/(2·1) = 1/2$, so $θ = 60°. A$ negative dot product would give 120°.","explanationZh":"$\\cos θ = (\\overrightarrow{a} · \\overrightarrow{b})/(|\\overrightarrow{a}||\\overrightarrow{b}|) = 1/(2·1) = 1/2，故 θ = 60°。若数量积为负才得 120°。$","tags":["medium","angle","dot-product"],"status":"published"},{"id":"math-vectors-012","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Let $\\overrightarrow{a} = (1, 2)$ and $\\overrightarrow{b} = (−2, 1)$. Find the angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$.","stemZh":"已知 $\\overrightarrow{a} = (1, 2)，\\overrightarrow{b} = (−2, 1)，求 \\overrightarrow{a} 与 \\overrightarrow{b}$ 的夹角。","options":[{"id":"A","text":"0°"},{"id":"B","text":"30°"},{"id":"C","text":"60°"},{"id":"D","text":"90°"}],"answer":"D","explanation":"$\\overrightarrow{a} · \\overrightarrow{b} = 1·(−2) + 2·1 = 0$, so $\\cos θ = 0$ and the vectors are $perpendicular: θ = 90°$.","explanationZh":"$\\overrightarrow{a} · \\overrightarrow{b} = 1·(−2) + 2·1 = 0，故 \\cos θ = 0，两向量垂直，夹角为 90°。$","tags":["medium","perpendicular","dot-product"],"status":"published"},{"id":"math-vectors-013","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Given $A(1, 1), B(3, 5)$ and $C(4, 7)$, which statement is correct?","stemZh":"已知 $A(1, 1)，B(3, 5)，C(4, 7)，下列结论正确的是？$","options":[{"id":"A","text":"A, B, C are collinear"},{"id":"B","text":"$|\\overrightarrow{AB}| = |\\overrightarrow{BC}|$"},{"id":"C","text":"B is the midpoint of AC"},{"id":"D","text":"$\\overrightarrow{AB} · \\overrightarrow{AC} = 0$"}],"answer":"A","explanation":"$\\overrightarrow{AB} = (2, 4)$ and $\\overrightarrow{AC} = (3, 6)$. Since $2·6 − 3·4 = 0$, the vectors are parallel, and they share the point A, so A, B, C are collinear. Also $|\\overrightarrow{AB}| = \\sqrt{20} ≠ \\sqrt{5} = |\\overrightarrow{BC}|$.","explanationZh":"$\\overrightarrow{AB} = (2, 4)，\\overrightarrow{AC} = (3, 6)，2·6 − 3·4 = 0，两向量平行且共点$ A，故 A、B、C 三点共线；且 $|\\overrightarrow{AB}| = \\sqrt{20} ≠ \\sqrt{5} = |\\overrightarrow{BC}|。$","tags":["medium","collinearity"],"status":"published"},{"id":"math-vectors-014","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Let $\\overrightarrow{a} = (2, 1)$ and $\\overrightarrow{b} = (1, λ)$. If $\\overrightarrow{a} ⟂ \\overrightarrow{b}$, find $λ$.","stemZh":"已知 $\\overrightarrow{a} = (2, 1)，\\overrightarrow{b} = (1, λ)，若 \\overrightarrow{a} ⟂ \\overrightarrow{b}，求 λ。$","options":[{"id":"A","text":"2"},{"id":"B","text":"−2"},{"id":"C","text":"1/2"},{"id":"D","text":"−1/2"}],"answer":"B","explanation":"$\\overrightarrow{a} ⟂ \\overrightarrow{b} ⇔ \\overrightarrow{a} · \\overrightarrow{b} = 2·1 + 1·λ = 0$, so $λ = −2 (the\\text{ sign is the }trap)$.","explanationZh":"$\\overrightarrow{a} ⟂ \\overrightarrow{b} ⇔ \\overrightarrow{a} · \\overrightarrow{b} = 2·1 + 1·λ = 0，故 λ = −2，注意符号。$","tags":["medium","perpendicular","dot-product"],"status":"published"},{"id":"math-vectors-015","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"In triangle ABC, M is the midpoint of BC. Express $\\overrightarrow{AM}$ in terms of $\\overrightarrow{AB}$ and $\\overrightarrow{AC}$.","stemZh":"在 △ABC 中，M 为 BC 的中点，用 $\\overrightarrow{AB}、\\overrightarrow{AC}$ 表示 $\\overrightarrow{AM}。$","options":[{"id":"A","text":"$\\overrightarrow{AB} + \\overrightarrow{AC}$"},{"id":"B","text":"$(1/2)(\\overrightarrow{AB} − \\overrightarrow{AC})$"},{"id":"C","text":"$(1/2)(\\overrightarrow{AB} + \\overrightarrow{AC})$"},{"id":"D","text":"$\\overrightarrow{AB} − \\overrightarrow{AC}$"}],"answer":"C","explanation":"By the parallelogram law, if D completes the parallelogram ABDC then the diagonal $\\overrightarrow{AD} = \\overrightarrow{AB} + \\overrightarrow{AC}$ passes through the midpoint M of BC, so $\\overrightarrow{AM} = (1/2)(\\overrightarrow{AB} + \\overrightarrow{AC})$.","explanationZh":"$由平行四边形法则，\\overrightarrow{AD} = \\overrightarrow{AB} + \\overrightarrow{AC}，其中$ AD 过 BC 中点 M，故 $\\overrightarrow{AM} = (1/2)(\\overrightarrow{AB} + \\overrightarrow{AC})。$","tags":["medium","midpoint","addition"],"status":"published"},{"id":"math-vectors-016","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Let $\\overrightarrow{a} = (1, −2)$ and $\\overrightarrow{b} = (3, 4)$. Find $|\\overrightarrow{2a} − \\overrightarrow{b}|$.","stemZh":"已知 $\\overrightarrow{a} = (1, −2)，\\overrightarrow{b} = (3, 4)，求 |\\overrightarrow{2a} − \\overrightarrow{b}|。$","options":[{"id":"A","text":"$\\sqrt{5}$"},{"id":"B","text":"5"},{"id":"C","text":"$\\sqrt{89}$"},{"id":"D","text":"$\\sqrt{65}$"}],"answer":"D","explanation":"$\\overrightarrow{2a} − \\overrightarrow{b} = (2, −4) − (3, 4) = (−1, −8)$, so $|\\overrightarrow{2a} − \\overrightarrow{b}| = \\sqrt{1 + 64} = \\sqrt{65}.$","explanationZh":"$\\overrightarrow{2a} − \\overrightarrow{b} = (2, −4) − (3, 4) = (−1, −8)，故 |\\overrightarrow{2a} − \\overrightarrow{b}| = \\sqrt{1 + 64} = \\sqrt{65}。}$","tags":["medium","magnitude","scalar-multiplication"],"status":"published"},{"id":"math-vectors-017","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Which of the following is true for all plane vectors $\\overrightarrow{a}$ and $\\overrightarrow{b}$?","stemZh":"对任意平面向量 $\\overrightarrow{a}、\\overrightarrow{b}，下列恒成立的式子是？$","options":[{"id":"A","text":"$|\\overrightarrow{a} + \\overrightarrow{b}| ≤ |\\overrightarrow{a}| + |\\overrightarrow{b}|$"},{"id":"B","text":"$|\\overrightarrow{a} + \\overrightarrow{b}| = |\\overrightarrow{a}| + |\\overrightarrow{b}|$"},{"id":"C","text":"$|\\overrightarrow{a} · \\overrightarrow{b}| = |\\overrightarrow{a}||\\overrightarrow{b}|$"},{"id":"D","text":"$\\overrightarrow{a} · \\overrightarrow{b} = 0$ implies $\\overrightarrow{a} = \\overrightarrow{0}$ or $\\overrightarrow{b} = \\overrightarrow{0}$"}],"answer":"A","explanation":"The triangle inequality $|\\overrightarrow{a} + \\overrightarrow{b}| ≤ |\\overrightarrow{a}| + |\\overrightarrow{b}|$ always holds. Equality in B needs the same $direction; |\\overrightarrow{a} · \\overrightarrow{b}| = |\\overrightarrow{a}||\\overrightarrow{b}||\\cos θ| ≤ |\\overrightarrow{a}||\\overrightarrow{b}|$; and $\\overrightarrow{a} · \\overrightarrow{b} = 0$ also happens for perpendicular nonzero vectors.","explanationZh":"三角不等式 $|\\overrightarrow{a} + \\overrightarrow{b}| ≤ |\\overrightarrow{a}| + |\\overrightarrow{b}|$ 恒成立。B $取等号需同向；|\\overrightarrow{a} · \\overrightarrow{b}| = |\\overrightarrow{a}||\\overrightarrow{b}||\\cos θ| ≤ |\\overrightarrow{a}||\\overrightarrow{b}|；\\overrightarrow{a} · \\overrightarrow{b} = 0$ 也可能是两非零向量垂直。","tags":["medium","properties"],"status":"published"},{"id":"math-vectors-018","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Let $\\overrightarrow{a} = (3, 4)$. Find the vector of magnitude 10 that has the same direction as $\\overrightarrow{a}$.","stemZh":"已知 $\\overrightarrow{a} = (3, 4)，求与 \\overrightarrow{a}$ 同向且模为 10 的向量。","options":[{"id":"A","text":"$(30, 40)$"},{"id":"B","text":"$(6, 8)$"},{"id":"C","text":"$(6/5, 8/5)$"},{"id":"D","text":"$(3, 4)$"}],"answer":"B","explanation":"$|\\overrightarrow{a}| = 5$, so the required vector is $(10/5)\\overrightarrow{a} = \\overrightarrow{2a} = (6, 8)$. Using $\\overrightarrow{10a}$ gives the wrong length 50.","explanationZh":"$|\\overrightarrow{a}| = 5，所求向量为 (10/5)\\overrightarrow{a} = \\overrightarrow{2a} = (6, 8)。若直接取 \\overrightarrow{10a}，模长为 50，错误。$","tags":["medium","unit-vector","scalar-multiplication"],"status":"published"},{"id":"math-vectors-019","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"In parallelogram $ABCD, \\overrightarrow{AB} = \\overrightarrow{a}$ and $\\overrightarrow{AD} = \\overrightarrow{b}$. Find $\\overrightarrow{BD}$.","stemZh":"在平行四边形 ABCD $中，\\overrightarrow{AB} = \\overrightarrow{a}，\\overrightarrow{AD} = \\overrightarrow{b}，求 \\overrightarrow{BD}。$","options":[{"id":"A","text":"$\\overrightarrow{a} + \\overrightarrow{b}$"},{"id":"B","text":"$\\overrightarrow{a} − \\overrightarrow{b}$"},{"id":"C","text":"$\\overrightarrow{b} − \\overrightarrow{a}$"},{"id":"D","text":"$−\\overrightarrow{a} − \\overrightarrow{b}$"}],"answer":"C","explanation":"$\\overrightarrow{BD} = \\overrightarrow{AD} − \\overrightarrow{AB} = \\overrightarrow{b} − \\overrightarrow{a}$. The sum $\\overrightarrow{a} + \\overrightarrow{b}$ is the other diagonal $\\overrightarrow{AC}$.","explanationZh":"$\\overrightarrow{BD} = \\overrightarrow{AD} − \\overrightarrow{AB} = \\overrightarrow{b} − \\overrightarrow{a}；\\overrightarrow{a} + \\overrightarrow{b}$ 是另一条对角线 $\\overrightarrow{AC}。$","tags":["medium","parallelogram","subtraction"],"status":"published"},{"id":"math-vectors-020","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Let $\\overrightarrow{a} = (1, 2)$ and $\\overrightarrow{b} = (−3, 4)$. Find $\\cos θ$, where $θ$ is the angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$.","stemZh":"已知 $\\overrightarrow{a} = (1, 2)，\\overrightarrow{b} = (−3, 4)，求 \\overrightarrow{a} 与 \\overrightarrow{b}$ 夹角 $θ$ 的余弦。","options":[{"id":"A","text":"1/5"},{"id":"B","text":"2/5"},{"id":"C","text":"$\\sqrt{5}$"},{"id":"D","text":"$\\sqrt{5}/5$"}],"answer":"D","explanation":"$\\overrightarrow{a} · \\overrightarrow{b} = 1·(−3) + 2·4 = 5, |\\overrightarrow{a}| = \\sqrt{5}, |\\overrightarrow{b}| = 5$, so $\\cos θ = 5/(5\\sqrt{5}) = 1/\\sqrt{5} = \\sqrt{5}/5$.","explanationZh":"$\\overrightarrow{a} · \\overrightarrow{b} = 1·(−3) + 2·4 = 5，|\\overrightarrow{a}| = \\sqrt{5}，|\\overrightarrow{b}| = 5，故 \\cos θ = 5/(5\\sqrt{5}) = \\sqrt{5}/5。$","tags":["medium","angle","dot-product"],"status":"published"},{"id":"math-vectors-021","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":4,"stem":"Let $\\overrightarrow{a} = (k, 1)$ and $\\overrightarrow{b} = (4, k)$. Find all real k for which $\\overrightarrow{a} ∥ \\overrightarrow{b}$.","stemZh":"已知 $\\overrightarrow{a} = (k, 1)，\\overrightarrow{b} = (4, k)，求使 \\overrightarrow{a} ∥ \\overrightarrow{b}$ 的所有实数 k。","options":[{"id":"A","text":"$\\pm2$"},{"id":"B","text":"2"},{"id":"C","text":"−2"},{"id":"D","text":"4"}],"answer":"A","explanation":"$x_{1}y_{2} − x_{2}y_{1} = k·k − 4·1 = k^{2} − 4 = 0$ gives $k = 2$ or $k = −2$. Taking only $k = 2$ misses the opposite-direction case.","explanationZh":"$x_{1}y_{2} − x_{2}y_{1} = k·k − 4·1 = k^{2} − 4 = 0，得 k = 2 或 k = −2；只取 k = 2$ 会漏掉反向平行的情形。","tags":["hard","collinear"],"status":"published"},{"id":"math-vectors-022","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":4,"stem":"For nonzero vectors $\\overrightarrow{a}$ and $\\overrightarrow{b}, \\overrightarrow{a} · \\overrightarrow{b} < 0$. What can be concluded about the angle $θ$ between them?","stemZh":"非零向量 $\\overrightarrow{a}、\\overrightarrow{b}$ 满足 $\\overrightarrow{a} · \\overrightarrow{b} < 0，则它们的夹角 θ$ 满足？","options":[{"id":"A","text":"$θ$ is acute $(0° ≤ θ < 90°)$"},{"id":"B","text":"$θ$ is obtuse $(90° < θ ≤ 180°)$"},{"id":"C","text":"$θ = 90°$"},{"id":"D","text":"$\\overrightarrow{a}$ and $\\overrightarrow{b}$ have opposite directions"}],"answer":"B","explanation":"$\\overrightarrow{a} · \\overrightarrow{b} = |\\overrightarrow{a}||\\overrightarrow{b}|\\cos θ$ and $|\\overrightarrow{a}||\\overrightarrow{b}| > 0$, so $\\overrightarrow{a} · \\overrightarrow{b} < 0$ means $\\cos θ < 0$, i.e. $90° < θ ≤ 180°$: an obtuse angle. Opposite directions $(θ = 180°)$ is only one special case.","explanationZh":"$\\overrightarrow{a} · \\overrightarrow{b} = |\\overrightarrow{a}||\\overrightarrow{b}|\\cos θ，且 |\\overrightarrow{a}||\\overrightarrow{b}| > 0，故 \\overrightarrow{a} · \\overrightarrow{b} < 0$ 等价于 $\\cos θ < 0，即 90° < θ ≤ 180°，夹角为钝角；反向（θ = 180°）只是其中一个特例。$","tags":["hard","dot-product","angle"],"status":"published"},{"id":"math-vectors-023","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":4,"stem":"Which statement about plane vectors is correct?","stemZh":"下列关于平面向量的命题正确的是？","options":[{"id":"A","text":"If $\\overrightarrow{a} ∥ \\overrightarrow{b}$ and $\\overrightarrow{b} ∥ \\overrightarrow{c}$, then $\\overrightarrow{a} ∥ \\overrightarrow{c}$"},{"id":"B","text":"If $\\overrightarrow{a} · \\overrightarrow{b} = \\overrightarrow{a} · \\overrightarrow{c}$, then $\\overrightarrow{b} = \\overrightarrow{c}$"},{"id":"C","text":"The zero vector $\\overrightarrow{0}$ is parallel to every vector"},{"id":"D","text":"If $|\\overrightarrow{a} + \\overrightarrow{b}| = |\\overrightarrow{a} − \\overrightarrow{b}|$, then $\\overrightarrow{a} = \\overrightarrow{b}$"}],"answer":"C","explanation":"By definition $\\overrightarrow{0}$ is parallel to every vector. A fails when $\\overrightarrow{b} = \\overrightarrow{0} (then \\overrightarrow{a}\\text{ and }\\overrightarrow{c}\\text{ may be }arbitrary); B$ fails because $\\overrightarrow{a} · (\\overrightarrow{b} − \\overrightarrow{c}) = 0$ only gives perpendicularity or $\\overrightarrow{a} = \\overrightarrow{0}; D$ gives $\\overrightarrow{a} · \\overrightarrow{b} = 0$, i.e. $\\overrightarrow{a} ⟂ \\overrightarrow{b}$.","explanationZh":"规定零向量与任意向量平行。A $在 \\overrightarrow{b} = \\overrightarrow{0}$ 时不成立；B $中 \\overrightarrow{a} · (\\overrightarrow{b} − \\overrightarrow{c}) = 0$ 只能说明垂直或 $\\overrightarrow{a} = \\overrightarrow{0}；D$ 实际推出 $\\overrightarrow{a} · \\overrightarrow{b} = 0，即 \\overrightarrow{a} ⟂ \\overrightarrow{b}。$","tags":["hard","zero-vector","properties"],"status":"published"},{"id":"math-vectors-024","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":4,"stem":"Let $|\\overrightarrow{a}| = 1, |\\overrightarrow{b}| = 2$ and $\\overrightarrow{a} ⟂ \\overrightarrow{b}$. Find $|\\overrightarrow{a} − \\overrightarrow{2b}|$.","stemZh":"已知 $|\\overrightarrow{a}| = 1，|\\overrightarrow{b}| = 2，且 \\overrightarrow{a} ⟂ \\overrightarrow{b}，求 |\\overrightarrow{a} − \\overrightarrow{2b}|。$","options":[{"id":"A","text":"3"},{"id":"B","text":"$\\sqrt{5}$"},{"id":"C","text":"5"},{"id":"D","text":"$\\sqrt{17}$"}],"answer":"D","explanation":"$|\\overrightarrow{a} − \\overrightarrow{2b}|^{2} = |\\overrightarrow{a}|^{2} − \\overrightarrow{4a} · \\overrightarrow{b} + 4|\\overrightarrow{b}|^{2} = 1 − 0 + 16 = 17$, so $|\\overrightarrow{a} − \\overrightarrow{2b}| = \\sqrt{17}$. Note that the factor 2 must be squared to 4.","explanationZh":"$|\\overrightarrow{a} − \\overrightarrow{2b}|^{2} = |\\overrightarrow{a}|^{2} − \\overrightarrow{4a} · \\overrightarrow{b} + 4|\\overrightarrow{b}|^{2} = 1 − 0 + 16 = 17，故 |\\overrightarrow{a} − \\overrightarrow{2b}| = \\sqrt{17}；注意系数 2$ 平方后为 4。","tags":["hard","magnitude","dot-product"],"status":"published"},{"id":"math-vectors-025","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":4,"stem":"The points $A(1, 2), B(2, 5)$ and $C(4, k)$ are collinear. Find k.","stemZh":"已知 $A(1, 2)，B(2, 5)，C(4, k)$ 三点共线，求 k。","options":[{"id":"A","text":"11"},{"id":"B","text":"9"},{"id":"C","text":"7"},{"id":"D","text":"−7"}],"answer":"A","explanation":"$\\overrightarrow{AB} = (1, 3)$ and $\\overrightarrow{AC} = (3, k − 2)$. Collinearity gives $1·(k − 2) − 3·3 = 0$, so $k − 2 = 9$ and $k = 11$.","explanationZh":"$\\overrightarrow{AB} = (1, 3)，\\overrightarrow{AC} = (3, k − 2)，共线得 1·(k − 2) − 3·3 = 0，即 k − 2 = 9，k = 11。$","tags":["hard","collinearity"],"status":"published"},{"id":"math-vectors-026","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":5,"stem":"In triangle ABC, a point P satisfies $\\overrightarrow{AP} = (1/3)\\overrightarrow{AB} + (2/3)\\overrightarrow{AC}$. Which statement is true?","stemZh":"在 △ABC 中，点 P 满足 $\\overrightarrow{AP} = (1/3)\\overrightarrow{AB} + (2/3)\\overrightarrow{AC}，则下列说法正确的是？$","options":[{"id":"A","text":"P is the midpoint of BC"},{"id":"B","text":"P lies on BC and BP : PC $= 2 : 1$"},{"id":"C","text":"P lies on AB"},{"id":"D","text":"P is the centroid of triangle ABC"}],"answer":"B","explanation":"The coefficients add to 1, so P lies on line BC. $\\overrightarrow{BP} = \\overrightarrow{AP} − \\overrightarrow{AB} = (2/3)(\\overrightarrow{AC} − \\overrightarrow{AB}) = (2/3)\\overrightarrow{BC}$, hence BP : PC $= 2 : 1$. The centroid would be $(1/3)(\\overrightarrow{AB} + \\overrightarrow{AC})$.","explanationZh":"系数之和为 1，故 P 在直线 BC $上。\\overrightarrow{BP} = \\overrightarrow{AP} − \\overrightarrow{AB} = (2/3)(\\overrightarrow{AC} − \\overrightarrow{AB}) = (2/3)\\overrightarrow{BC}，所以$ BP : PC $= 2 : 1；重心对应的是 (1/3)(\\overrightarrow{AB} + \\overrightarrow{AC})。$","tags":["hard","collinearity","geometry-application"],"status":"published"},{"id":"math-vectors-027","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":5,"stem":"Nonzero vectors $\\overrightarrow{a}$ and $\\overrightarrow{b}$ satisfy $|\\overrightarrow{a}| = |\\overrightarrow{b}| = |\\overrightarrow{a} − \\overrightarrow{b}|$. Find the angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$.","stemZh":"非零向量 $\\overrightarrow{a}、\\overrightarrow{b}$ 满足 $|\\overrightarrow{a}| = |\\overrightarrow{b}| = |\\overrightarrow{a} − \\overrightarrow{b}|，求 \\overrightarrow{a} 与 \\overrightarrow{b}$ 的夹角。","options":[{"id":"A","text":"30°"},{"id":"B","text":"45°"},{"id":"C","text":"60°"},{"id":"D","text":"120°"}],"answer":"C","explanation":"Let $|\\overrightarrow{a}| = |\\overrightarrow{b}| =$ r. Squaring $|\\overrightarrow{a} − \\overrightarrow{b}| = r$ gives $r^{2} + r^{2} − 2r^{2}\\cos θ = r^{2}$, so $\\cos θ = 1/2$ and $θ = 60° (the$ three vectors form an equilateral triangle).","explanationZh":"$设 |\\overrightarrow{a}| = |\\overrightarrow{b}| =$ r，由 $|\\overrightarrow{a} − \\overrightarrow{b}| = r$ 平方得 $r^{2} + r^{2} − 2r^{2}\\cos θ = r^{2}，故 \\cos θ = 1/2，θ = 60°（三边相等构成等边三角形）。$","tags":["hard","angle","magnitude"],"status":"published"},{"id":"math-vectors-028","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":4,"stem":"Let $\\overrightarrow{a} = (2, 1)$ and $\\overrightarrow{b} = (1, 3)$. Find the projection vector of $\\overrightarrow{a}$ in the direction of $\\overrightarrow{b}$.","stemZh":"已知 $\\overrightarrow{a} = (2, 1)，\\overrightarrow{b} = (1, 3)，求 \\overrightarrow{a} 在 \\overrightarrow{b}$ 方向上的投影向量。","options":[{"id":"A","text":"$(1, 3)$"},{"id":"B","text":"$(5, 15)$"},{"id":"C","text":"$(1/2, 3)$"},{"id":"D","text":"$(1/2, 3/2)$"}],"answer":"D","explanation":"The projection is $((\\overrightarrow{a} · \\overrightarrow{b})/|\\overrightarrow{b}|^{2}) \\overrightarrow{b} = (5/10)(1, 3) = (1/2, 3/2)$. Forgetting to divide by $|\\overrightarrow{b}|^{2} = 10$ gives $(5, 15)$.","explanationZh":"投影向量为 $((\\overrightarrow{a} · \\overrightarrow{b})/|\\overrightarrow{b}|^{2}) \\overrightarrow{b} = (5/10)(1, 3) = (1/2, 3/2)；若漏除以 |\\overrightarrow{b}|^{2} = 10$ 会得到 $(5, 15)。$","tags":["hard","projection","dot-product"],"status":"published"},{"id":"math-vectors-029","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":5,"stem":"Let $|\\overrightarrow{a}| = 2, |\\overrightarrow{b}| = 3$ and $(\\overrightarrow{a} + \\overrightarrow{b}) · (\\overrightarrow{a} − \\overrightarrow{2b}) = −17$. Find the angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$.","stemZh":"已知 $|\\overrightarrow{a}| = 2，|\\overrightarrow{b}| = 3，(\\overrightarrow{a} + \\overrightarrow{b}) · (\\overrightarrow{a} − \\overrightarrow{2b}) = −17，求 \\overrightarrow{a} 与 \\overrightarrow{b}$ 的夹角。","options":[{"id":"A","text":"60°"},{"id":"B","text":"30°"},{"id":"C","text":"120°"},{"id":"D","text":"90°"}],"answer":"A","explanation":"$Expand: |\\overrightarrow{a}|^{2} − \\overrightarrow{a} · \\overrightarrow{b} − 2|\\overrightarrow{b}|^{2} = 4 − \\overrightarrow{a} · \\overrightarrow{b} − 18 = −14 − \\overrightarrow{a} · \\overrightarrow{b} = −17$, so $\\overrightarrow{a} · \\overrightarrow{b} = 3$. Then $\\cos θ = 3/(2·3) = 1/2$ and $θ = 60°$.","explanationZh":"展开得 $|\\overrightarrow{a}|^{2} − \\overrightarrow{a} · \\overrightarrow{b} − 2|\\overrightarrow{b}|^{2} = 4 − \\overrightarrow{a} · \\overrightarrow{b} − 18 = −14 − \\overrightarrow{a} · \\overrightarrow{b} = −17，故 \\overrightarrow{a} · \\overrightarrow{b} = 3，\\cos θ = 3/(2·3) = 1/2，θ = 60°。$","tags":["hard","dot-product","angle"],"status":"published"},{"id":"math-vectors-030","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":5,"stem":"Nonzero vectors $\\overrightarrow{a}$ and $\\overrightarrow{b}$ satisfy $|\\overrightarrow{a} + \\overrightarrow{b}| = |\\overrightarrow{a}| + |\\overrightarrow{b}|$. What must be true?","stemZh":"非零向量 $\\overrightarrow{a}、\\overrightarrow{b}$ 满足 $|\\overrightarrow{a} + \\overrightarrow{b}| = |\\overrightarrow{a}| + |\\overrightarrow{b}|，则必有？$","options":[{"id":"A","text":"$\\overrightarrow{a} ⟂ \\overrightarrow{b}$"},{"id":"B","text":"$\\overrightarrow{a}$ and $\\overrightarrow{b}$ have the same direction"},{"id":"C","text":"$|\\overrightarrow{a}| = |\\overrightarrow{b}|$"},{"id":"D","text":"$\\overrightarrow{a} = \\overrightarrow{b}$"}],"answer":"B","explanation":"Equality holds in the triangle inequality exactly when the angle is 0°, i.e. the vectors have the same direction. Equal length (C) or even equality (D) is not required.","explanationZh":"三角不等式取等号的条件是夹角为 0°，即两向量同向；并不要求长度相等（C），更不要求两向量相等（D）。","tags":["hard","properties","magnitude"],"status":"published"},{"id":"math-complex-001","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":1,"stem":"Let $z = 3 − 4i$. Find the real part of z.","stemZh":"已知 $z = 3 − 4i，求 z$ 的实部。","options":[{"id":"A","text":"3"},{"id":"B","text":"−4"},{"id":"C","text":"4"},{"id":"D","text":"−3"}],"answer":"A","explanation":"For $z = a +$ bi the real part is $a = 3$ and the imaginary part is $b = −4$.","explanationZh":"$z = a +$ bi 的实部为 $a = 3，虚部为 b = −4。$","tags":["easy","real-imaginary-parts"],"status":"published"},{"id":"math-complex-002","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":1,"stem":"What is $i^{2}$ equal to?","stemZh":"$i^{2}$ 等于多少？","options":[{"id":"A","text":"1"},{"id":"B","text":"−1"},{"id":"C","text":"i"},{"id":"D","text":"−i"}],"answer":"B","explanation":"By definition i is the imaginary unit with $i^{2} = −1$.","explanationZh":"虚数单位 i 满足 $i^{2} = −1。$","tags":["easy","powers-of-i"],"status":"published"},{"id":"math-complex-003","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":1,"stem":"Let $z = 2 + 3i$. Find the conjugate $\\bar{z}$.","stemZh":"已知 $z = 2 + 3i，求其共轭复数 \\bar{z}。$","options":[{"id":"A","text":"−2 + 3i"},{"id":"B","text":"2 + 3i"},{"id":"C","text":"2 − 3i"},{"id":"D","text":"−2 − 3i"}],"answer":"C","explanation":"The conjugate keeps the real part and flips the sign of the imaginary $part: \\bar{z} = 2 − 3i$.","explanationZh":"$共轭复数实部不变、虚部变号：\\bar{z} = 2 − 3i。$","tags":["easy","conjugate"],"status":"published"},{"id":"math-complex-004","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":1,"stem":"Let $z = 3 + 4i$. Find $|z|$.","stemZh":"已知 $z = 3 + 4i，求 |z|。$","options":[{"id":"A","text":"7"},{"id":"B","text":"1"},{"id":"C","text":"25"},{"id":"D","text":"5"}],"answer":"D","explanation":"$|z| = \\sqrt{a^{2} + b^{2}} = \\sqrt{9 + 16} = 5$. Note that 25 is $|z|^{2}$, not $|z|$.","explanationZh":"$|z| = \\sqrt{a^{2} + b^{2}} = \\sqrt{9 + 16} = 5；注意 25 是 |z|^{2}，不是 |z|。$","tags":["easy","modulus"],"status":"published"},{"id":"math-complex-005","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":2,"stem":"Compute $(2 + 3i) + (1 − 4i)$.","stemZh":"计算 $(2 + 3i) + (1 − 4i)。$","options":[{"id":"A","text":"3 − i"},{"id":"B","text":"3 + i"},{"id":"C","text":"1 + 7i"},{"id":"D","text":"3 − 7i"}],"answer":"A","explanation":"Add real and imaginary parts $separately: (2 + 1) + (3 − 4)i = 3 −$ i.","explanationZh":"$实部与虚部分别相加：(2 + 1) + (3 − 4)i = 3 −$ i。","tags":["easy","addition"],"status":"published"},{"id":"math-complex-006","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":2,"stem":"Compute $(2 + 3i) − (1 − 4i)$.","stemZh":"计算 $(2 + 3i) − (1 − 4i)。$","options":[{"id":"A","text":"1 − i"},{"id":"B","text":"1 + 7i"},{"id":"C","text":"3 − i"},{"id":"D","text":"1 − 7i"}],"answer":"B","explanation":"$(2 − 1) + (3 − (−4))i = 1 + 7i$; the imaginary part picks up a plus sign.","explanationZh":"$(2 − 1) + (3 − (−4))i = 1 + 7i，虚部要注意变号。$","tags":["easy","subtraction"],"status":"published"},{"id":"math-complex-007","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":2,"stem":"Compute $i^{5}$.","stemZh":"计算 $i^{5}。$","options":[{"id":"A","text":"1"},{"id":"B","text":"−1"},{"id":"C","text":"i"},{"id":"D","text":"−i"}],"answer":"C","explanation":"$i^{4} = 1$, so $i^{5} = i^{4} · i =$ i.","explanationZh":"$i^{4} = 1，故 i^{5} = i^{4} · i =$ i。","tags":["easy","powers-of-i"],"status":"published"},{"id":"math-complex-008","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":2,"stem":"Compute $(1 + i)(1 − i)$.","stemZh":"计算 $(1 + i)(1 − i)。$","options":[{"id":"A","text":"0"},{"id":"B","text":"1"},{"id":"C","text":"$\\sqrt{2}$"},{"id":"D","text":"2"}],"answer":"D","explanation":"$(1 + i)(1 − i) = 1 − i^{2} = 1 − (−1) = 2$. Note $\\sqrt{2}$ is $|1 + i|$, not the product.","explanationZh":"$(1 + i)(1 − i) = 1 − i^{2} = 1 + 1 = 2；\\sqrt{2} 是 |1 + i|，不是乘积。$","tags":["easy","multiplication"],"status":"published"},{"id":"math-complex-009","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":1,"stem":"Let $z = a +$ bi with $a, b ∈ ℝ$. When is z a real number?","stemZh":"$设 z = a + bi（a, b ∈ ℝ），z$ 为实数的条件是？","options":[{"id":"A","text":"$b = 0$"},{"id":"B","text":"$a = 0$"},{"id":"C","text":"$a = b = 0$"},{"id":"D","text":"$a > 0$"}],"answer":"A","explanation":"z is real exactly when its imaginary part $b = 0; a = 0 (with b ≠ 0)$ gives a pure imaginary number.","explanationZh":"z 为实数当且仅当虚部 $b = 0；a = 0（且 b ≠ 0）时 z$ 为纯虚数。","tags":["easy","real-imaginary-parts"],"status":"published"},{"id":"math-complex-010","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":2,"stem":"Let $z = −2 + 5i$. Find the imaginary part of z.","stemZh":"已知 $z = −2 + 5i，求 z$ 的虚部。","options":[{"id":"A","text":"5i"},{"id":"B","text":"5"},{"id":"C","text":"−2"},{"id":"D","text":"−2i"}],"answer":"B","explanation":"The imaginary part is the real coefficient $b = 5$, not $5i. (The$ real part is $−2.)$","explanationZh":"虚部是实数 $b = 5，而不是 5i（实部为 −2）。$","tags":["easy","real-imaginary-parts"],"status":"published"},{"id":"math-complex-011","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Let $z = (m − 1) + (m + 2)i$ with m real. For which m is z a pure imaginary number?","stemZh":"$设 z = (m − 1) + (m + 2)i（m ∈ ℝ），z$ 为纯虚数时 $m = ？$","options":[{"id":"A","text":"−2"},{"id":"B","text":"2"},{"id":"C","text":"1"},{"id":"D","text":"0"}],"answer":"C","explanation":"A pure imaginary number needs real part 0 and imaginary part $nonzero: m − 1 = 0$ gives $m = 1$, and then $m + 2 = 3 ≠ 0$. The value $m = −2$ makes the imaginary part 0, so $z = −3$ is real.","explanationZh":"纯虚数要求实部为 0 且虚部不为 $0：m − 1 = 0 得 m = 1，此时 m + 2 = 3 ≠ 0；m = −2$ 时虚部为 $0，z = −3$ 是实数。","tags":["medium","pure-imaginary"],"status":"published"},{"id":"math-complex-012","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Compute $(3 + 2i)/(1 − i)$.","stemZh":"计算 $(3 + 2i)/(1 − i)。$","options":[{"id":"A","text":"1 + 5i"},{"id":"B","text":"$1/2 − (5/2)i$"},{"id":"C","text":"$5/2 + (1/2)i$"},{"id":"D","text":"$1/2 + (5/2)i$"}],"answer":"D","explanation":"Multiply by the $conjugate: (3 + 2i)(1 + i)/((1 − i)(1 + i)) = (1 + 5i)/2 = 1/2 + (5/2)i$. The denominator is 2, not 1.","explanationZh":"$分子分母同乘共轭：(3 + 2i)(1 + i)/((1 − i)(1 + i)) = (1 + 5i)/2 = 1/2 + (5/2)i；分母是 2$ 不能漏除。","tags":["medium","division"],"status":"published"},{"id":"math-complex-013","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Let $z = (1 + i)^{3}$. Find $|z|$.","stemZh":"已知 $z = (1 + i)^{3}，求 |z|。$","options":[{"id":"A","text":"$2\\sqrt{2}$"},{"id":"B","text":"2"},{"id":"C","text":"$\\sqrt{2}$"},{"id":"D","text":"4"}],"answer":"A","explanation":"$|z| = |1 + i|^{3} = (\\sqrt{2})^{3} = 2\\sqrt{2}$. Direct $check: (1 + i)^{3} = 2i(1 + i) = −2 + 2i$, and $|−2 + 2i| = \\sqrt{8} = 2\\sqrt{2}$.","explanationZh":"$|z| = |1 + i|^{3} = (\\sqrt{2})^{3} = 2\\sqrt{2}；也可直接算：(1 + i)^{3} = 2i(1 + i) = −2 + 2i，模为 \\sqrt{8} = 2\\sqrt{2}。$","tags":["medium","modulus"],"status":"published"},{"id":"math-complex-014","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Let $z = 2 +$ i. Compute $z · \\bar{z}$.","stemZh":"已知 $z = 2 +$ i，计算 $z · \\bar{z}。$","options":[{"id":"A","text":"3 + 4i"},{"id":"B","text":"5"},{"id":"C","text":"$\\sqrt{5}$"},{"id":"D","text":"3"}],"answer":"B","explanation":"$z · \\bar{z} = (2 + i)(2 − i) = 4 + 1 = 5 = |z|^{2}$, always a real number. Note 3 + 4i is $z^{2}$, not $z · \\bar{z}$.","explanationZh":"$z · \\bar{z} = (2 + i)(2 − i) = 5 = |z|^{2}，结果一定是实数；3 + 4i 是 z^{2}，不是 z · \\bar{z}。$","tags":["medium","modulus","conjugate"],"status":"published"},{"id":"math-complex-015","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Compute $(1 + i)^{4}$.","stemZh":"计算 $(1 + i)^{4}。$","options":[{"id":"A","text":"4"},{"id":"B","text":"4i"},{"id":"C","text":"−4"},{"id":"D","text":"−4i"}],"answer":"C","explanation":"$(1 + i)^{2} = 2i$, so $(1 + i)^{4} = (2i)^{2} = 4i^{2} = −4$.","explanationZh":"$(1 + i)^{2} = 2i，故 (1 + i)^{4} = (2i)^{2} = 4i^{2} = −4。$","tags":["medium","powers-of-i","multiplication"],"status":"published"},{"id":"math-complex-016","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Let x, y be real and $(x + yi) + (2 − 3i) = 5 +$ i. Find $(x, y)$.","stemZh":"$设 x, y ∈ ℝ 且 (x + yi) + (2 − 3i) = 5 +$ i，求 $(x, y)。$","options":[{"id":"A","text":"$(3, −4)$"},{"id":"B","text":"$(−3, 4)$"},{"id":"C","text":"$(7, −2)$"},{"id":"D","text":"$(3, 4)$"}],"answer":"D","explanation":"Equality of complex numbers means equal real and equal imaginary $parts: x + 2 = 5$ and $y − 3 = 1$, so $x = 3, y = 4$.","explanationZh":"复数相等即实部、虚部分别相等：x $+ 2 = 5，y − 3 = 1，得 x = 3，y = 4。$","tags":["medium","equality"],"status":"published"},{"id":"math-complex-017","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Let $z = (m^{2} − 1) + (m − 1)i$ with m real. For which m is z a real number?","stemZh":"$设 z = (m^{2} − 1) + (m − 1)i（m ∈ ℝ），z$ 为实数时 $m = ？$","options":[{"id":"A","text":"1"},{"id":"B","text":"−1"},{"id":"C","text":"$\\pm1$"},{"id":"D","text":"0"}],"answer":"A","explanation":"The imaginary part must $vanish: m − 1 = 0$, so $m = 1 (then z = 0,\\text{ which is }real)$. For $m = −1$ the real part is 0 but the imaginary part is −2, so $z = −2i$ is not real.","explanationZh":"虚部为 $0：m − 1 = 0，得 m = 1（此时 z = 0$ 是实数）。m $= −1$ 时实部为 0 但虚部为 $−2，z = −2i$ 不是实数。","tags":["medium","real-imaginary-parts"],"status":"published"},{"id":"math-complex-018","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Compute $i + i^{2} + i^{3} + i^{4}$.","stemZh":"计算 $i + i^{2} + i^{3} + i^{4}。$","options":[{"id":"A","text":"1"},{"id":"B","text":"0"},{"id":"C","text":"i"},{"id":"D","text":"−1"}],"answer":"B","explanation":"$i + (−1) + (−i) + 1 = 0$; any four consecutive powers of i add up to 0.","explanationZh":"$i + (−1) + (−i) + 1 = 0，任意连续四个 i$ 的幂之和都为 0。","tags":["medium","powers-of-i"],"status":"published"},{"id":"math-complex-019","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Compute $(2 − i)^{2}$.","stemZh":"计算 $(2 − i)^{2}。$","options":[{"id":"A","text":"5"},{"id":"B","text":"3 + 4i"},{"id":"C","text":"3 − 4i"},{"id":"D","text":"5 − 4i"}],"answer":"C","explanation":"$(2 − i)^{2} = 4 − 4i + i^{2} = 4 − 4i − 1 = 3 − 4i$. Using $i^{2} = 1$ instead of −1 gives 5 − 4i.","explanationZh":"$(2 − i)^{2} = 4 − 4i + i^{2} = 4 − 4i − 1 = 3 − 4i；若误用 i^{2} = 1$ 会得到 5 − 4i。","tags":["medium","multiplication"],"status":"published"},{"id":"math-complex-020","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Let $z = 3 +$ bi with b real and $b < 0$. If $|z| = 5$, find b.","stemZh":"$设 z = 3 +$ bi（b $∈ ℝ，b < 0），若 |z| = 5，求$ b。","options":[{"id":"A","text":"4"},{"id":"B","text":"16"},{"id":"C","text":"−16"},{"id":"D","text":"−4"}],"answer":"D","explanation":"$9 + b^{2} = 25$ gives $b^{2} = 16$, so $b = \\pm4$; the condition $b < 0$ selects $b = −4$.","explanationZh":"$9 + b^{2} = 25 得 b^{2} = 16，b = \\pm4；由 b < 0 取 b = −4。$","tags":["medium","modulus"],"status":"published"},{"id":"math-complex-021","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":4,"stem":"Let $z = (m^{2} − m − 2) + (m^{2} − 1)i$ with m real. For which m is $z = 0$?","stemZh":"$设 z = (m^{2} − m − 2) + (m^{2} − 1)i（m ∈ ℝ），z = 0 时 m = ？$","options":[{"id":"A","text":"−1"},{"id":"B","text":"1"},{"id":"C","text":"2"},{"id":"D","text":"−2"}],"answer":"A","explanation":"$z = 0$ needs both parts $zero: m^{2} − m − 2 = 0$ gives $m = 2$ or $m = −1$, and $m^{2} − 1 = 0$ gives $m = \\pm1$. The common value is $m = −1$.","explanationZh":"$z = 0$ 需实部、虚部同时为 $0：m^{2} − m − 2 = 0 得 m = 2 或 −1，m^{2} − 1 = 0 得 m = \\pm1，公共解为 m = −1。$","tags":["hard","equality","real-imaginary-parts"],"status":"published"},{"id":"math-complex-022","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":4,"stem":"A complex number z satisfies $z · \\bar{z} = 25$ and $z + \\bar{z} = 6$. Find z.","stemZh":"复数 z 满足 $z · \\bar{z} = 25，z + \\bar{z} = 6，求$ z。","options":[{"id":"A","text":"3 + 4i"},{"id":"B","text":"$3 \\pm 4i$"},{"id":"C","text":"$4 \\pm 3i$"},{"id":"D","text":"$\\pm5$"}],"answer":"B","explanation":"Write $z = a +$ bi. Then $z + \\bar{z} = 2a = 6$, so $a = 3$, and $z · \\bar{z} = a^{2} + b^{2} = 25$ gives $b^{2} = 16, b = \\pm4$. Both $z = 3 + 4i$ and $z = 3 − 4i$ work.","explanationZh":"$设 z = a +$ bi，则 $z + \\bar{z} = 2a = 6，a = 3；z · \\bar{z} = a^{2} + b^{2} = 25 得 b^{2} = 16，b = \\pm4，故 z = 3 \\pm 4i。$","tags":["hard","modulus","conjugate"],"status":"published"},{"id":"math-complex-023","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":4,"stem":"Compute $i^{1} + i^{2} + i^{3} + … + i^{2024}$.","stemZh":"计算 $i^{1} + i^{2} + i^{3} + … + i^{2024}。$","options":[{"id":"A","text":"1"},{"id":"B","text":"i"},{"id":"C","text":"0"},{"id":"D","text":"−1"}],"answer":"C","explanation":"Powers of i repeat with period 4 and each block $i + i^{2} + i^{3} + i^{4} = i − 1 − i + 1 = 0$. Since $2024 = 4 \\times 506$, the sum is 0.","explanationZh":"i 的幂以 4 为周期，每四项之和 $i + i^{2} + i^{3} + i^{4} = 0；2024 = 4 \\times 506，故总和为 0。$","tags":["hard","powers-of-i"],"status":"published"},{"id":"math-complex-024","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":4,"stem":"Let $z = (1 + i)/(1 − i)$. Compute $z^{2024}$.","stemZh":"$设 z = (1 + i)/(1 − i)，计算 z^{2024}。$","options":[{"id":"A","text":"i"},{"id":"B","text":"−1"},{"id":"C","text":"−i"},{"id":"D","text":"1"}],"answer":"D","explanation":"$z = (1 + i)^{2}/((1 − i)(1 + i)) = 2i/2 = i$, so $z^{2024} = i^{2024} = (i^{4})^{506} = 1$.","explanationZh":"$z = (1 + i)^{2}/((1 − i)(1 + i)) = 2i/2 =$ i，故 $z^{2024} = i^{2024} = (i^{4})^{506} = 1。$","tags":["hard","division","powers-of-i"],"status":"published"},{"id":"math-complex-025","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":4,"stem":"Let $z = a +$ bi with $a, b ∈ ℝ$. Which expression always equals $|z|^{2}$?","stemZh":"$设 z = a + bi（a, b ∈ ℝ），下列哪个式子恒等于 |z|^{2}？$","options":[{"id":"A","text":"$z · \\bar{z}$"},{"id":"B","text":"$z^{2}$"},{"id":"C","text":"$a^{2} − b^{2}$"},{"id":"D","text":"$(a + b)^{2}$"}],"answer":"A","explanation":"$z · \\bar{z} = (a + bi)(a − bi) = a^{2} + b^{2} = |z|^{2}$, always a nonnegative real number. $z^{2} = a^{2} − b^{2} + 2abi$ is generally not real.","explanationZh":"$z · \\bar{z} = (a + bi)(a − bi) = a^{2} + b^{2} = |z|^{2}，恒为非负实数；而 z^{2} = a^{2} − b^{2} + 2abi$ 一般不是实数。","tags":["hard","modulus","conjugate"],"status":"published"},{"id":"math-complex-026","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":5,"stem":"Let $z = (m^{2} − 4) + (m^{2} − 3m + 2)i$ with m real. For which m is z a pure imaginary number?","stemZh":"$设 z = (m^{2} − 4) + (m^{2} − 3m + 2)i（m ∈ ℝ），z$ 为纯虚数时 $m = ？$","options":[{"id":"A","text":"2"},{"id":"B","text":"−2"},{"id":"C","text":"$\\pm2$"},{"id":"D","text":"1"}],"answer":"B","explanation":"Real part 0 gives $m^{2} − 4 = 0$, so $m = \\pm2$. Check the imaginary part: at $m = 2$ it is $4 − 6 + 2 = 0$, giving $z = 0 (not$ pure imaginary); at $m = −2$ it is $4 + 6 + 2 = 12 ≠ 0$, giving $z = 12i$.","explanationZh":"实部为 $0 得 m^{2} − 4 = 0，m = \\pm2。再检验虚部：m = 2$ 时虚部 $4 − 6 + 2 = 0，z = 0$ 不是纯虚数；m $= −2$ 时虚部 $4 + 6 + 2 = 12 ≠ 0，z = 12i。$","tags":["hard","pure-imaginary"],"status":"published"},{"id":"math-complex-027","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":5,"stem":"Let $z = (a^{2} − 1) + 2ai$ with a real. Find the minimum value of $|z|$.","stemZh":"$设 z = (a^{2} − 1) + 2ai（a ∈ ℝ），求 |z|$ 的最小值。","options":[{"id":"A","text":"0"},{"id":"B","text":"1/2"},{"id":"C","text":"1"},{"id":"D","text":"2"}],"answer":"C","explanation":"$|z|^{2} = (a^{2} − 1)^{2} + 4a^{2} = a^{4} + 2a^{2} + 1 = (a^{2} + 1)^{2}$, so $|z| = a^{2} + 1 ≥ 1$, with equality at $a = 0 (then z = −1)$.","explanationZh":"$|z|^{2} = (a^{2} − 1)^{2} + 4a^{2} = a^{4} + 2a^{2} + 1 = (a^{2} + 1)^{2}，故 |z| = a^{2} + 1 ≥ 1，当 a = 0$ 时取等号（此时 $z = −1）。$","tags":["hard","modulus"],"status":"published"},{"id":"math-complex-028","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":4,"stem":"Let $z_{1} = 3 + 4i$ and $z_{2} = 4 − 3i$. Find $|z_{1} · z_{2}|$.","stemZh":"已知 $z_{1} = 3 + 4i，z_{2} = 4 − 3i，求 |z_{1} · z_{2}|。$","options":[{"id":"A","text":"7"},{"id":"B","text":"1"},{"id":"C","text":"5"},{"id":"D","text":"25"}],"answer":"D","explanation":"$|z_1 · z_2| = |z_1| · |z_2| = 5 · 5 = 25$. Direct $check: z_1z_2 = (3 + 4i)(4 − 3i) = 24 + 7i$ and $|24 + 7i| = \\sqrt{576 + 49} = 25$.","explanationZh":"$|z_1 · z_2| = |z_1| · |z_2| = 5 · 5 = 25；直接计算 z_1z_2 = 24 + 7i，|24 + 7i| = \\sqrt{576 + 49} = 25。$","tags":["hard","multiplication","modulus"],"status":"published"},{"id":"math-complex-029","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":5,"stem":"Let $z = 1 +$ i. Compute $z + z^{2} + z^{3} + z^{4}$.","stemZh":"$设 z = 1 +$ i，计算 $z + z^{2} + z^{3} + z^{4}。$","options":[{"id":"A","text":"−5 + 5i"},{"id":"B","text":"5 + 5i"},{"id":"C","text":"−5 − 5i"},{"id":"D","text":"5 − 5i"}],"answer":"A","explanation":"$z^{2} = 2i, z^{3} = 2i(1 + i) = −2 + 2i, z^{4} = (z^{2})^{2} = −4. Summing: (1 + 0 − 2 − 4) + (1 + 2 + 2 + 0)i = −5 + 5i$.","explanationZh":"$z^{2} = 2i，z^{3} = 2i(1 + i) = −2 + 2i，z^{4} = (z^{2})^{2} = −4，相加得 (1 + 0 − 2 − 4) + (1 + 2 + 2 + 0)i = −5 + 5i。$","tags":["hard","powers","addition"],"status":"published"},{"id":"math-complex-030","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":5,"stem":"Let $z = a +$ bi with $a, b ∈ ℝ$. If $z^{2}$ is a real number, what must hold?","stemZh":"$设 z = a + bi（a, b ∈ ℝ），若 z^{2}$ 是实数，则必有？","options":[{"id":"A","text":"$a = 0$"},{"id":"B","text":"$a = 0$ or $b = 0$"},{"id":"C","text":"$b = 0$"},{"id":"D","text":"$a = b = 0$"}],"answer":"B","explanation":"$z^{2} = (a^{2} − b^{2}) + 2abi$ is real exactly when $2ab = 0$, i.e. $a = 0$ or $b = 0$. Either condition alone is sufficient, so the answer must allow both.","explanationZh":"$z^{2} = (a^{2} − b^{2}) + 2abi$ 为实数当且仅当 $2ab = 0，即 a = 0 或 b = 0；二者都足以保证结论，故答案须包含两种情况。$","tags":["hard","real-imaginary-parts","multiplication"],"status":"published"},{"id":"math-space-001","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":1,"stem":"A cube has edge length 2. Its volume is","stemZh":"棱长为 2 的正方体，其体积为","options":[{"id":"A","text":"4"},{"id":"B","text":"6"},{"id":"C","text":"8"},{"id":"D","text":"12"}],"answer":"C","explanation":"Volume of $a cube: V = a^{3}$. With $a = 2$ we get $V = 2^{3} = 8. 6a^{2}$ would be the surface area, 12a is the total edge length.","explanationZh":"正方体体积公式 $V = a^{3}，取 a = 2 得 V = 2^{3} = 8。6a^{2} 是表面积，12a$ 是棱长总和。","tags":["easy","cube","volume"],"status":"published"},{"id":"math-space-002","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":1,"stem":"The volume of a cube is 27. Its edge length is","stemZh":"正方体的体积为 27，其棱长为","options":[{"id":"A","text":"3"},{"id":"B","text":"9"},{"id":"C","text":"4"},{"id":"D","text":"$\\sqrt{3}$"}],"answer":"A","explanation":"$V = a^{3} = 27$, so $a = \\sqrt[3]{}27 = 3$. The cube root is required — do not divide by $3 (that$ would give $9)$.","explanationZh":"$由 V = a^{3} = 27 得 a = \\sqrt[3]{}27 = 3。这里要开三次方，不能除以 3（否则得 9）。$","tags":["easy","cube","volume"],"status":"published"},{"id":"math-space-003","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":1,"stem":"A sphere has radius 3. Its volume is","stemZh":"半径为 3 的球，其体积为","options":[{"id":"A","text":"$36π$"},{"id":"B","text":"$12π$"},{"id":"C","text":"$108π$"},{"id":"D","text":"$9π$"}],"answer":"A","explanation":"$V = (4/3)πr^{3} = (4/3)π·27 = 36π$. Note $36π = 4π·9 = 4πr^{2}$ is the surface area only if you forget the factor r/3; the correct volume keeps the 4/3 factor.","explanationZh":"球体积公式 $V = (4/3)πr^{3}，取 r = 3 得 V = (4/3)π·27 = 36π。注意不要漏掉系数 4/3。$","tags":["easy","sphere","volume"],"status":"published"},{"id":"math-space-004","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":1,"stem":"A sphere has radius 2. Its surface area is","stemZh":"半径为 2 的球，其表面积为","options":[{"id":"A","text":"$4π$"},{"id":"B","text":"$16π$"},{"id":"C","text":"$8π$"},{"id":"D","text":"$32π$"}],"answer":"B","explanation":"$S = 4πr^{2} = 4π·4 = 16π$. Squaring the radius comes $first: 8π$ would come from using 2r, and $32π$ from using $r^{3}$.","explanationZh":"球表面积公式 $S = 4πr^{2}，取 r = 2 得 S = 4π·4 = 16π。要先把半径平方，2r 与 r^{3}$ 的用法都是错的。","tags":["easy","sphere","surface-area"],"status":"published"},{"id":"math-space-005","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":1,"stem":"A cube has edge length a. Its surface area is","stemZh":"棱长为 a 的正方体，其表面积为","options":[{"id":"A","text":"$a^{2}$"},{"id":"B","text":"$4a^{2}$"},{"id":"C","text":"$12a^{2}$"},{"id":"D","text":"$6a^{2}$"}],"answer":"D","explanation":"A cube has 6 congruent square faces, each of area $a^{2}$, so $S = 6a^{2}. a^{2}$ is one face only and $12a^{2}$ double-counts.","explanationZh":"正方体有 6 个全等的正方形面，每个面积为 $a^{2}，故 S = 6a^{2}。a^{2}$ 只是一个面的面积。","tags":["easy","cube","surface-area"],"status":"published"},{"id":"math-space-006","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":2,"stem":"A rectangular box measures 3 by 4 by 5. Its volume is","stemZh":"长、宽、高分别为 3、4、5 的长方体，其体积为","options":[{"id":"A","text":"12"},{"id":"B","text":"20"},{"id":"C","text":"94"},{"id":"D","text":"60"}],"answer":"D","explanation":"$V = 3·4·5 = 60$. The value $94 = 2(3·4 + 4·5 + 3·5)$ is the surface area, not the volume.","explanationZh":"长方体体积 $V = 3·4·5 = 60。94 = 2(3·4 + 4·5 + 3·5)$ 是表面积，不是体积。","tags":["easy","cuboid","volume"],"status":"published"},{"id":"math-space-007","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":2,"stem":"A cube has edge length 1. Its space diagonal has length","stemZh":"棱长为 1 的正方体，其体对角线长为","options":[{"id":"A","text":"1"},{"id":"B","text":"$\\sqrt{3}$"},{"id":"C","text":"$\\sqrt{2}$"},{"id":"D","text":"2"}],"answer":"B","explanation":"The space diagonal is $\\sqrt{1^{2} + 1^{2} + 1^{2}} = \\sqrt{3}. \\sqrt{2}$ is a face diagonal, obtained by using only two dimensions.","explanationZh":"体对角线长 $= \\sqrt{1^{2} + 1^{2} + 1^{2}} = \\sqrt{3}。\\sqrt{2}$ 是面对角线，只用了两个方向的长度。","tags":["easy","cube","space-diagonal"],"status":"published"},{"id":"math-space-008","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":2,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$, which pair of lines is skew (neither parallel nor intersecting)?","stemZh":"在正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，哪一对直线是异面直线（既不平行也不相交）？","options":[{"id":"A","text":"AB and CD"},{"id":"B","text":"AB and $A_{1}B_{1}$"},{"id":"C","text":"AB and AD"},{"id":"D","text":"AB and $CC_{1}$"}],"answer":"D","explanation":"AB lies in the bottom face while the vertical edge $CC_{1}$ passes through C, which is not on AB; the two lines are neither parallel nor intersecting, so they are skew. AB ∥ CD, AB $∥ A_{1}B_{1}$, and AB meets AD at A.","explanationZh":"AB 在底面内，而竖棱 $CC_{1}$ 经过不在线 AB 上的点 C，两直线既不平行也不相交，因此是异面直线。AB ∥ CD，AB $∥ A_{1}B_{1}，AB 与$ AD 相交于 A。","tags":["easy","skew-lines","line-position"],"status":"published"},{"id":"math-space-009","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":2,"stem":"A cylinder has base radius 1 and height 2. Its lateral (side) surface area is","stemZh":"底面半径为 1、高为 2 的圆柱，其侧面积为","options":[{"id":"A","text":"$4π$"},{"id":"B","text":"$2π$"},{"id":"C","text":"$6π$"},{"id":"D","text":"$8π$"}],"answer":"A","explanation":"Lateral area $= 2πrh = 2π·1·2 = 4π$. The value $2π$ is the volume $πr^{2}h$, and $8π$ comes from replacing r by the diameter 2.","explanationZh":"圆柱侧面积 $S = 2πrh = 2π·1·2 = 4π。2π$ 是体积 $πr^{2}h，8π$ 是误把直径当半径代入的结果。","tags":["easy","cylinder","surface-area"],"status":"published"},{"id":"math-space-010","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":2,"stem":"A solid stands on a horizontal table. Its front view is a rectangle, its side view is a rectangle, and its top view is a circle. The solid is","stemZh":"一个几何体放在水平桌面上，它的正视图是矩形，侧视图是矩形，俯视图是圆。这个几何体是","options":[{"id":"A","text":"a cone"},{"id":"B","text":"a sphere"},{"id":"C","text":"a cylinder"},{"id":"D","text":"a cube"}],"answer":"C","explanation":"Only a cylinder gives two rectangular side views and a circular top view. A sphere would show circles in all three views, a cone shows a triangle from the side, and a cube never shows a circle.","explanationZh":"只有圆柱的正视图、侧视图都是矩形而俯视图是圆。球的三视图都是圆，圆锥的侧视是三角形，正方体的视图不会出现圆。","tags":["easy","three-view","cylinder"],"status":"published"},{"id":"math-space-011","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$ of edge 1, the angle between the skew lines $AB_{1}$ and $BC_{1}$ is","stemZh":"在棱长为 1 的正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，异面直线 $AB_{1} 与 BC_{1}$ 所成的角是","options":[{"id":"A","text":"30°"},{"id":"B","text":"45°"},{"id":"C","text":"60°"},{"id":"D","text":"90°"}],"answer":"C","explanation":"Translate $BC_1$ to $AD_1 (they$ are parallel). Then the required angle is $∠B_1AD_1$, and $AB_1, AD_1, B_1D_1$ are all face diagonals of length $\\sqrt{2}$, so triangle $AB_1D_1$ is equilateral and the angle is 60°.","explanationZh":"$将 BC_1$ 平移到 $AD_1（二者平行），所求角即为 ∠B_1AD_1。AB_1、AD_1、B_1D_1$ 都是长为 $\\sqrt{2}$ 的面对角线，故 $△AB_1D_1$ 是等边三角形，夹角为 60°。","tags":["medium","skew-lines","angle"],"status":"published"},{"id":"math-space-012","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$ of edge 1, the angle between the line $A_{1}D$ and the base plane ABCD is","stemZh":"在棱长为 1 的正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，直线 $A_{1}D$ 与底面 ABCD 所成的角是","options":[{"id":"A","text":"30°"},{"id":"B","text":"60°"},{"id":"C","text":"90°"},{"id":"D","text":"45°"}],"answer":"D","explanation":"$A_{1}$ projects to A on the base plane, so AD is the projection of $A_{1}D$ and the required angle is $∠A_{1}DA$. Since $AA_{1} =$ AD $= 1, \\tan∠A_{1}DA = AA_{1}/AD = 1$, giving 45°.","explanationZh":"$A_{1}$ 在底面上的射影是 A，故 AD $是 A_{1}D$ 的射影，所求角为 $∠A_{1}DA。因为 AA_{1} =$ AD $= 1，\\tan∠A_{1}DA = AA_{1}/AD = 1，所以夹角为 45°。$","tags":["medium","line-plane-angle","cube"],"status":"published"},{"id":"math-space-013","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"To conclude that a line l is perpendicular to a plane $α$, it is enough to know that l is perpendicular to","stemZh":"要判定直线 l 垂直于平面 $α，只需知道 l$ 垂直于","options":[{"id":"A","text":"one line lying in $α$"},{"id":"B","text":"two lines lying in $α$"},{"id":"C","text":"two intersecting lines lying in $α$"},{"id":"D","text":"infinitely many lines lying in $α$"}],"answer":"C","explanation":"The criterion requires two INTERSECTING lines in the plane. Two parallel lines are not enough (a line can be perpendicular to a whole family of parallel lines without being perpendicular to the $plane)$, and neither is a single line.","explanationZh":"判定定理要求直线垂直于平面内两条相交直线。只垂直于两条平行线是不够的（直线可以垂直于平面内一族平行线而不垂直于平面），只垂直于一条直线更不行。","tags":["medium","line-plane-perpendicular","criterion"],"status":"published"},{"id":"math-space-014","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"A cone has base radius 3 and height 4. Its volume is","stemZh":"底面半径为 3、高为 4 的圆锥，其体积为","options":[{"id":"A","text":"$12π$"},{"id":"B","text":"$36π$"},{"id":"C","text":"$15π$"},{"id":"D","text":"$24π$"}],"answer":"A","explanation":"$V = (1/3)πr^{2}h = (1/3)π·9·4 = 12π$. Dropping the factor 1/3 gives $36π; 15π = πrl$ is the lateral area (the slant height is $l = \\sqrt{3^{2}+4^{2}} = 5)$.","explanationZh":"$V = (1/3)πr^{2}h = (1/3)π·9·4 = 12π。漏掉系数 1/3$ 会得到 $36π；15π = πrl$ 是侧面积（母线 $l = \\sqrt{3^{2}+4^{2}} = 5）。$","tags":["medium","cone","volume"],"status":"published"},{"id":"math-space-015","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"A cone has base radius 3 and slant height 5. Its total surface area is","stemZh":"底面半径为 3、母线长为 5 的圆锥，其全面积（表面积）为","options":[{"id":"A","text":"$15π$"},{"id":"B","text":"$24π$"},{"id":"C","text":"$9π$"},{"id":"D","text":"$30π$"}],"answer":"B","explanation":"Total surface area $=$ lateral area + base area $= πrl + πr^{2} = π·3·5 + π·9 = 15π + 9π = 24π$. The value $15π$ is the lateral area alone and $9π$ is the base alone.","explanationZh":"表面积 $=$ 侧面积 + 底面积 $= πrl + πr^{2} = π·3·5 + π·9 = 15π + 9π = 24π。15π 只是侧面积，9π$ 只是底面积。","tags":["medium","cone","surface-area"],"status":"published"},{"id":"math-space-016","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"A cube has edge length 2. The volume of the sphere inscribed in the cube (touching all six faces) is","stemZh":"棱长为 2 的正方体，其内切球（与六个面都相切）的体积为","options":[{"id":"A","text":"$4π/3$"},{"id":"B","text":"$4π$"},{"id":"C","text":"$32π/3$"},{"id":"D","text":"$4\\sqrt{3}π$"}],"answer":"A","explanation":"The inscribed sphere has diameter equal to the edge, so $r = 1$ and $V = (4/3)πr^{3} = 4π/3$. Here $4π$ is the surface $area, 32π/3$ comes from using $r = 2$, and $4\\sqrt{3}π$ is the circumscribed sphere $(R = \\sqrt{3})$.","explanationZh":"内切球的直径等于棱长，故 $r = 1，V = (4/3)πr^{3} = 4π/3。4π 是表面积，32π/3$ 是误取 $r = 2，4\\sqrt{3}π$ 是外接球（R $= \\sqrt{3}）的体积。$","tags":["medium","sphere","inscribed-sphere"],"status":"published"},{"id":"math-space-017","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"A rectangular box measures 1 by 2 by 2. Its space diagonal has length","stemZh":"长、宽、高分别为 1、2、2 的长方体，其体对角线长为","options":[{"id":"A","text":"3"},{"id":"B","text":"$\\sqrt{5}$"},{"id":"C","text":"$\\sqrt{6}$"},{"id":"D","text":"$2\\sqrt{2}$"}],"answer":"A","explanation":"Space diagonal $= \\sqrt{1^{2} + 2^{2} + 2^{2}} = \\sqrt{9} = 3. \\sqrt{5}$ and $2\\sqrt{2} = \\sqrt{8}$ come from omitting one $dimension, \\sqrt{6}$ from using 1, 1, 2.","explanationZh":"体对角线长 $= \\sqrt{1^{2} + 2^{2} + 2^{2}} = \\sqrt{9} = 3。\\sqrt{5} 与 2\\sqrt{2} = \\sqrt{8} 都是漏掉一个方向的结果，\\sqrt{6}$ 是误用 1、1、2 计算的。","tags":["medium","cuboid","space-diagonal"],"status":"published"},{"id":"math-space-018","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"Which condition guarantees that plane $α$ is parallel to plane $β$?","stemZh":"下列哪个条件可以保证平面 $α$ 平行于平面 $β？$","options":[{"id":"A","text":"Two lines in $α$ are parallel to $β$"},{"id":"B","text":"Two intersecting lines in $α$ are parallel to $β$"},{"id":"C","text":"One line in $α$ is parallel to one line in $β$"},{"id":"D","text":"Both $α$ and $β$ are perpendicular to a third plane $γ$"}],"answer":"B","explanation":"The criterion needs two INTERSECTING lines of one plane parallel to the other plane. Two parallel lines in $α$ are not enough, and two planes both perpendicular to $γ$ may still intersect each other.","explanationZh":"判定定理要求一个平面内有两条相交直线都平行于另一个平面。两条平行线不够；两个平面都垂直于第三个平面时，它们仍可能相交。","tags":["medium","plane-plane-parallel","criterion"],"status":"published"},{"id":"math-space-019","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"An ant sits at vertex A of a cube of edge 1 and walks on the surface to the opposite vertex $C_{1}$. The shortest possible distance is","stemZh":"一只蚂蚁在棱长为 1 的正方体的顶点 A 处，沿表面爬到相对的顶点 $C_{1}，最短路程为$","options":[{"id":"A","text":"$\\sqrt{3}$"},{"id":"B","text":"3"},{"id":"C","text":"$1 + \\sqrt{2}$"},{"id":"D","text":"$\\sqrt{5}$"}],"answer":"D","explanation":"Unfold two adjacent faces into a 2 by 1 rectangle; the straight segment then has length $\\sqrt{2^{2} + 1^{2}} = \\sqrt{5}$. The value $\\sqrt{3}$ is the straight space diagonal (which leaves the surface) and $1 + \\sqrt{2}$ is a longer edge-plus-diagonal route.","explanationZh":"把相邻两个面展开成 $2\\times1$ 的矩形，最短路线长为 $\\sqrt{2^{2} + 1^{2}} = \\sqrt{5}。\\sqrt{3} 是体对角线（离开了表面），1 + \\sqrt{2}$ 是先走棱再走面对角线的较长路线。","tags":["medium","net","shortest-path"],"status":"published"},{"id":"math-space-020","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$, the plane through $A_{1}, B, C$ and $D_{1}$ makes with the base plane ABCD a dihedral angle of","stemZh":"在正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，过 $A_{1}、B、C、D_{1}$ 的平面与底面 ABCD 所成的二面角为","options":[{"id":"A","text":"30°"},{"id":"B","text":"45°"},{"id":"C","text":"60°"},{"id":"D","text":"90°"}],"answer":"B","explanation":"The two planes meet along BC. Cut by the plane through $A, B, A_{1}, B_{1}$: the section of the base is BA and the section of the oblique plane is $BA_{1}$, and $∠ABA_{1} = 45°$ because $AA_{1} =$ AB. So the dihedral angle is 45°.","explanationZh":"两平面交于 BC。用过 $A、B、A_{1}、B_{1}$ 的平面去截：底面截线为 BA，斜面截线为 $BA_{1}。由于 AA_{1} = AB，∠ABA_{1} = 45°，故二面角为 45°。$","tags":["medium","dihedral-angle","cube"],"status":"published"},{"id":"math-space-021","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":4,"stem":"A cube has edge length 2. Find the surface area of the sphere that passes through all eight vertices of the cube.","stemZh":"棱长为 2 的正方体，求经过它八个顶点的球的表面积。","options":[{"id":"A","text":"$12π$"},{"id":"B","text":"$4π$"},{"id":"C","text":"$48π$"},{"id":"D","text":"$16π$"}],"answer":"A","explanation":"The circumscribed sphere has diameter equal to the space diagonal $2\\sqrt{3}$, so $R = \\sqrt{3}$ and $S = 4πR^{2} = 4π·3 = 12π$. Here $4π$ is the inscribed sphere $(r = 1), 48π$ uses the full diagonal as the radius, and $16π$ uses $R = 2$.","explanationZh":"外接球的直径等于体对角线 $2\\sqrt{3}，故 R = \\sqrt{3}，S = 4πR^{2} = 4π·3 = 12π。4π$ 是内切球（r $= 1），48π 是把整条体对角线当半径，16π$ 是误取 $R = 2。$","tags":["hard","sphere","circumscribed-sphere"],"status":"published"},{"id":"math-space-022","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":5,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$ of edge 1, the distance from A to the plane $A_{1}BD$ is","stemZh":"在棱长为 1 的正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，点 A 到平面 $A_{1}BD$ 的距离为","options":[{"id":"A","text":"$\\sqrt{3}$"},{"id":"B","text":"$\\sqrt{3}/2$"},{"id":"C","text":"1/2"},{"id":"D","text":"$\\sqrt{3}/3$"}],"answer":"D","explanation":"Use the plane $x + y + z = 1 (it$ contains $A_1, B$ and D). The distance from $A(0,0,0)$ to it is $|0 − 1|/\\sqrt{1^{2}+1^{2}+1^{2}} = 1/\\sqrt{3} = \\sqrt{3}/3. Equivalently$, compare volumes of the tetrahedron $A-A_1BD: (1/3)·(\\sqrt{3}/2)·d = 1/6$.","explanationZh":"平面 $A_1BD$ 的方程为 $x + y + z = 1，点 A(0,0,0)$ 到它的距离 $d = |0 − 1|/\\sqrt{1^{2}+1^{2}+1^{2}} = 1/\\sqrt{3} = \\sqrt{3}/3。也可用四面体 A-A_1BD 的体积法：(1/3)·(\\sqrt{3}/2)·d = 1/6。$","tags":["hard","point-to-plane-distance","cube"],"status":"published"},{"id":"math-space-023","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":4,"stem":"A cone has base radius 4 and height 3. Its lateral surface area is","stemZh":"底面半径为 4、高为 3 的圆锥，其侧面积为","options":[{"id":"A","text":"$12π$"},{"id":"B","text":"$20π$"},{"id":"C","text":"$16π$"},{"id":"D","text":"$25π$"}],"answer":"B","explanation":"The slant height is $l = \\sqrt{r^{2} + h^{2}} = \\sqrt{16 + 9} = 5$, so the lateral area is $πrl = π·4·5 = 20π$. Using the height 3 instead of the slant height gives $12π; 16π$ is the base area and $25π = πl^{2}$.","explanationZh":"母线 $l = \\sqrt{r^{2} + h^{2}} = \\sqrt{16 + 9} = 5，故侧面积 S = πrl = π·4·5 = 20π。把高 3$ 当成母线会得到 $12π；16π 是底面积，25π = πl^{2}。$","tags":["hard","cone","surface-area"],"status":"published"},{"id":"math-space-024","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":5,"stem":"An ant crawls on the surface of a rectangular box measuring 1 by 2 by 2 from one vertex to the opposite vertex. The shortest possible distance is","stemZh":"一只蚂蚁沿长、宽、高分别为 1、2、2 的长方体表面，从一个顶点爬到相对的顶点，最短路程为","options":[{"id":"A","text":"3"},{"id":"B","text":"$\\sqrt{13}$"},{"id":"C","text":"$2\\sqrt{2}$"},{"id":"D","text":"$\\sqrt{17}$"}],"answer":"B","explanation":"Unfolding gives a rectangle whose sides are $(1 + 2)$ and 2, so the distance is $\\sqrt{3^{2} + 2^{2}} = \\sqrt{13}$. The other $unfolding, (2 + 2)$ by 1, gives $\\sqrt{17}$, which is longer; 3 is the straight space diagonal, which leaves the surface.","explanationZh":"展开后得到边长为 $(1 + 2) 与 2$ 的矩形，最短路程为 $\\sqrt{3^{2} + 2^{2}} = \\sqrt{13}。另一种展开（2 + 2）与 1$ 给出 $\\sqrt{17}，更长；3$ 是体对角线，离开了表面。","tags":["hard","net","shortest-path"],"status":"published"},{"id":"math-space-025","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":4,"stem":"A cube is inscribed in a sphere of radius $1 (all$ eight vertices of the cube lie on the sphere). The volume of the cube is","stemZh":"一个正方体内接于半径为 1 的球（正方体八个顶点都在球面上），正方体的体积为","options":[{"id":"A","text":"$8\\sqrt{3}/9$"},{"id":"B","text":"8"},{"id":"C","text":"$2\\sqrt{3}/3$"},{"id":"D","text":"8/9"}],"answer":"A","explanation":"The space diagonal of the cube equals the diameter 2, so $a\\sqrt{3} = 2$ and $a = 2\\sqrt{3}/3$. Then $V = a^{3} = 8·3\\sqrt{3}/27 = 8\\sqrt{3}/9$. The value 8 uses the diameter as the edge, while $2\\sqrt{3}/3$ is the edge length itself, not the volume.","explanationZh":"正方体的体对角线等于球的直径 $2，即 a\\sqrt{3} = 2，得 a = 2\\sqrt{3}/3。于是 V = a^{3} = 8·3\\sqrt{3}/27 = 8\\sqrt{3}/9。取 8 是把直径当棱长，2\\sqrt{3}/3$ 是棱长本身而非体积。","tags":["hard","cube","inscribed-sphere"],"status":"published"},{"id":"math-space-026","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":4,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$, which of the following statements is FALSE?","stemZh":"在正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，下列哪个结论是错误的？","options":[{"id":"A","text":"AB ⊥ plane $ADD_{1}A_{1}$"},{"id":"B","text":"$A_{1}A ⊥$ plane ABCD"},{"id":"C","text":"AB ⊥ plane ABCD"},{"id":"D","text":"AC ⊥ plane $BB_{1}D_{1}D$"}],"answer":"C","explanation":"AB lies IN the base plane ABCD, so it cannot be perpendicular to it. The others hold: AB ⊥ AD and AB $⊥ AA_{1} (two\\text{ intersecting lines of plane }ADD_{1}A_{1}); AA_{1}$ is an edge perpendicular to the base; and AC ⊥ BD together with AC $⊥ BB_{1}$ gives AC ⊥ plane $BB_{1}D_{1}D$.","explanationZh":"AB 本身在底面 ABCD 内，不可能垂直于该平面。其余都成立：AB ⊥ AD 且 AB $⊥ AA_{1}（平面 ADD_{1}A_{1} 内两条相交直线）；AA_{1}$ 是垂直于底面的棱；由 AC ⊥ BD 且 AC $⊥ BB_{1} 得$ AC ⊥ 平面 $BB_{1}D_{1}D。$","tags":["hard","line-plane-perpendicular","cube"],"status":"published"},{"id":"math-space-027","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":5,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$, which line is parallel to the plane $A_{1}BD$?","stemZh":"在正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，哪条直线平行于平面 $A_{1}BD？$","options":[{"id":"A","text":"AB"},{"id":"B","text":"$C_{1}D_{1}$"},{"id":"C","text":"$CD_{1}$"},{"id":"D","text":"$A_{1}C_{1}$"}],"answer":"C","explanation":"With $A(0,0,0), B(1,0,0), D(0,1,0), A_{1}(0,0,1)$ the plane $A_{1}BD$ is $x + y + z = 1$, with normal $(1,1,1)$. The direction of $CD_{1}$ is $(−1,0,1)$, whose dot product with the normal is 0, and C does not lie on the plane, so $CD_{1} ∥$ plane $A_{1}BD$. The other three lines each meet the plane $(AB\\text{ at B, }A_{1}C_{1}\\text{ at }A_{1},\\text{ and the line }C_{1}D_{1}\\text{ meets the extended }plane)$.","explanationZh":"$设 A(0,0,0)、B(1,0,0)、D(0,1,0)、A_{1}(0,0,1)，则平面 A_{1}BD 为 x + y + z = 1，法向量为 (1,1,1)。CD_{1}$ 的方向向量为 $(−1,0,1)，与法向量数量积为 0，且 C$ 不在该平面内，故 $CD_{1} ∥$ 平面 $A_{1}BD。其余三条直线都与该平面相交（AB$ 交于 $B，A_{1}C_{1}$ 交于 $A_{1}，C_{1}D_{1}$ 所在直线与平面相交）。","tags":["hard","line-plane-parallel","cube"],"status":"published"},{"id":"math-space-028","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":5,"stem":"A sphere is inscribed in a right circular cylinder, touching both bases and the lateral surface. The ratio V(sphere) : V(cylinder) is","stemZh":"一个球内切于一个直圆柱（与两个底面和侧面都相切），则 V(球) : V(圆柱) 等于","options":[{"id":"A","text":"1 : 2"},{"id":"B","text":"3 : 4"},{"id":"C","text":"2 : 3"},{"id":"D","text":"1 : 3"}],"answer":"C","explanation":"The cylinder has radius r and height 2r. So $V(sphere) = (4/3)πr^{3}$ and $V(cylinder) = πr^{2}·2r = 2πr^{3}$, giving the ratio $(4/3) : 2 = 2 : 3$.","explanationZh":"圆柱底面半径为 r、高为 $2r，故 V(球) = (4/3)πr^{3}，V(圆柱) = πr^{2}·2r = 2πr^{3}，二者之比为 (4/3) : 2 = 2 : 3。$","tags":["hard","sphere","cylinder","volume-ratio"],"status":"published"},{"id":"math-space-029","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":4,"stem":"The lateral surface of a cylinder with base radius 1 and height 2 is cut along a generator and flattened. The resulting rectangle measures","stemZh":"把底面半径为 1、高为 2 的圆柱沿一条母线剪开并展平，所得矩形的两边长为","options":[{"id":"A","text":"$π$ and 2"},{"id":"B","text":"$2π$ and 2"},{"id":"C","text":"$2π$ and 4"},{"id":"D","text":"$4π$ and 2"}],"answer":"B","explanation":"One side of the rectangle is the height 2, the other is the circumference of the $base, 2πr = 2π$. Using $πr$ would give $π$, and using $2π(2r) = 4π$ would give a strip that is twice as wide as the actual lateral surface.","explanationZh":"矩形的一边是高 $2，另一边是底面周长 2πr = 2π。用 πr$ 会得到 $π，用 2π(2r) = 4π$ 则把侧面宽度算成了实际的两倍。","tags":["hard","net","cylinder"],"status":"published"},{"id":"math-space-030","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":5,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$ of edge 1, the tetrahedron with vertices $A_{1}, B, C_{1}$ and D has volume","stemZh":"在棱长为 1 的正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，以 $A_{1}、B、C_{1}、D$ 为顶点的四面体的体积为","options":[{"id":"A","text":"1/6"},{"id":"B","text":"2/3"},{"id":"C","text":"1/2"},{"id":"D","text":"1/3"}],"answer":"D","explanation":"Its six edges are face diagonals of length $\\sqrt{2}$, so it is a regular tetrahedron, obtained from the cube by removing four congruent corner tetrahedra $(A-A_1BD, B-A_1BC_1, C_1-BC_1D, D-A_1C_1D)$, each of volume 1/6. Hence $V = 1 − 4·(1/6) = 1/3$.","explanationZh":"该四面体的六条棱都是长为 $\\sqrt{2} 的面对角线，是正四面体。它由正方体挖去四个全等的角上四面体（A-A_1BD、B-A_1BC_1、C_1-BC_1D、D-A_1C_1D）得到，每个体积为 1/6，故 V = 1 − 4·(1/6) = 1/3。$","tags":["hard","tetrahedron","volume","cube"],"status":"published"},{"id":"math-counting-001","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":1,"stem":"A student owns 3 different shirts and 4 different pairs of trousers. How many different outfits can be made by choosing one shirt and one pair of trousers?","stemZh":"某学生有 3 件不同的衬衫和 4 条不同的裤子，选一件衬衫和一条裤子搭配，共有多少种不同穿法？","options":[{"id":"A","text":"7"},{"id":"B","text":"24"},{"id":"C","text":"12"},{"id":"D","text":"10"}],"answer":"C","explanation":"This needs two steps $(shirt$, then $trousers)$, so use the multiplication $principle: 3 \\times 4 = 12$. Adding $(3 + 4 = 7)$ answers a different question — choosing a shirt OR a pair of trousers.","explanationZh":"$这件事分两步完成（先选衬衫，再选裤子），用分步乘法计数原理：3 \\times 4 = 12。若用加法 3 + 4 = 7，回答的是‘选一件衬衫或一条裤子’的问题。$","tags":["easy","multiplication-principle"],"status":"published"},{"id":"math-counting-002","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":2,"stem":"How many three-digit numbers can be formed from the digits 1, 2, 3, 4 if no digit may be repeated?","stemZh":"用数字 1、2、3、4 组成没有重复数字的三位数，共可组成多少个？","options":[{"id":"A","text":"24"},{"id":"B","text":"64"},{"id":"C","text":"12"},{"id":"D","text":"6"}],"answer":"A","explanation":"We arrange 3 of the 4 digits, and order $matters: A(4,3) = 4 \\times 3 \\times 2 = 24$. The value $4^{3} = 64$ allows repetition, and $C(4,3) = 12$ ignores the order of the digits.","explanationZh":"从 4 个数字中取 $3 个排成一列，与顺序有关：A(4,3) = 4 \\times 3 \\times 2 = 24。4^{3} = 64 是允许重复的情形，C(4,3) = 12$ 则忽略了数字的顺序。","tags":["easy","permutation"],"status":"published"},{"id":"math-counting-003","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":1,"stem":"In how many ways can 5 different books be arranged in a row on a shelf?","stemZh":"5 本不同的书排成一排放在书架上，共有多少种排法？","options":[{"id":"A","text":"25"},{"id":"B","text":"120"},{"id":"C","text":"60"},{"id":"D","text":"24"}],"answer":"B","explanation":"Arranging all 5 objects is a full $permutation: 5! = 5 \\times 4 \\times 3 \\times 2 \\times 1 = 120$. The slips $5^{2} = 25$ and $4! = 24$ are common but wrong.","explanationZh":"$5 个不同元素的全排列：5! = 5 \\times 4 \\times 3 \\times 2 \\times 1 = 120。常见错误是写成 5^{2} = 25 或 4! = 24。$","tags":["easy","factorial","permutation"],"status":"published"},{"id":"math-counting-004","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":1,"stem":"The value of $C(6,2)$ is","stemZh":"$C(6,2)$ 的值为","options":[{"id":"A","text":"30"},{"id":"B","text":"36"},{"id":"C","text":"12"},{"id":"D","text":"15"}],"answer":"D","explanation":"$C(6,2) = 6!/(2!·4!) = (6 \\times 5)/2 = 15$. The value 30 is $A(6,2)$, which counts each pair twice because it keeps the order.","explanationZh":"$C(6,2) = 6!/(2!·4!) = (6 \\times 5)/2 = 15。30 是 A(6,2)，它保留了顺序，把同一组重复计算了两次。$","tags":["easy","combination"],"status":"published"},{"id":"math-counting-005","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":2,"stem":"The value of $A(5,3)$ is","stemZh":"$A(5,3)$ 的值为","options":[{"id":"A","text":"15"},{"id":"B","text":"20"},{"id":"C","text":"60"},{"id":"D","text":"125"}],"answer":"C","explanation":"$A(5,3) = 5 \\times 4 \\times 3 = 60$. Using $C(5,3) = 15$ ignores the order, and $5^{3} = 125$ is the case where repetition is allowed.","explanationZh":"$A(5,3) = 5 \\times 4 \\times 3 = 60。C(5,3) = 15 忽略了顺序，5^{3} = 125$ 是允许重复选取的情形。","tags":["easy","permutation"],"status":"published"},{"id":"math-counting-006","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":2,"stem":"Using the identity $C(n,m) = C(n,n−m)$, the value of $C(8,6)$ is","stemZh":"利用公式 $C(n,m) = C(n,n−m)，C(8,6)$ 的值为","options":[{"id":"A","text":"28"},{"id":"B","text":"56"},{"id":"C","text":"48"},{"id":"D","text":"40320"}],"answer":"A","explanation":"$C(8,6) = C(8,2) = (8 \\times 7)/2 = 28$. The value $56 = A(8,2)$ counts $order, 48 = 8 \\times 6$ is plain multiplication, and $40320 = 8$! is a full permutation.","explanationZh":"$C(8,6) = C(8,2) = (8 \\times 7)/2 = 28。56 = A(8,2) 把顺序算进去了，48 = 8 \\times 6 只是简单相乘，40320 = 8$! 是全排列。","tags":["easy","combination","symmetry"],"status":"published"},{"id":"math-counting-007","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":1,"stem":"To travel from town A to town B one may take one of 3 buses or one of 2 trains. How many different ways are there to make the trip?","stemZh":"从 A 城到 B 城可以乘 3 路公交车中的任意一班，或乘 2 列火车中的任意一列，共有多少种走法？","options":[{"id":"A","text":"6"},{"id":"B","text":"5"},{"id":"C","text":"3"},{"id":"D","text":"2"}],"answer":"B","explanation":"The two types of transport are alternatives, so use the addition $principle: 3 + 2 = 5$. Multiplying $(3 \\times 2 = 6)$ would only be right if the trip required a bus AND a train.","explanationZh":"$公交与火车是彼此独立的两类方案，任选其一即可，用分类加法计数原理：3 + 2 = 5。只有‘先乘公交再乘火车’这类分步问题才用乘法 3 \\times 2 = 6。$","tags":["easy","addition-principle"],"status":"published"},{"id":"math-counting-008","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":1,"stem":"From 5 students, in how many ways can 2 of them be chosen to form a committee?","stemZh":"从 5 名学生中选出 2 人组成一个小组，共有多少种选法？","options":[{"id":"A","text":"20"},{"id":"B","text":"25"},{"id":"C","text":"7"},{"id":"D","text":"10"}],"answer":"D","explanation":"The two members have no distinct roles, so order does not $matter: C(5,2) = (5 \\times 4)/2 = 10. A(5,2) = 20$ counts every pair twice.","explanationZh":"$小组成员没有职务差别，与顺序无关：C(5,2) = (5 \\times 4)/2 = 10。A(5,2) = 20$ 会把同一组重复计算两次。","tags":["easy","combination"],"status":"published"},{"id":"math-counting-009","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":2,"stem":"How many two-digit numbers can be formed from the digits 1, 2, 3, 4, 5 if digits may be repeated?","stemZh":"用数字 1、2、3、4、5 组成两位数，数字允许重复使用，共可组成多少个？","options":[{"id":"A","text":"20"},{"id":"B","text":"10"},{"id":"C","text":"25"},{"id":"D","text":"32"}],"answer":"C","explanation":"Each of the two positions can be filled in 5 ways, so $5 \\times 5 = 25$. The value $A(5,2) = 20$ is the answer when repetition is NOT allowed.","explanationZh":"十位和个位各有 5 种填法，故 $5 \\times 5 = 25。A(5,2) = 20$ 是不允许重复数字时的答案。","tags":["easy","multiplication-principle","repetition"],"status":"published"},{"id":"math-counting-010","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":2,"stem":"The coefficient of $x^{2}$ in the expansion of $(1 + x)^{4}$ is","stemZh":"$(1 + x)^{4}$ 的展开式中 $x^{2}$ 的系数是","options":[{"id":"A","text":"6"},{"id":"B","text":"4"},{"id":"C","text":"12"},{"id":"D","text":"16"}],"answer":"A","explanation":"By the binomial theorem the term is $C(4,2)·1^{2}·x^{2} = 6x^{2}$, so the coefficient is 6. The exponent 4 and the value $4^{2} = 16$ are not coefficients.","explanationZh":"由二项式定理，该项为 $C(4,2)·1^{2}·x^{2} = 6x^{2}，故系数为 6。指数 4$ 以及 $4^{2} = 16$ 都不是系数。","tags":["easy","binomial-theorem"],"status":"published"},{"id":"math-counting-011","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"Six people stand in a row. If Ann and Ben must stand next to each other, how many arrangements are possible?","stemZh":"6 人站成一排，若甲、乙两人必须相邻，共有多少种排法？","options":[{"id":"A","text":"720"},{"id":"B","text":"120"},{"id":"C","text":"48"},{"id":"D","text":"240"}],"answer":"D","explanation":"Bind Ann and Ben into one block: 5 objects give $5! = 120 orders$, and inside the block they can swap $(2! = 2)$. Total $= 2 \\times 120 = 240$.","explanationZh":"把甲、乙捆成一个整体：5 个对象全排列有 $5! = 120 种，整体内部两人可交换位置（2! = 2）。故共 2 \\times 120 = 240$ 种。","tags":["medium","permutation","adjacent"],"status":"published"},{"id":"math-counting-012","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"Six people stand in a row and Ann must stand in the first position. How many arrangements are possible?","stemZh":"6 人站成一排，若甲必须站在第一个位置，共有多少种排法？","options":[{"id":"A","text":"720"},{"id":"B","text":"120"},{"id":"C","text":"600"},{"id":"D","text":"24"}],"answer":"B","explanation":"Ann occupies a fixed place, so only the other 5 people are $arranged: 5! = 120$. The value $6! = 720$ ignores the restriction.","explanationZh":"甲的位置固定，只需排列其余 $5 人：5! = 120。6! = 720$ 是没有考虑限制条件的排法数。","tags":["medium","permutation","restriction"],"status":"published"},{"id":"math-counting-013","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"In how many ways can 5 people be seated around a round table, if arrangements that differ only by rotation count as the same?","stemZh":"5 人围坐在一张圆桌旁，若仅由旋转得到的坐法视为相同，共有多少种坐法？","options":[{"id":"A","text":"120"},{"id":"B","text":"20"},{"id":"C","text":"24"},{"id":"D","text":"5"}],"answer":"C","explanation":"Fix one person to remove the rotational symmetry, then arrange the remaining $4: (5 − 1)! = 4! = 24$. The value $5! = 120$ counts each seating 5 times.","explanationZh":"先固定一人以消除旋转带来的重复，再排列其余 $4 人：(5 − 1)! = 4! = 24。5! = 120$ 会把每种坐法重复计算 5 次。","tags":["medium","circular-permutation"],"status":"published"},{"id":"math-counting-014","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"How many different arrangements can be made using all the letters of the word BANANA?","stemZh":"用单词 BANANA 的全部字母进行排列，共有多少种不同的排法？","options":[{"id":"A","text":"60"},{"id":"B","text":"720"},{"id":"C","text":"120"},{"id":"D","text":"30"}],"answer":"A","explanation":"BANANA has 6 letters, with A appearing 3 times and N twice. Divide by the $repeats: 6!/(3!·2!) = 720/12 = 60$.","explanationZh":"BANANA 共 6 个字母，其中 A 出现 3 次、N 出现 $2 次，需除以重复元素的全排列：6!/(3!·2!) = 720/12 = 60。$","tags":["medium","repeated-elements","permutation"],"status":"published"},{"id":"math-counting-015","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"A group of 3 is chosen from 4 boys and 3 girls and must contain exactly 1 girl. How many such groups are there?","stemZh":"从 4 名男生和 3 名女生中选出 3 人，要求恰好有 1 名女生，共有多少种选法？","options":[{"id":"A","text":"35"},{"id":"B","text":"30"},{"id":"C","text":"9"},{"id":"D","text":"18"}],"answer":"D","explanation":"Choose the girl $(C(3,1) = 3)$ and the two boys $(C(4,2) = 6)$, then $multiply: 3 \\times 6 = 18$. The value $35 = C(7,3)$ is the total without the restriction.","explanationZh":"先选女生 $C(3,1) = 3$ 种，再选两名男生 $C(4,2) = 6$ 种，相乘得 $3 \\times 6 = 18。35 = C(7,3)$ 是没有限制条件时的总数。","tags":["medium","combination","classification"],"status":"published"},{"id":"math-counting-016","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"How many diagonals does a convex hexagon have?","stemZh":"一个凸六边形共有多少条对角线？","options":[{"id":"A","text":"15"},{"id":"B","text":"9"},{"id":"C","text":"12"},{"id":"D","text":"6"}],"answer":"B","explanation":"$C(6,2) = 15$ counts all segments joining two vertices; 6 of these are sides of the hexagon, leaving $15 − 6 = 9$ diagonals.","explanationZh":"$C(6,2) = 15$ 表示任取两个顶点连成的线段总数，其中 6 条是六边形的边，故对角线为 $15 − 6 = 9$ 条。","tags":["medium","combination","geometry"],"status":"published"},{"id":"math-counting-017","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"How many three-digit numbers with no repeated digits can be formed from the digits 0, 1, 2, 3, 4?","stemZh":"用数字 0、1、2、3、4 组成没有重复数字的三位数，共可组成多少个？","options":[{"id":"A","text":"60"},{"id":"B","text":"36"},{"id":"C","text":"48"},{"id":"D","text":"24"}],"answer":"C","explanation":"The hundreds digit cannot be 0, so it has 4 choices. Then 4 digits remain for the tens place and 3 for the $units: 4 \\times 4 \\times 3 = 48$. Using $A(5,3) = 60$ wrongly allows a leading zero.","explanationZh":"百位不能为 0，有 4 种选法；十位从剩下 4 个数字中选，个位从剩下 $3 个中选：4 \\times 4 \\times 3 = 48。A(5,3) = 60$ 错误地允许了首位为 0。","tags":["medium","permutation","restriction"],"status":"published"},{"id":"math-counting-018","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"From 10 candidates, in how many ways can a president and a vice-president be selected?","stemZh":"从 10 名候选人中选出一名主席和一名副主席，共有多少种选法？","options":[{"id":"A","text":"45"},{"id":"B","text":"100"},{"id":"C","text":"20"},{"id":"D","text":"90"}],"answer":"D","explanation":"The two posts are different, so order $matters: A(10,2) = 10 \\times 9 = 90$. Using $C(10,2) = 45$ would treat the two posts as identical.","explanationZh":"$主席与副主席是两个不同的职位，与顺序有关：A(10,2) = 10 \\times 9 = 90。若用 C(10,2) = 45，相当于把两个职位看成相同的。$","tags":["medium","permutation"],"status":"published"},{"id":"math-counting-019","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"In how many ways can 5 different gifts be given to 3 children, if each gift goes to exactly one child?","stemZh":"5 件不同的礼物分给 3 个小孩，每件礼物恰好分给一个小孩，共有多少种分法？","options":[{"id":"A","text":"243"},{"id":"B","text":"125"},{"id":"C","text":"60"},{"id":"D","text":"15"}],"answer":"A","explanation":"Each of the 5 gifts independently has 3 possible recipients, so $3^{5} = 243$. The values $5^{3} = 125$ and $3 \\times 5 = 15$ swap the roles of gifts and children.","explanationZh":"每件礼物都有 $3 种分配对象，5 件礼物相互独立：3^{5} = 243。5^{3} = 125 与 3 \\times 5 = 15$ 都弄混了礼物与小孩的角色。","tags":["medium","multiplication-principle"],"status":"published"},{"id":"math-counting-020","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"The coefficient of $x^{3}$ in the expansion of $(1 + x)^{7}$ is","stemZh":"$(1 + x)^{7}$ 的展开式中 $x^{3}$ 的系数是","options":[{"id":"A","text":"21"},{"id":"B","text":"35"},{"id":"C","text":"210"},{"id":"D","text":"343"}],"answer":"B","explanation":"The general term is $C(7,k)·x^k$, so the coefficient of $x^{3}$ is $C(7,3) = 35$. Using $A(7,3) = 210$ wrongly treats the coefficient as an arrangement count.","explanationZh":"通项为 $C(7,k)·x^k，故 x^{3}$ 的系数为 $C(7,3) = 35。A(7,3) = 210$ 是错把系数当成了排列数。","tags":["medium","binomial-theorem"],"status":"published"},{"id":"math-counting-021","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":4,"stem":"Seven people stand in a row and Ann, Ben and Carl must all stand together. How many arrangements are possible?","stemZh":"7 人站成一排，若甲、乙、丙三人必须站在一起，共有多少种排法？","options":[{"id":"A","text":"5040"},{"id":"B","text":"120"},{"id":"C","text":"720"},{"id":"D","text":"1440"}],"answer":"C","explanation":"Bind the three into a single block: 5 objects give $5! = 120 orders$, and inside the block the three can be ordered in $3! = 6$ ways. Total $= 6 \\times 120 = 720$.","explanationZh":"把三人捆成一个整体：5 个对象全排列有 $5! = 120$ 种，整体内部三人有 $3! = 6$ 种排法。故共 $6 \\times 120 = 720$ 种。","tags":["hard","permutation","adjacent"],"status":"published"},{"id":"math-counting-022","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":4,"stem":"Six people stand in a row and Ann and Ben must NOT stand next to each other. How many arrangements are possible?","stemZh":"6 人站成一排，若甲、乙两人不能相邻，共有多少种排法？","options":[{"id":"A","text":"240"},{"id":"B","text":"360"},{"id":"C","text":"600"},{"id":"D","text":"480"}],"answer":"D","explanation":"Use the complement: all $6! = 720$ orders minus the $2!·5! = 240$ orders in which Ann and Ben are adjacent, giving $720 − 240 = 480$.","explanationZh":"用补集法：全体排法 $6! = 720，减去甲、乙相邻的 2!·5! = 240$ 种，得 $720 − 240 = 480$ 种。","tags":["hard","permutation","complement"],"status":"published"},{"id":"math-counting-023","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":4,"stem":"A team of 4 is chosen from 5 boys and 4 girls and must contain at least one girl. How many such teams are there?","stemZh":"从 5 名男生和 4 名女生中选出 4 人，要求至少有一名女生，共有多少种选法？","options":[{"id":"A","text":"126"},{"id":"B","text":"121"},{"id":"C","text":"120"},{"id":"D","text":"84"}],"answer":"B","explanation":"Take all $C(9,4) = 126$ possible teams and remove the all-boy teams $C(5,4) = 5: 126 − 5 = 121$. Directly counting 1, 2, 3 or 4 girls gives the same result but takes longer.","explanationZh":"先取全体 $C(9,4) = 126$ 种，再减去全是男生的 $C(5,4) = 5 种：126 − 5 = 121。分类数 1、2、3、4$ 名女生也可得到同样结果，但计算更繁琐。","tags":["hard","combination","at-least-one"],"status":"published"},{"id":"math-counting-024","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":5,"stem":"How many four-digit even numbers with no repeated digits can be formed from the digits 0, 1, 2, 3, 4, 5?","stemZh":"用数字 0、1、2、3、4、5 组成没有重复数字的四位偶数，共可组成多少个？","options":[{"id":"A","text":"156"},{"id":"B","text":"300"},{"id":"C","text":"180"},{"id":"D","text":"108"}],"answer":"A","explanation":"Split on the units digit. If it is 0, the other three places are filled from the remaining $5 digits: A(5,3) = 60$. If it is 2 or $4 (2 choices)$, the thousands digit has 4 choices $(not 0$ and not the units digit) and the middle two places give $A(4,2) = 12: 2 \\times 4 \\times 12 = 96$. Total $= 60 + 96 = 156$.","explanationZh":"按个位分类。个位为 0 时，其余三位从剩下 $5 个数字中选：A(5,3) = 60；个位为 2 或 4 时（2$ 种），千位有 4 种选法（不能为 $0，也不能与个位重复），中间两位有 A(4,2) = 12$ 种，共 $2 \\times 4 \\times 12 = 96。合计 60 + 96 = 156。$","tags":["hard","permutation","restriction"],"status":"published"},{"id":"math-counting-025","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":5,"stem":"Six people stand in a row. In how many arrangements is Ann to the left of Ben (not necessarily adjacent)?","stemZh":"6 人站成一排，甲站在乙的左边（不要求相邻）的排法共有多少种？","options":[{"id":"A","text":"720"},{"id":"B","text":"120"},{"id":"C","text":"360"},{"id":"D","text":"240"}],"answer":"C","explanation":"Of the $6! = 720 orders$, exactly half have Ann left of Ben and half have her right of him (swap their names to pair them $up)$, so $720/2 = 360$.","explanationZh":"$在 6! = 720$ 种排法中，甲在乙左与甲在乙右的情况一一对应、各占一半，故 $720/2 = 360$ 种。","tags":["hard","permutation","symmetry"],"status":"published"},{"id":"math-counting-026","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":4,"stem":"How many different arrangements can be made using all the letters of the word SUCCESS?","stemZh":"用单词 SUCCESS 的全部字母进行排列，共有多少种不同的排法？","options":[{"id":"A","text":"5040"},{"id":"B","text":"840"},{"id":"C","text":"210"},{"id":"D","text":"420"}],"answer":"D","explanation":"SUCCESS has 7 letters, with S appearing 3 times and $C twice: 7!/(3!·2!) = 5040/12 = 420$. Dividing by only one of the repeats gives 840 or 1680.","explanationZh":"SUCCESS 共 7 个字母，其中 S 出现 3 次、C 出现 $2 次：7!/(3!·2!) = 5040/12 = 420。只除以其中一个重复因子会得到 840$ 等错误结果。","tags":["hard","repeated-elements","permutation"],"status":"published"},{"id":"math-counting-027","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":5,"stem":"In how many ways can 4 different books be distributed to 3 students so that every student receives at least one book?","stemZh":"4 本不同的书分给 3 名学生，要求每名学生至少分到一本书，共有多少种分法？","options":[{"id":"A","text":"36"},{"id":"B","text":"81"},{"id":"C","text":"72"},{"id":"D","text":"24"}],"answer":"A","explanation":"Without the condition there are $3^{4} = 81$ ways. Subtract the ways with a chosen student $empty, C(3,1)·2^{4} = 48$, then add back those with two chosen students $empty, C(3,2)·1^{4} = 3: 81 − 48 + 3 = 36. (Equivalently$: the split must be 2+1+1 — choose who gets 2 books $(3 ways)$, choose the pair $(C(4,2) = 6)$ and hand out the rest $(2!): 3 \\times 6 \\times 2 = 36.)$","explanationZh":"不加限制有 $3^{4} = 81$ 种；减去指定一名学生为空的 $C(3,1)·2^{4} = 48$ 种，再加回指定两名学生为空的 $C(3,2)·1^{4} = 3 种：81 − 48 + 3 = 36。（也可按 2+1+1$ 分配：选拿到 2 本的人 3 种，选这 2 本书 $C(4,2) = 6$ 种，其余分法 $2! 种，3 \\times 6 \\times 2 = 36。）$","tags":["hard","inclusion-exclusion","distribution"],"status":"published"},{"id":"math-counting-028","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":4,"stem":"The coefficient of $x^{2}$ in the expansion of $(2 + x)^{5}$ is","stemZh":"$(2 + x)^{5}$ 的展开式中 $x^{2}$ 的系数是","options":[{"id":"A","text":"40"},{"id":"B","text":"80"},{"id":"C","text":"10"},{"id":"D","text":"32"}],"answer":"B","explanation":"The term is $C(5,2)·2^{3}·x^{2} = 10 \\times 8 \\times x^{2} = 80x^{2}$, so the coefficient is 80. Taking only $C(5,2) = 10$ forgets the power of the constant 2.","explanationZh":"该项为 $C(5,2)·2^{3}·x^{2} = 10 \\times 8 \\times x^{2} = 80x^{2}，故系数为 80。只取 C(5,2) = 10$ 是漏掉了常数 2 的幂次。","tags":["hard","binomial-theorem"],"status":"published"},{"id":"math-counting-029","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":5,"stem":"In how many ways can 6 people be split into 3 unordered pairs?","stemZh":"把 6 个人分成 3 组，每组 2 人且组与组之间不分顺序，共有多少种分法？","options":[{"id":"A","text":"120"},{"id":"B","text":"45"},{"id":"C","text":"15"},{"id":"D","text":"30"}],"answer":"C","explanation":"Line the 6 people up $(6!)$, then divide by the order inside each pair $(2!·2!·2!)$ and by the order of the 3 pairs $(3!): 720/48 = 15$.","explanationZh":"先把 $6 人排成一列（6!），再除以每组内部的顺序（2!·2!·2!）和 3 组之间的顺序（3!）：720/48 = 15。$","tags":["hard","grouping","combination"],"status":"published"},{"id":"math-counting-030","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":4,"stem":"How many non-empty subsets does a set with 5 elements have?","stemZh":"含有 5 个元素的集合共有多少个非空子集？","options":[{"id":"A","text":"32"},{"id":"B","text":"25"},{"id":"C","text":"10"},{"id":"D","text":"31"}],"answer":"D","explanation":"Each of the 5 elements is independently either in or out, giving $2^{5} = 32 subsets$; deleting the empty set leaves 31.","explanationZh":"5 个元素各自都有‘取’与‘不取’两种选择，共有 $2^{5} = 32$ 个子集；去掉空集后为 31 个。","tags":["hard","subsets","counting"],"status":"published"},{"id":"math-classical-probability-001","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":1,"stem":"A fair die is rolled once. What is the probability of getting an even score?","stemZh":"掷一枚均匀的骰子一次，出现偶数点的概率是多少？","options":[{"id":"A","text":"1/2"},{"id":"B","text":"1/6"},{"id":"C","text":"1/3"},{"id":"D","text":"2/3"}],"answer":"A","explanation":"Three of the six equally likely outcomes $(2, 4, 6)$ are even, so $P = 3/6 = 1/2$.","explanationZh":"6 个等可能结果中，2、4、6 共 3 个是偶数，故 $P = 3/6 = 1/2。$","tags":["easy","classical-probability","dice"],"status":"published"},{"id":"math-classical-probability-002","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":2,"stem":"A fair coin is tossed twice. What is the probability of getting at least one head?","stemZh":"一枚均匀硬币连掷两次，至少出现一次正面的概率是多少？","options":[{"id":"A","text":"1/4"},{"id":"B","text":"1/2"},{"id":"C","text":"3/4"},{"id":"D","text":"1"}],"answer":"C","explanation":"The four equally likely outcomes are HH, HT, TH, TT; only TT has no head, so $P = 3/4 (equivalently 1 − 1/4 = 3/4)$.","explanationZh":"4 个等可能结果为 HH、HT、TH、TT，只有 TT 没有正面，故 $P = 3/4（即 1 − 1/4 = 3/4）。$","tags":["easy","complement","coins"],"status":"published"},{"id":"math-classical-probability-003","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":1,"stem":"A bag contains 3 red balls and 2 blue balls. One ball is drawn at random. What is the probability that it is red?","stemZh":"袋中有 3 个红球和 2 个蓝球，随机取出一个，取到红球的概率是多少？","options":[{"id":"A","text":"2/5"},{"id":"B","text":"3/5"},{"id":"C","text":"1/5"},{"id":"D","text":"3/2"}],"answer":"B","explanation":"All 5 balls are equally likely to be drawn and 3 of them are red, so $P = 3/5$.","explanationZh":"5 个球被取到的机会相同，其中红球 3 个，故 $P = 3/5。$","tags":["easy","classical-probability","balls"],"status":"published"},{"id":"math-classical-probability-004","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":1,"stem":"One card is drawn at random from a standard 52-card deck. What is the probability that it is a heart?","stemZh":"从一副 52 张的扑克牌中随机抽取一张，抽到红桃的概率是多少？","options":[{"id":"A","text":"1/52"},{"id":"B","text":"1/13"},{"id":"C","text":"4/13"},{"id":"D","text":"1/4"}],"answer":"D","explanation":"13 of the 52 cards are hearts, so $P = 13/52 = 1/4$. The value 1/13 is the probability of one particular rank being drawn.","explanationZh":"52 张牌中有 13 张红桃，故 $P = 13/52 = 1/4。1/13$ 是抽到某一特定点数的概率。","tags":["easy","classical-probability","cards"],"status":"published"},{"id":"math-classical-probability-005","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":2,"stem":"A fair coin is tossed three times. What is the probability that all three tosses are heads?","stemZh":"一枚均匀硬币连掷三次，三次都是正面的概率是多少？","options":[{"id":"A","text":"1/8"},{"id":"B","text":"1/6"},{"id":"C","text":"1/4"},{"id":"D","text":"3/8"}],"answer":"A","explanation":"There are $2^{3} = 8$ equally likely outcomes and only HHH qualifies, so $P = 1/8$; by independence this is also $(1/2)^{3} = 1/8$.","explanationZh":"共有 $2^{3} = 8$ 个等可能结果，只有 HHH 符合，故 $P = 1/8；由独立性也可得 (1/2)^{3} = 1/8。$","tags":["easy","coins","independent-events"],"status":"published"},{"id":"math-classical-probability-006","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":1,"stem":"A number is chosen at random from 1, 2, …, 10. What is the probability that it is a multiple of 3?","stemZh":"从 1, 2, …, 10 中随机取一个数，它是 3 的倍数的概率是多少？","options":[{"id":"A","text":"1/3"},{"id":"B","text":"1/10"},{"id":"C","text":"3/10"},{"id":"D","text":"3/9"}],"answer":"C","explanation":"The multiples of 3 in the list are 3, 6 and 9, so 3 of the 10 numbers are $favourable: P = 3/10$.","explanationZh":"1 到 10 中 3 的倍数有 3、6、9，共 3 个数，故 $P = 3/10。$","tags":["easy","classical-probability"],"status":"published"},{"id":"math-classical-probability-007","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":1,"stem":"A box contains 4 white balls and 6 black balls. One ball is drawn at random. What is the probability that it is white?","stemZh":"盒中有 4 个白球和 6 个黑球，随机取出一个，取到白球的概率是多少？","options":[{"id":"A","text":"4/6"},{"id":"B","text":"2/5"},{"id":"C","text":"1/5"},{"id":"D","text":"3/5"}],"answer":"B","explanation":"There are 10 balls in total, of which 4 are white, so $P = 4/10 = 2/5$. The ratio 4/6 compares white with black instead of with the total.","explanationZh":"共 10 个球，其中白球 4 个，故 $P = 4/10 = 2/5。4/6$ 是白球与黑球之比，分母不是总数。","tags":["easy","classical-probability","balls"],"status":"published"},{"id":"math-classical-probability-008","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":1,"stem":"A fair die is rolled once. What is the probability that the score is greater than 4?","stemZh":"掷一枚均匀的骰子一次，点数大于 4 的概率是多少？","options":[{"id":"A","text":"1/6"},{"id":"B","text":"1/2"},{"id":"C","text":"2/3"},{"id":"D","text":"1/3"}],"answer":"D","explanation":"Only the scores 5 and 6 qualify, so $P = 2/6 = 1/3$. Using a strict inequality matters: 4 itself does not count.","explanationZh":"只有 5 点和 6 点符合，故 $P = 2/6 = 1/3。注意是严格大于 4，点数 4$ 本身不算。","tags":["easy","dice","classical-probability"],"status":"published"},{"id":"math-classical-probability-009","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":2,"stem":"A bag contains 2 red balls and 3 blue balls. Two balls are drawn at random without replacement. What is the probability that both are red?","stemZh":"袋中有 2 个红球和 3 个蓝球，不放回地随机取出 2 个，两个都是红球的概率是多少？","options":[{"id":"A","text":"1/10"},{"id":"B","text":"4/25"},{"id":"C","text":"1/5"},{"id":"D","text":"2/5"}],"answer":"A","explanation":"$P = C(2,2)/C(5,2) = 1/10$, or step by step $(2/5) \\times (1/4) = 1/10$. The value $4/25 = (2/5)^{2}$ is the answer when the first ball is replaced.","explanationZh":"$P = C(2,2)/C(5,2) = 1/10，或分步计算 (2/5) \\times (1/4) = 1/10。4/25 = (2/5)^{2}$ 是有放回抽取时的答案。","tags":["easy","without-replacement","balls"],"status":"published"},{"id":"math-classical-probability-010","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":1,"stem":"A family has two children. Assuming a boy and a girl are equally likely and the births are independent, what is the probability that both children are boys?","stemZh":"某家庭有两个孩子，假设生男生女等可能且相互独立，两个孩子都是男孩的概率是多少？","options":[{"id":"A","text":"1/2"},{"id":"B","text":"1/3"},{"id":"C","text":"1/4"},{"id":"D","text":"3/4"}],"answer":"C","explanation":"The four outcomes BB, BG, GB, GG are equally likely and only BB qualifies, so $P = 1/4$. Treating BG and GB as one case would wrongly give 1/3.","explanationZh":"BB、BG、GB、GG 四种结果等可能，只有 BB 符合，故 $P = 1/4。若把$ BG 与 GB 合并成一种情况，会错得 1/3。","tags":["easy","classical-probability","equiprobable"],"status":"published"},{"id":"math-classical-probability-011","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"Two fair dice are rolled. What is the probability that the sum of the scores is 7?","stemZh":"同时掷两枚均匀的骰子，点数之和为 7 的概率是多少？","options":[{"id":"A","text":"1/12"},{"id":"B","text":"1/6"},{"id":"C","text":"5/36"},{"id":"D","text":"1/9"}],"answer":"B","explanation":"There are $6 \\times 6 = 36$ equally likely ordered pairs, and 6 of them give the sum $7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$. So $P = 6/36 = 1/6$.","explanationZh":"共有 $6 \\times 6 = 36$ 个等可能的有序数对，其中和为 7 的有 $(1,6)、(2,5)、(3,4)、(4,3)、(5,2)、(6,1) 共 6$ 个，故 $P = 6/36 = 1/6。$","tags":["medium","dice","classical-probability"],"status":"published"},{"id":"math-classical-probability-012","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"A bag contains 5 red balls and 3 white balls. Two balls are drawn at random without replacement. What is the probability that both are white?","stemZh":"袋中有 5 个红球和 3 个白球，不放回地随机取出 2 个，两个都是白球的概率是多少？","options":[{"id":"A","text":"9/64"},{"id":"B","text":"1/4"},{"id":"C","text":"3/8"},{"id":"D","text":"3/28"}],"answer":"D","explanation":"$P = C(3,2)/C(8,2) = 3/28$, or $(3/8) \\times (2/7) = 6/56 = 3/28$. The value $9/64 = (3/8)^{2}$ assumes the first ball is put back.","explanationZh":"$P = C(3,2)/C(8,2) = 3/28，或分步计算 (3/8) \\times (2/7) = 6/56 = 3/28。9/64 = (3/8)^{2}$ 是把第一个球放回后的结果。","tags":["medium","without-replacement","balls"],"status":"published"},{"id":"math-classical-probability-013","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"Two cards are drawn at random from a standard 52-card deck without replacement. What is the probability that both are aces?","stemZh":"从一副 52 张的扑克牌中不放回地随机抽取 2 张，两张都是 A 的概率是多少？","options":[{"id":"A","text":"1/221"},{"id":"B","text":"1/169"},{"id":"C","text":"1/13"},{"id":"D","text":"1/52"}],"answer":"A","explanation":"$P = C(4,2)/C(52,2) = 6/1326 = 1/221$, or $(4/52) \\times (3/51) = 1/221$. The value $1/169 = (4/52)^{2}$ treats the two draws as independent (with replacement).","explanationZh":"$P = C(4,2)/C(52,2) = 6/1326 = 1/221，或分步计算 (4/52) \\times (3/51) = 1/221。1/169 = (4/52)^{2}$ 是把两次抽取当作相互独立（有放回）的结果。","tags":["medium","without-replacement","cards"],"status":"published"},{"id":"math-classical-probability-014","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"A fair coin is tossed four times. What is the probability of getting exactly two heads?","stemZh":"一枚均匀硬币连掷四次，恰好出现两次正面的概率是多少？","options":[{"id":"A","text":"1/2"},{"id":"B","text":"1/4"},{"id":"C","text":"3/8"},{"id":"D","text":"5/16"}],"answer":"C","explanation":"There are $2^{4} = 16$ equally likely outcomes, of which $C(4,2) = 6$ have exactly two heads, so $P = 6/16 = 3/8$.","explanationZh":"共有 $2^{4} = 16$ 个等可能结果，其中恰好两次正面的是从 4 次中选 $2 次：C(4,2) = 6，故 P = 6/16 = 3/8。$","tags":["medium","coins","combination"],"status":"published"},{"id":"math-classical-probability-015","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"A bag contains 4 red, 3 blue and 2 green balls. One ball is drawn at random. What is the probability that it is red or blue?","stemZh":"袋中有 4 个红球、3 个蓝球和 2 个绿球，随机取出一个，取到红球或蓝球的概率是多少？","options":[{"id":"A","text":"4/9"},{"id":"B","text":"7/9"},{"id":"C","text":"2/9"},{"id":"D","text":"5/9"}],"answer":"B","explanation":"Red and blue cannot occur together, so the events are mutually exclusive and the probabilities $add: 4/9 + 3/9 = 7/9 (i.e. 7\\text{ of the }9 balls)$.","explanationZh":"$红球与蓝球不可能同时取到，两事件互斥，概率相加：4/9 + 3/9 = 7/9（即 9$ 个球中有 7 个符合）。","tags":["medium","mutually-exclusive","addition-rule"],"status":"published"},{"id":"math-classical-probability-016","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"Events A and B are mutually exclusive, with $P(A) = 0.3$ and $P(B) = 0.4$. What is $P(A ∪ B)$?","stemZh":"事件 A 与 B 互斥，且 $P(A) = 0.3，P(B) = 0.4，则 P(A ∪ B)$ 等于多少？","options":[{"id":"A","text":"0.12"},{"id":"B","text":"0.1"},{"id":"C","text":"1"},{"id":"D","text":"0.7"}],"answer":"D","explanation":"For mutually exclusive events the probabilities simply $add: P(A ∪ B) = 0.3 + 0.4 = 0.7$. Multiplying $(0.12)$ applies to independent events, not to a union.","explanationZh":"$互斥事件的概率可以直接相加：P(A ∪ B) = 0.3 + 0.4 = 0.7。相乘得 0.12$ 是独立事件同时发生的算法，不适用于求并事件。","tags":["medium","mutually-exclusive","addition-rule"],"status":"published"},{"id":"math-classical-probability-017","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"A shooter hits the target with probability 0.8 on each shot, and shots are independent. What is the probability that he misses at least once in two shots?","stemZh":"某射手每发命中目标的概率为 0.8，各发相互独立，则他两发中至少有一发脱靶的概率是多少？","options":[{"id":"A","text":"0.36"},{"id":"B","text":"0.04"},{"id":"C","text":"0.64"},{"id":"D","text":"0.2"}],"answer":"A","explanation":"Use the complement: P(at least one $miss) = 1 − P(both hit) = 1 − 0.8^{2} = 1 − 0.64 = 0.36$. The value $0.04 = 0.2^{2}$ is the probability of missing both shots.","explanationZh":"$用对立事件：P(至少一发脱靶) = 1 − P(两发都命中) = 1 − 0.8^{2} = 1 − 0.64 = 0.36。0.04 = 0.2^{2}$ 是两发都脱靶的概率。","tags":["medium","complement","independent-events"],"status":"published"},{"id":"math-classical-probability-018","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"A and B are independent events with $P(A) = 1/2$ and $P(B) = 1/3$. What is $P(A ∩ B)$?","stemZh":"事件 A 与 B 相互独立，且 $P(A) = 1/2，P(B) = 1/3，则 P(A ∩ B)$ 等于多少？","options":[{"id":"A","text":"5/6"},{"id":"B","text":"1/3"},{"id":"C","text":"1/6"},{"id":"D","text":"2/3"}],"answer":"C","explanation":"For independent events the probabilities $multiply: P(A ∩ B) = (1/2) \\times (1/3) = 1/6$. The value 5/6 is $P(A ∪ B)$, which is a different question.","explanationZh":"$独立事件同时发生的概率相乘：P(A ∩ B) = (1/2) \\times (1/3) = 1/6。5/6 是 P(A ∪ B)，问的是另一个问题。$","tags":["medium","independent-events","multiplication-rule"],"status":"published"},{"id":"math-classical-probability-019","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"A three-digit number is formed using only the digits 1 to 9, with repetition allowed. What is the probability that all three digits are odd?","stemZh":"只用数字 1 到 9 组成一个三位数（数字可重复），三个数字全为奇数的概率是多少？","options":[{"id":"A","text":"5/9"},{"id":"B","text":"25/81"},{"id":"C","text":"1/8"},{"id":"D","text":"125/729"}],"answer":"D","explanation":"There are $9^{3} = 729$ possible numbers and $5^{3} = 125$ in which every digit is odd (chosen from $1, 3, 5, 7, 9)$, so $P = 125/729 = (5/9)^{3}$. The value $25/81 = (5/9)^{2}$ uses only two positions.","explanationZh":"共有 $9^{3} = 729$ 个三位数，每位都从 1、3、5、7、9 中选的三位数有 $5^{3} = 125$ 个，故 $P = 125/729 = (5/9)^{3}。25/81 = (5/9)^{2}$ 只算了两位的情形。","tags":["medium","classical-probability","multiplication-principle"],"status":"published"},{"id":"math-classical-probability-020","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"Two students are chosen at random from a group of 5 boys and 4 girls. What is the probability that both are girls?","stemZh":"从 5 名男生和 4 名女生中随机选取 2 人，两人都是女生的概率是多少？","options":[{"id":"A","text":"16/81"},{"id":"B","text":"1/6"},{"id":"C","text":"2/9"},{"id":"D","text":"4/9"}],"answer":"B","explanation":"$P = C(4,2)/C(9,2) = 6/36 = 1/6$. The value $16/81 = (4/9)^{2}$ would be correct only if the first student were replaced before the second was chosen.","explanationZh":"$P = C(4,2)/C(9,2) = 6/36 = 1/6。16/81 = (4/9)^{2}$ 只有在‘选完第一人后再放回’时才成立。","tags":["medium","without-replacement","combination"],"status":"published"},{"id":"math-classical-probability-021","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":4,"stem":"A bag contains 6 red balls and 4 white balls. Three balls are drawn at random without replacement. What is the probability that at least one is white?","stemZh":"袋中有 6 个红球和 4 个白球，不放回地随机取出 3 个，至少有一个白球的概率是多少？","options":[{"id":"A","text":"1/6"},{"id":"B","text":"2/5"},{"id":"C","text":"5/6"},{"id":"D","text":"1/2"}],"answer":"C","explanation":"Use the $complement: P(no white) = C(6,3)/C(10,3) = 20/120 = 1/6$, so P(at least one $white) = 1 − 1/6 = 5/6$. Counting 1, 2 or 3 white balls directly gives the same value.","explanationZh":"$用对立事件：P(全是红球) = C(6,3)/C(10,3) = 20/120 = 1/6，故 P(至少一个白球) = 1 − 1/6 = 5/6。直接分 1、2、3$ 个白球计算也得到同样结果。","tags":["hard","complement","at-least-one"],"status":"published"},{"id":"math-classical-probability-022","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":4,"stem":"A and B are independent events with $P(A) = 0.6$ and $P(B) = 0.5$. What is $P(A ∪ B)$?","stemZh":"事件 A 与 B 相互独立，且 $P(A) = 0.6，P(B) = 0.5，则 P(A ∪ B)$ 等于多少？","options":[{"id":"A","text":"0.8"},{"id":"B","text":"1.1"},{"id":"C","text":"0.3"},{"id":"D","text":"0.9"}],"answer":"A","explanation":"$P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.5 − (0.6 \\times 0.5) = 1.1 − 0.3 = 0.8$. Simply adding gives 1.1, which forgets to remove the overlap and is impossible for a probability.","explanationZh":"$P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.5 − 0.6 \\times 0.5 = 1.1 − 0.3 = 0.8。直接相加得 1.1，既漏掉了重叠部分，也超过了概率的取值范围。$","tags":["hard","independent-events","addition-rule"],"status":"published"},{"id":"math-classical-probability-023","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":4,"stem":"Two fair dice are rolled. What is the probability that the product of the two scores is 12?","stemZh":"同时掷两枚均匀的骰子，点数之积为 12 的概率是多少？","options":[{"id":"A","text":"1/6"},{"id":"B","text":"1/12"},{"id":"C","text":"1/18"},{"id":"D","text":"1/9"}],"answer":"D","explanation":"The ordered pairs with product 12 are $(2,6), (6,2), (3,4)$ and $(4,3) — 4$ out of 36 — so $P = 4/36 = 1/9$. Counting only unordered pairs would give $2/36 = 1/18$.","explanationZh":"积为 12 的有序数对为 $(2,6)、(6,2)、(3,4)、(4,3)，共 4$ 个，故 $P = 4/36 = 1/9。若只按无序数对计算会得到 2/36 = 1/18。$","tags":["hard","dice","classical-probability"],"status":"published"},{"id":"math-classical-probability-024","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":5,"stem":"A box contains 3 red, 2 white and 1 black ball. Two balls are drawn at random without replacement. What is the probability that they have different colours?","stemZh":"盒中有 3 个红球、2 个白球和 1 个黑球，不放回地随机取出 2 个，两个球颜色不同的概率是多少？","options":[{"id":"A","text":"4/15"},{"id":"B","text":"11/15"},{"id":"C","text":"1/3"},{"id":"D","text":"2/5"}],"answer":"B","explanation":"$P(same colour) = [C(3,2) + C(2,2)]/C(6,2) = (3 + 1)/15 = 4/15 (black\\text{ cannot pair with }itself)$, so $P(different colours) = 1 − 4/15 = 11/15$.","explanationZh":"$P(同色) = [C(3,2) + C(2,2)]/C(6,2) = (3 + 1)/15 = 4/15（黑球只有 1$ 个，无法同色配对），故 $P(异色) = 1 − 4/15 = 11/15。$","tags":["hard","complement","balls"],"status":"published"},{"id":"math-classical-probability-025","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":4,"stem":"In a class of 40 students, 22 study French, 15 study German and 7 study both. What is the probability that a randomly chosen student studies at least one of the two languages?","stemZh":"某班 40 名学生中，22 人学法语，15 人学德语，7 人两门都学。随机抽取一名学生，他至少学一门外语的概率是多少？","options":[{"id":"A","text":"37/40"},{"id":"B","text":"1/4"},{"id":"C","text":"3/4"},{"id":"D","text":"7/40"}],"answer":"C","explanation":"By $inclusion–exclusion, 22 + 15 − 7 = 30$ students study at least one language, so $P = 30/40 = 3/4$. Adding $22 + 15 = 37$ double-counts the 7 students who study both.","explanationZh":"由容斥原理，至少学一门的人数为 $22 + 15 − 7 = 30，故 P = 30/40 = 3/4。直接算 22 + 15 = 37$ 会把两门都学的 7 人重复计算一次。","tags":["hard","inclusion-exclusion","addition-rule"],"status":"published"},{"id":"math-classical-probability-026","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":4,"stem":"A fair die is rolled once. Given that the score is even, what is the probability that it is greater than 4?","stemZh":"掷一枚均匀的骰子一次，已知点数为偶数，则点数大于 4 的概率是多少？","options":[{"id":"A","text":"1/3"},{"id":"B","text":"1/6"},{"id":"C","text":"1/2"},{"id":"D","text":"2/3"}],"answer":"A","explanation":"Given that the score is even, the reduced sample space is {2, 4, 6} and only 6 exceeds 4, so $P = 1/3$. By the $formula, P(>4 | even) = P(>4\\text{ and }even)/P(even) = (1/6)/(1/2) = 1/3$.","explanationZh":"已知点数为偶数，样本空间缩减为 ${2, 4, 6}，其中只有 6$ 大于 $4，故 P = 1/3。用条件概率公式：P(>4 | 偶数) = P(大于 4 且为偶数)/P(偶数) = (1/6)/(1/2) = 1/3。$","tags":["hard","conditional-probability","dice"],"status":"published"},{"id":"math-classical-probability-027","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":5,"stem":"Two cards are drawn at random from a standard 52-card deck without replacement. What is the probability that both are red?","stemZh":"从一副 52 张的扑克牌中不放回地随机抽取 2 张，两张都是红色牌的概率是多少？","options":[{"id":"A","text":"1/4"},{"id":"B","text":"25/51"},{"id":"C","text":"13/51"},{"id":"D","text":"25/102"}],"answer":"D","explanation":"$P = (26/52) \\times (25/51) = (1/2) \\times (25/51) = 25/102$, or $C(26,2)/C(52,2) = 325/1326 = 25/102$. The value $1/4 = (26/52)^{2}$ is the answer with replacement, and 25/51 forgets the first factor.","explanationZh":"$P = (26/52) \\times (25/51) = (1/2) \\times (25/51) = 25/102，或 C(26,2)/C(52,2) = 325/1326 = 25/102。1/4 = (26/52)^{2} 是有放回抽取的结果，25/51$ 漏掉了第一个因子。","tags":["hard","without-replacement","cards"],"status":"published"},{"id":"math-classical-probability-028","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":4,"stem":"A box contains 4 good items and 2 defective items. Two items are drawn at random without replacement. What is the probability that exactly one is defective?","stemZh":"盒中有 4 件正品和 2 件次品，不放回地随机取出 2 件，恰好有 1 件次品的概率是多少？","options":[{"id":"A","text":"4/15"},{"id":"B","text":"8/15"},{"id":"C","text":"1/3"},{"id":"D","text":"1/5"}],"answer":"B","explanation":"$P = C(2,1)·C(4,1)/C(6,2) = 8/15$. Computing only $(2/6) \\times (4/5) = 4/15$ forgets that the defective item can be drawn first or second; both orders give $8/30 + 8/30 = 8/15$.","explanationZh":"$P = C(2,1)·C(4,1)/C(6,2) = 8/15。只算 (2/6) \\times (4/5) = 4/15，会漏掉次品在第二次被取到的情形；两种顺序相加为 8/30 + 8/30 = 8/15。$","tags":["hard","without-replacement","balls"],"status":"published"},{"id":"math-classical-probability-029","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":4,"stem":"A fair coin is tossed five times. What is the probability of getting exactly three heads?","stemZh":"一枚均匀硬币连掷五次，恰好出现三次正面的概率是多少？","options":[{"id":"A","text":"3/8"},{"id":"B","text":"1/2"},{"id":"C","text":"5/16"},{"id":"D","text":"5/32"}],"answer":"C","explanation":"There are $2^{5} = 32$ equally likely outcomes, of which $C(5,3) = 10$ contain exactly three heads, so $P = 10/32 = 5/16$.","explanationZh":"共有 $2^{5} = 32$ 个等可能结果，其中恰好三次正面的为从 5 次中选 $3 次：C(5,3) = 10，故 P = 10/32 = 5/16。$","tags":["hard","coins","combination"],"status":"published"},{"id":"math-classical-probability-030","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":5,"stem":"Bag 1 contains 3 red and 2 blue balls; bag 2 contains 1 red and 4 blue balls. A bag is chosen at random and then one ball is drawn from it. What is the probability that the ball is red?","stemZh":"1 号袋中有 3 个红球和 2 个蓝球，2 号袋中有 1 个红球和 4 个蓝球。先随机选一个袋子，再从该袋中取出一个球，取到红球的概率是多少？","options":[{"id":"A","text":"2/5"},{"id":"B","text":"3/5"},{"id":"C","text":"1/5"},{"id":"D","text":"3/10"}],"answer":"A","explanation":"By the law of total $probability, P = (1/2)(3/5) + (1/2)(1/5) = 3/10 + 1/10 = 4/10 = 2/5$. Taking 3/5 or 1/5 uses only one bag, and 3/10 counts just the first branch.","explanationZh":"由全概率公式：P $= (1/2)(3/5) + (1/2)(1/5) = 3/10 + 1/10 = 4/10 = 2/5。只取 3/5 或 1/5 是只考虑了一个袋子，3/10$ 只算了第一条分支。","tags":["hard","total-probability","conditional-probability"],"status":"published"},{"id":"math-descriptive-001","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":1,"stem":"Find the mean of the data set 2, 4, 6, 8, 10.","stemZh":"求数据 2, 4, 6, 8, 10 的平均数。","options":[{"id":"A","text":"4"},{"id":"B","text":"6"},{"id":"C","text":"7"},{"id":"D","text":"8"}],"answer":"B","explanation":"Sum $= 2 + 4 + 6 + 8 + 10 = 30$ and $n = 5$, so $\\bar{x} = 30/5 = 6$.","explanationZh":"总和 $2+4+6+8+10 = 30，n = 5，所以 \\bar{x} = 30/5 = 6。$","tags":["easy","mean"],"status":"published"},{"id":"math-descriptive-002","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":1,"stem":"Find the median of the data set 3, 1, 7, 5, 9.","stemZh":"求数据 3, 1, 7, 5, 9 的中位数。","options":[{"id":"A","text":"5"},{"id":"B","text":"7"},{"id":"C","text":"1"},{"id":"D","text":"3"}],"answer":"A","explanation":"Sort the data first: 1, 3, 5, 7, 9. With $n = 5$ the median is the 3rd value, namely 5. Taking 7 means reading off the middle term without sorting.","explanationZh":"$先排序：1, 3, 5, 7, 9。n = 5，中位数是第 3$ 个数，即 5。选 7 是没有排序直接取中间项。","tags":["easy","median","sorting"],"status":"published"},{"id":"math-descriptive-003","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":1,"stem":"Find the mode of the data set 2, 3, 3, 5, 7, 3, 8.","stemZh":"求数据 2, 3, 3, 5, 7, 3, 8 的众数。","options":[{"id":"A","text":"2"},{"id":"B","text":"5"},{"id":"C","text":"3"},{"id":"D","text":"7"}],"answer":"C","explanation":"The value 3 occurs three times, more often than any other value, so the mode is 3.","explanationZh":"3 出现 3 次，次数最多，所以众数是 3。","tags":["easy","mode"],"status":"published"},{"id":"math-descriptive-004","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":1,"stem":"Find the range of the data set 4, 9, 2, 11, 6.","stemZh":"求数据 4, 9, 2, 11, 6 的极差。","options":[{"id":"A","text":"11"},{"id":"B","text":"7"},{"id":"C","text":"2"},{"id":"D","text":"9"}],"answer":"D","explanation":"Range $=$ maximum − minimum $= 11 − 2 = 9$. Option A only quotes the maximum; option B subtracts the first value $(11 − 4)$ instead of the minimum.","explanationZh":"极差 $=$ 最大值 − 最小值 $= 11 − 2 = 9。A$ 只取了最大值；B 用第一个数 4 代替了最小值。","tags":["easy","range"],"status":"published"},{"id":"math-descriptive-005","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":2,"stem":"Find the variance $s^{2}$ of the data set 1, 2, 3, 4, 5.","stemZh":"求数据 1, 2, 3, 4, 5 的方差 $s^{2}。$","options":[{"id":"A","text":"2"},{"id":"B","text":"$\\sqrt{2}$"},{"id":"C","text":"10"},{"id":"D","text":"2.5"}],"answer":"A","explanation":"$\\bar{x} = 15/5 = 3. Deviations: −2, −1, 0, 1, 2$, so $Σ(x_{i} − \\bar{x})^{2} = 4 + 1 + 0 + 1 + 4 = 10$ and $s^{2} = 10/5 = 2. \\sqrt{2}$ is the standard deviation, and 2.5 comes from dividing by $n − 1 = 4$ instead of n.","explanationZh":"$\\bar{x} = 15/5 = 3，离差为 −2, −1, 0, 1, 2，Σ(x_{i} − \\bar{x})^{2} = 10，方差 s^{2} = 10/5 = 2。\\sqrt{2} 是标准差；2.5$ 是误除以 $n − 1 = 4。$","tags":["easy","variance"],"status":"published"},{"id":"math-descriptive-006","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":2,"stem":"Find the standard deviation s of the data set 1, 2, 3, 4, 5.","stemZh":"求数据 1, 2, 3, 4, 5 的标准差 s。","options":[{"id":"A","text":"2"},{"id":"B","text":"$\\sqrt{2}$"},{"id":"C","text":"1"},{"id":"D","text":"$\\sqrt{10}$"}],"answer":"B","explanation":"$\\bar{x} = 3$ and $Σ(x_{i} − \\bar{x})^{2} = 10$, so $s^{2} = 10/5 = 2$ and $s = \\sqrt{2}$. The value 2 is the variance, not the standard deviation.","explanationZh":"$\\bar{x} = 3，Σ(x_{i} − \\bar{x})^{2} = 10，所以 s^{2} = 10/5 = 2，s = \\sqrt{2}。2$ 是方差而不是标准差。","tags":["easy","standard-deviation"],"status":"published"},{"id":"math-descriptive-007","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":2,"stem":"A data set has mean $\\bar{x} = 5$. Every value x is replaced by $y = 2x + 3$. Find the mean of the new data set.","stemZh":"一组数据的平均数 $\\bar{x} = 5，把每个数据 x$ 换成 $y = 2x + 3，求新数据的平均数。$","options":[{"id":"A","text":"8"},{"id":"B","text":"10"},{"id":"C","text":"13"},{"id":"D","text":"16"}],"answer":"C","explanation":"Under $y =$ ax + b the mean becomes $ȳ = a·\\bar{x} + b = 2·5 + 3 = 13$. Adding 3 only $(8)$, doubling only $(10)$ or doubling after adding $(16)$ are all wrong.","explanationZh":"$y =$ ax + b 时平均数变为 $ȳ = a·\\bar{x} + b = 2\\times5 + 3 = 13。只加 3、只乘 2$ 或先加后乘都不对。","tags":["easy","transformation","mean"],"status":"published"},{"id":"math-descriptive-008","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":2,"stem":"Find the median of the data set 8, 2, 6, 4.","stemZh":"求数据 8, 2, 6, 4 的中位数。","options":[{"id":"A","text":"4"},{"id":"B","text":"6"},{"id":"C","text":"7"},{"id":"D","text":"5"}],"answer":"D","explanation":"Sort first: 2, 4, 6, 8. With $n = 4$ the median is the average of the two middle $values: (4 + 6)/2 = 5$. Taking 7 averages the middle two of the unsorted list $(8$ and $6)$.","explanationZh":"$先排序：2, 4, 6, 8。n = 4，中位数是中间两个数的平均数：(4 + 6)/2 = 5。取 7$ 是没有排序就把中间两项 8 和 6 求平均。","tags":["easy","median","sorting"],"status":"published"},{"id":"math-descriptive-009","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":2,"stem":"A sample of size 10 is summarised as: value $1 (frequency 2)$, value $2 (frequency 3)$, value $3 (frequency 5)$. Find the mean of the sample.","stemZh":"容量为 10 的样本：数值 1（频数 2）、数值 2（频数 3）、数值 3（频数 5），求样本平均数。","options":[{"id":"A","text":"2"},{"id":"B","text":"2.3"},{"id":"C","text":"7.7"},{"id":"D","text":"23"}],"answer":"B","explanation":"$\\bar{x} = (1·2 + 2·3 + 3·5)/10 = (2 + 6 + 15)/10 = 23/10 = 2.3$. Option A averages only the three distinct $values, (1 + 2 + 3)/3 = 2$, ignoring the frequencies.","explanationZh":"$\\bar{x} = (1\\times2 + 2\\times3 + 3\\times5)/10 = 23/10 = 2.3。A$ 是只对三个不同数值求平均 $(1+2+3)/3 = 2，忽略了频数。$","tags":["easy","frequency-table","mean"],"status":"published"},{"id":"math-descriptive-010","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":1,"stem":"Which of the following quantities measures how spread out a data set is?","stemZh":"下列哪个量用来刻画一组数据的离散（波动）程度？","options":[{"id":"A","text":"Standard deviation"},{"id":"B","text":"Mean"},{"id":"C","text":"Median"},{"id":"D","text":"Mode"}],"answer":"A","explanation":"The standard deviation (and the variance) measure spread. The mean, median and mode measure the centre or the most typical value of the data.","explanationZh":"标准差（以及方差）刻画离散程度；平均数、中位数、众数刻画数据的集中趋势。","tags":["easy","standard-deviation","concept"],"status":"published"},{"id":"math-descriptive-011","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"Find the variance $s^{2}$ of the data set 2, 4, 6, 8.","stemZh":"求数据 2, 4, 6, 8 的方差 $s^{2}。$","options":[{"id":"A","text":"$\\sqrt{5}$"},{"id":"B","text":"20"},{"id":"C","text":"5"},{"id":"D","text":"20/3"}],"answer":"C","explanation":"$\\bar{x} = 20/4 = 5. Deviations: −3, −1, 1, 3$, so $Σ(x_{i} − \\bar{x})^{2} = 9 + 1 + 1 + 9 = 20$ and $s^{2} = 20/4 = 5. \\sqrt{5}$ is the standard deviation, 20 forgets the division, and 20/3 divides by n − 1.","explanationZh":"$\\bar{x} = 20/4 = 5，Σ(x_{i} − \\bar{x})^{2} = 9+1+1+9 = 20，s^{2} = 20/4 = 5。\\sqrt{5} 是标准差，20$ 忘了除以 n，20/3 是除以 n − 1。","tags":["medium","variance"],"status":"published"},{"id":"math-descriptive-012","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"A data set has mean 4 and variance 9. Every value x is replaced by $y = 3x − 2$. Find the variance of the new data set.","stemZh":"一组数据平均数 4、方差 9，把每个 x 换成 $y = 3x − 2，求新数据的方差。$","options":[{"id":"A","text":"9"},{"id":"B","text":"25"},{"id":"C","text":"27"},{"id":"D","text":"81"}],"answer":"D","explanation":"Under $y =$ ax + b the variance is multiplied by $a^{2}$ and the shift b has no $effect: s^{2}_y = 3^{2}·9 = 81$. Applying the linear rule to the variance $(3·9 − 2 = 25)$ or multiplying by a instead of $a^{2} (27)$ are common errors.","explanationZh":"$y =$ ax + b 时方差乘以 $a^{2}，平移 b 不影响方差：s^{2}_y = 3^{2}\\times9 = 81。把线性关系直接套到方差上（3\\times9 − 2 = 25）或只乘 a$ 不乘 $a^{2}（27）都是常见错误。$","tags":["medium","transformation","variance"],"status":"published"},{"id":"math-descriptive-013","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"A data set has standard deviation 5. Every value x is replaced by $y = −2x + 1$. Find the standard deviation of the new data set.","stemZh":"一组数据标准差为 5，把每个 x 换成 $y = −2x + 1，求新数据的标准差。$","options":[{"id":"A","text":"10"},{"id":"B","text":"−10"},{"id":"C","text":"5"},{"id":"D","text":"20"}],"answer":"A","explanation":"Under $y =$ ax + b the standard deviation is scaled by $|a|: s_y = |−2|·5 = 10. A$ standard deviation is never negative, and adding 1 does not change the spread.","explanationZh":"$y =$ ax + b 时标准差变为 $s_y = |a|·s = |−2|\\times5 = 10。标准差不可能为负，加 1$ 不改变离散程度。","tags":["medium","transformation","standard-deviation"],"status":"published"},{"id":"math-descriptive-014","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"A frequency table has classes $[0, 10)$ with frequency $3, [10, 20)$ with frequency 7 and $[20, 30)$ with frequency 10. Using class midpoints, estimate the mean.","stemZh":"$频数分布表：[0,10)$ 频数 $3，[10,20)$ 频数 $7，[20,30)$ 频数 10。用组中值估计平均数。","options":[{"id":"A","text":"15"},{"id":"B","text":"18.5"},{"id":"C","text":"20"},{"id":"D","text":"25"}],"answer":"B","explanation":"Midpoints are 5, 15, 25 and $n = 3 + 7 + 10 = 20$, so $\\bar{x} ≈ (5·3 + 15·7 + 25·10)/20 = (15 + 105 + 250)/20 = 370/20 = 18.5$. Option A averages the three midpoints and ignores the frequencies.","explanationZh":"组中值为 $5, 15, 25，n = 3+7+10 = 20，\\bar{x} ≈ (5\\times3 + 15\\times7 + 25\\times10)/20 = 370/20 = 18.5。A$ 只对三个组中值求平均，忽略了频数。","tags":["medium","grouped-data","mean"],"status":"published"},{"id":"math-descriptive-015","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"Student A scores 78, 82, 80, 80, 80 and student B scores 70, 90, 75, 85, 80 on five tests. Whose results are more stable?","stemZh":"甲五次成绩为 78, 82, 80, 80, 80，乙五次成绩为 70, 90, 75, 85, 80，谁的成绩更稳定？","options":[{"id":"A","text":"Student B"},{"id":"B","text":"It cannot be determined"},{"id":"C","text":"Student A"},{"id":"D","text":"They are equally stable"}],"answer":"C","explanation":"Both means are 80, so compare variances. $A: Σ(x_{i} − 80)^{2} = 4 + 4 + 0 + 0 + 0 = 8, s^{2} = 8/5 = 1.6. B: Σ(x_{i} − 80)^{2} = 100 + 100 + 25 + 25 + 0 = 250, s^{2} = 250/5 = 50$. The smaller variance belongs to A, so A is more stable.","explanationZh":"两人平均数都是 $80，比较方差即可。甲：Σ(x_{i} − 80)^{2} = 8，s^{2} = 1.6；乙：Σ(x_{i} − 80)^{2} = 250，s^{2} = 50。甲方差更小，成绩更稳定。$","tags":["medium","stability","variance"],"status":"published"},{"id":"math-descriptive-016","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"The scores of 10 students are summarised $as: 60 (2 students), 70 (3 students), 80 (4 students), 90 (1 student)$. Find the median score.","stemZh":"10 名学生的成绩：60 分 2 人，70 分 3 人，80 分 4 人，90 分 1 人，求中位数。","options":[{"id":"A","text":"70"},{"id":"B","text":"80"},{"id":"C","text":"74"},{"id":"D","text":"75"}],"answer":"D","explanation":"In order the scores are 60, 60, 70, 70, 70, 80, 80, 80, 80, 90. With $n = 10$ the median is the average of the 5th and $6th values: (70 + 80)/2 = 75$. The value 74 is the $mean, (60·2 + 70·3 + 80·4 + 90·1)/10 = 740/10$.","explanationZh":"排序后为 $60, 60, 70, 70, 70, 80, 80, 80, 80, 90。n = 10，中位数是第 5、6 个数的平均数：(70 + 80)/2 = 75。74$ 是平均数。","tags":["medium","median","frequency-table"],"status":"published"},{"id":"math-descriptive-017","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"A data set has 5 values with mean 8. Four of the values are 6, 8, 10 and 7. Find the missing value.","stemZh":"一组 5 个数据的平均数为 8，其中四个是 6, 8, 10, 7，求第五个数。","options":[{"id":"A","text":"9"},{"id":"B","text":"8"},{"id":"C","text":"10"},{"id":"D","text":"7"}],"answer":"A","explanation":"The total is $5·8 = 40$, and the four known values sum to 31, so the missing value is $40 − 31 = 9$.","explanationZh":"总和为 $5\\times8 = 40，已知四个数之和为 31，故第五个数为 40 − 31 = 9。$","tags":["medium","mean","missing-value"],"status":"published"},{"id":"math-descriptive-018","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"A histogram has three classes of equal width whose bar heights are in the ratio 3 : 7 : 5. If the total frequency is 60, what is the frequency of the first class?","stemZh":"某频率分布直方图三个组距相同，条形高之比为 3 : 7 : 5，总频数为 60，求第一组的频数。","options":[{"id":"A","text":"20"},{"id":"B","text":"12"},{"id":"C","text":"15"},{"id":"D","text":"28"}],"answer":"B","explanation":"With equal class widths the frequency is proportional to the bar height, so the frequencies are in the ratio 3 : 7 : 5 and the total is $3 + 7 + 5 = 15$ parts. The first class gets $60·(3/15) = 12$.","explanationZh":"组距相同时频数与条形高成正比，频数之比为 3 : 7 : 5，共 15 份，第一组频数 $= 60 \\times 3/15 = 12。$","tags":["medium","histogram","frequency"],"status":"published"},{"id":"math-descriptive-019","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"Find the variance $s^{2}$ of the data set 2, 5, 5, 8.","stemZh":"求数据 2, 5, 5, 8 的方差 $s^{2}。$","options":[{"id":"A","text":"$\\sqrt{4.5}$"},{"id":"B","text":"18"},{"id":"C","text":"4.5"},{"id":"D","text":"6"}],"answer":"C","explanation":"$\\bar{x} = 20/4 = 5. Deviations: −3, 0, 0, 3$, so $Σ(x_{i} − \\bar{x})^{2} = 9 + 0 + 0 + 9 = 18$ and $s^{2} = 18/4 = 4.5. \\sqrt{4.5}$ is the standard deviation, and 6 comes from dividing by $n − 1 = 3$.","explanationZh":"$\\bar{x} = 20/4 = 5，Σ(x_{i} − \\bar{x})^{2} = 9+0+0+9 = 18，s^{2} = 18/4 = 4.5。\\sqrt{4.5} 是标准差，6$ 是除以 $n − 1 = 3$ 的结果。","tags":["medium","variance"],"status":"published"},{"id":"math-descriptive-020","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"Class A has 30 students with mean score 70 and class B has 20 students with mean score 85. Find the mean score of all 50 students.","stemZh":"A 班 30 人平均 70 分，B 班 20 人平均 85 分，求全体 50 人的平均分。","options":[{"id":"A","text":"77.5"},{"id":"B","text":"70"},{"id":"C","text":"85"},{"id":"D","text":"76"}],"answer":"D","explanation":"The combined mean is $(30·70 + 20·85)/50 = (2100 + 1700)/50 = 3800/50 = 76$. It is not $(70 + 85)/2 = 77.5$, because the two classes have different sizes.","explanationZh":"总平均 $= (30\\times70 + 20\\times85)/50 = 3800/50 = 76。不能直接用 (70 + 85)/2 = 77.5，因为两班人数不同。$","tags":["medium","weighted-mean","mean"],"status":"published"},{"id":"math-descriptive-021","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":4,"stem":"A data set has mean 10 and variance 4. Every value x is replaced by $y = −3x + 5$. Find the standard deviation of the new data set.","stemZh":"一组数据平均数 10、方差 4，把每个 x 换成 $y = −3x + 5，求新数据的标准差。$","options":[{"id":"A","text":"6"},{"id":"B","text":"36"},{"id":"C","text":"−6"},{"id":"D","text":"12"}],"answer":"A","explanation":"The original standard deviation is $s = \\sqrt{4} = 2$. Under $y =$ ax $+ b, s_y = |a|·s = |−3|·2 = 6$. The value 36 is the new variance $(a^{2}·s^{2} = 9·4)$, and a standard deviation can never be negative.","explanationZh":"原标准差 $s = \\sqrt{4} = 2，y =$ ax $+ b 时 s_y = |a|·s = |−3|\\times2 = 6。36 是新方差（a^{2}·s^{2} = 9\\times4），标准差不可能为负。$","tags":["hard","transformation","standard-deviation"],"status":"published"},{"id":"math-descriptive-022","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":4,"stem":"For the data set 1, 3, 5, 7, find the value of $(1/n)·Σ(x_{i} − \\bar{x})^{2}$, where n is the number of data values.","stemZh":"对数据 $1, 3, 5, 7，求 (1/n)·Σ(x_{i} − \\bar{x})^{2}$ 的值（n 为数据个数）。","options":[{"id":"A","text":"5"},{"id":"B","text":"20/3"},{"id":"C","text":"$\\sqrt{5}$"},{"id":"D","text":"20"}],"answer":"A","explanation":"$\\bar{x} = 16/4 = 4$, so $Σ(x_{i} − \\bar{x})^{2} = 9 + 1 + 1 + 9 = 20$ and $(1/n)·20 = 20/4 = 5$. The value 20/3 would come from dividing by n − 1, which is not the convention used here: the divisor is n.","explanationZh":"$\\bar{x} = 16/4 = 4，Σ(x_{i} − \\bar{x})^{2} = 9+1+1+9 = 20，(1/n)\\times20 = 20/4 = 5。20/3$ 是除以 n − 1 的结果，本题除以 n。","tags":["hard","variance","convention"],"status":"published"},{"id":"math-descriptive-023","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":5,"stem":"For a data set of $5 values, Σx = 40$ and $Σx^{2} = 340$. Find the variance $s^{2}$.","stemZh":"一组 5 个数据满足 $Σx = 40，Σx^{2} = 340，求方差 s^{2}。$","options":[{"id":"A","text":"4"},{"id":"B","text":"68"},{"id":"C","text":"2"},{"id":"D","text":"64"}],"answer":"A","explanation":"Use $s^{2} = (Σx^{2})/n − \\bar{x}^{2}$. Here $\\bar{x} = 40/5 = 8$, so $s^{2} = 340/5 − 8^{2} = 68 − 64 = 4$. Option B stops at $Σx^{2}/n$ and option D is just $\\bar{x}^{2}$, both forgetting to subtract.","explanationZh":"用公式 $s^{2} = (Σx^{2})/n − \\bar{x}^{2}。\\bar{x} = 40/5 = 8，所以 s^{2} = 340/5 − 8^{2} = 68 − 64 = 4。B$ 只算到 $Σx^{2}/n，D$ 只是 $\\bar{x}^{2}，都没有相减。$","tags":["hard","variance","computational-formula"],"status":"published"},{"id":"math-descriptive-024","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":4,"stem":"Find the median of the data set 12, 5, 9, 5, 20, 9.","stemZh":"求数据 12, 5, 9, 5, 20, 9 的中位数。","options":[{"id":"A","text":"7"},{"id":"B","text":"9"},{"id":"C","text":"10"},{"id":"D","text":"5"}],"answer":"B","explanation":"Sorted: 5, 5, 9, 9, 12, 20. With $n = 6$ the median is $(9 + 9)/2 = 9$. The value 7 averages the middle two of the unsorted list $(9$ and $5), 10$ is the mean, and 5 is the mode.","explanationZh":"$排序：5, 5, 9, 9, 12, 20。n = 6，中位数为 (9 + 9)/2 = 9。7$ 是未排序时中间两项 9 与 5 的平均；10 是平均数；5 是众数。","tags":["hard","median","sorting"],"status":"published"},{"id":"math-descriptive-025","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":5,"stem":"Class A has 20 students with mean score 80, and class B has mean score 90. If the mean of the two classes combined is 85, how many students are in class B?","stemZh":"A 班 20 人平均 80 分，B 班平均 90 分，两班合并平均 85 分，求 B 班人数。","options":[{"id":"A","text":"10"},{"id":"B","text":"20"},{"id":"C","text":"25"},{"id":"D","text":"30"}],"answer":"B","explanation":"Let class B have n students. Then $(20·80 + 90n)/(20 + n) = 85$, i.e. $1600 + 90n = 1700 + 85n$, so $5n = 100$ and $n = 20$.","explanationZh":"设 B 班 n 人，则 $(20\\times80 + 90n)/(20 + n) = 85，即 1600 + 90n = 1700 + 85n，得 5n = 100，n = 20。$","tags":["hard","weighted-mean","equation"],"status":"published"},{"id":"math-descriptive-026","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":4,"stem":"In the data set 4, 5, 6, 7, 8 the largest value 8 is replaced by 48. By how much does the median change?","stemZh":"数据 4, 5, 6, 7, 8 中最大值 8 被改成 48，中位数变化了多少？","options":[{"id":"A","text":"Increases by 8"},{"id":"B","text":"It does not change"},{"id":"C","text":"Increases to 14"},{"id":"D","text":"Increases by 6"}],"answer":"B","explanation":"The new data are 4, 5, 6, 7, 48, whose median is still the 3rd value, 6 — unchanged. The mean does move, from 6 to $70/5 = 14$, an increase of 8, but the median is insensitive to this extreme value.","explanationZh":"新数据为 4, 5, 6, 7, 48，中位数仍是第 3 个数 6，没有变化。平均数由 6 变为 $70/5 = 14（增加 8），但中位数不受极端值影响。$","tags":["hard","median","outlier"],"status":"published"},{"id":"math-descriptive-027","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":4,"stem":"A data set has standard deviation 5. Every value x is replaced by $y = 4x − 1$. Find the standard deviation of the new data set.","stemZh":"一组数据标准差为 5，把每个 x 换成 $y = 4x − 1，求新数据的标准差。$","options":[{"id":"A","text":"19"},{"id":"B","text":"25"},{"id":"C","text":"20"},{"id":"D","text":"5"}],"answer":"C","explanation":"Only the multiplier changes the $spread: s_y = |4|·5 = 20$. The shift −1 has no effect on the standard deviation, so neither $4·5 − 1 = 19$ nor the unchanged value 5 is right.","explanationZh":"$只有倍数改变离散程度：s_y = |4|\\times5 = 20。平移 −1$ 不影响标准差，所以 $4\\times5 − 1 = 19$ 和不变的 5 都不对。","tags":["hard","transformation","standard-deviation"],"status":"published"},{"id":"math-descriptive-028","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":4,"stem":"The mean of the four numbers 1, 5, x, 7 is 4. Find the variance of the completed data set.","stemZh":"四个数 1, 5, x, 7 的平均数为 4，求这组完整数据的方差。","options":[{"id":"A","text":"$\\sqrt{5}$"},{"id":"B","text":"4"},{"id":"C","text":"5"},{"id":"D","text":"20"}],"answer":"C","explanation":"The total is $4·4 = 16$, so $x = 16 − (1 + 5 + 7) = 3$. The data are 1, 3, 5, 7 with $\\bar{x} = 4$, so $Σ(x_{i} − \\bar{x})^{2} = 9 + 1 + 1 + 9 = 20$ and $s^{2} = 20/4 = 5$.","explanationZh":"总和为 $4\\times4 = 16，故 x = 16 − (1 + 5 + 7) = 3。数据为 1, 3, 5, 7，\\bar{x} = 4，Σ(x_{i} − \\bar{x})^{2} = 20，方差 s^{2} = 20/4 = 5。$","tags":["hard","missing-value","variance"],"status":"published"},{"id":"math-descriptive-029","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":5,"stem":"A data set has mean $\\bar{x}$ and variance $s^{2}$. Every value x is replaced by $y = 2x − 5$. Which statement about the new data set is correct?","stemZh":"一组数据平均数 $\\bar{x}、方差 s^{2}，把每个 x$ 换成 $y = 2x − 5，关于新数据的说法正确的是？$","options":[{"id":"A","text":"Mean $2\\bar{x} − 5$, variance $2s^{2} − 5$"},{"id":"B","text":"Mean $\\bar{x}$, variance $4s^{2}$"},{"id":"C","text":"Mean $2\\bar{x}$, variance $2s^{2}$"},{"id":"D","text":"Mean $2\\bar{x} − 5$, variance $4s^{2}$"}],"answer":"D","explanation":"Both parameters feel the same linear rule for the $centre: ȳ = 2\\bar{x} − 5$. But spread is scaled by the square of the multiplier and is unaffected by the $shift: s^{2}_y = 2^{2}·s^{2} = 4s^{2}$. The variance does not obey $y = 2x − 5$ itself.","explanationZh":"平均数按同样的线性关系变化：ȳ $= 2\\bar{x} − 5；离散程度只与倍数的平方有关，平移不影响：s^{2}_y = 2^{2}·s^{2} = 4s^{2}。方差本身不能代入 y = 2x − 5。$","tags":["hard","transformation","variance"],"status":"published"},{"id":"math-descriptive-030","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":5,"stem":"A data set of 10 values has mean 20 and variance 9. Two new values, both equal to 20, are added to the data set. What are the new mean and variance?","stemZh":"一组 10 个数据的平均数为 20、方差为 9，再加入两个都等于 20 的数据，新的平均数和方差是多少？","options":[{"id":"A","text":"Mean 20, variance 9"},{"id":"B","text":"Mean 21.8, variance 7.5"},{"id":"C","text":"Mean 20, variance 10.8"},{"id":"D","text":"Mean 20, variance 7.5"}],"answer":"D","explanation":"The new total is $200 + 40 = 240$, so the mean is $240/12 = 20 (unchanged)$. The two added values have deviation 0, so $Σ(x_{i} − 20)^{2}$ is still $10·9 = 90$ and the new variance is $90/12 = 7.5$.","explanationZh":"新总和 $200 + 40 = 240，平均数 240/12 = 20（不变）。新增两个数离差为 0，Σ(x_{i} − 20)^{2}$ 仍为 $10\\times9 = 90，新方差 = 90/12 = 7.5。$","tags":["hard","mean","variance"],"status":"published"},{"id":"math-normal-001","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":1,"stem":"For $X ~ N(3, 4)$, what is the mean $μ$?","stemZh":"已知 $X ~ N(3, 4)，均值 μ$ 是多少？","options":[{"id":"A","text":"3"},{"id":"B","text":"4"},{"id":"C","text":"2"},{"id":"D","text":"5"}],"answer":"A","explanation":"In $X ~ N(μ, σ^{2})$ the first parameter is the mean, so $μ = 3$. The second parameter is the $variance: σ^{2} = 4$, hence $σ = 2$, and $μ + σ = 5$.","explanationZh":"$X ~ N(μ, σ^{2})$ 中第一个参数是均值，故 $μ = 3；第二个参数是方差 σ^{2} = 4，所以 σ = 2，μ + σ = 5。$","tags":["easy","parameters","mean"],"status":"published"},{"id":"math-normal-002","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":1,"stem":"For $X ~ N(5, 9)$, what is the standard deviation $σ$?","stemZh":"已知 $X ~ N(5, 9)，标准差 σ$ 是多少？","options":[{"id":"A","text":"9"},{"id":"B","text":"3"},{"id":"C","text":"5"},{"id":"D","text":"81"}],"answer":"B","explanation":"The second parameter is the variance, so $σ^{2} = 9$ and $σ = 3$. The value 9 is $σ^{2}$ and 5 is the mean $μ$.","explanationZh":"第二个参数是方差，故 $σ^{2} = 9，σ = 3。9 是方差，5$ 是均值 $μ。$","tags":["easy","parameters","standard-deviation"],"status":"published"},{"id":"math-normal-003","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":1,"stem":"The density curve of a normal distribution $X ~ N(μ, σ^{2})$ is symmetric about which vertical line?","stemZh":"正态分布 $X ~ N(μ, σ^{2})$ 的密度曲线关于哪条竖直直线对称？","options":[{"id":"A","text":"$x = 0$"},{"id":"B","text":"$x = σ$"},{"id":"C","text":"$x = μ$"},{"id":"D","text":"$x = μ + σ$"}],"answer":"C","explanation":"The normal curve is bell-shaped and symmetric about the vertical line $x = μ$, the mean of the distribution.","explanationZh":"正态曲线呈钟形，关于直线 $x = μ（均值所在处）对称。$","tags":["easy","symmetry","curve"],"status":"published"},{"id":"math-normal-004","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":1,"stem":"What is the total area under a normal density curve?","stemZh":"正态密度曲线下方的总面积是多少？","options":[{"id":"A","text":"0.5"},{"id":"B","text":"0.6826"},{"id":"C","text":"2"},{"id":"D","text":"1"}],"answer":"D","explanation":"The area under any probability density curve equals 1; this is what makes the curve a probability distribution.","explanationZh":"任何概率密度曲线下方面积都等于 1，这正是概率分布的要求。","tags":["easy","area","concept"],"status":"published"},{"id":"math-normal-005","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":2,"stem":"For $X ~ N(μ, σ^{2})$, approximately what proportion of the values lie within one standard deviation of the mean, i.e. $P(μ − σ < X ≤ μ + σ)$?","stemZh":"$对 X ~ N(μ, σ^{2})，落在均值一个标准差范围内的比例 P(μ − σ < X ≤ μ + σ)$ 约为多少？","options":[{"id":"A","text":"0.9544"},{"id":"B","text":"0.6826"},{"id":"C","text":"0.9974"},{"id":"D","text":"0.3413"}],"answer":"B","explanation":"The empirical rule gives $P(μ − σ < X ≤ μ + σ) ≈ 0.6826$. The values 0.9544 and 0.9974 correspond to two and three standard deviations, and 0.3413 is only the right-hand half of the interval.","explanationZh":"由经验法则 $P(μ − σ < X ≤ μ + σ) ≈ 0.6826。0.9544 与 0.9974 分别对应两个、三个标准差，0.3413$ 只是区间右半部分。","tags":["easy","empirical-rule"],"status":"published"},{"id":"math-normal-006","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":2,"stem":"For $X ~ N(μ, σ^{2})$, evaluate $P(μ − 2σ < X ≤ μ + 2σ)$.","stemZh":"$对 X ~ N(μ, σ^{2})，求 P(μ − 2σ < X ≤ μ + 2σ)。$","options":[{"id":"A","text":"0.9544"},{"id":"B","text":"0.6826"},{"id":"C","text":"0.9974"},{"id":"D","text":"0.4772"}],"answer":"A","explanation":"The empirical rule gives $P(μ − 2σ < X ≤ μ + 2σ) ≈ 0.9544; 0.4772$ is only the half of that interval on one side of $μ$.","explanationZh":"经验法则给出 $P(μ − 2σ < X ≤ μ + 2σ) ≈ 0.9544；0.4772$ 只是 $μ$ 一侧的一半。","tags":["easy","empirical-rule"],"status":"published"},{"id":"math-normal-007","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":2,"stem":"For $X ~ N(10, 4)$, what is $P(X > 10)$?","stemZh":"已知 $X ~ N(10, 4)，求 P(X > 10)。$","options":[{"id":"A","text":"0.6826"},{"id":"B","text":"0.25"},{"id":"C","text":"0.5"},{"id":"D","text":"0.3174"}],"answer":"C","explanation":"The curve is symmetric about $x = μ = 10$, so half of the area lies on each $side: P(X > 10) = 0.5$.","explanationZh":"曲线关于 $x = μ = 10$ 对称，两侧面积各占一半，所以 $P(X > 10) = 0.5。$","tags":["easy","symmetry"],"status":"published"},{"id":"math-normal-008","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":2,"stem":"For $X ~ N(μ, σ^{2})$, by symmetry $P(X < μ − σ)$ equals $P(X > ?)$","stemZh":"$对 X ~ N(μ, σ^{2})，由对称性 P(X < μ − σ)$ 等于 $P(X > ?)$","options":[{"id":"A","text":"$μ$"},{"id":"B","text":"$μ − σ$"},{"id":"C","text":"$μ + 2σ$"},{"id":"D","text":"$μ + σ$"}],"answer":"D","explanation":"The points $μ − σ$ and $μ + σ$ are mirror images about $x = μ$, so the two tails $P(X < μ − σ)$ and $P(X > μ + σ)$ are equal.","explanationZh":"$μ − σ 与 μ + σ$ 关于 $x = μ$ 对称，所以两侧尾部概率 $P(X < μ − σ) 与 P(X > μ + σ)$ 相等。","tags":["easy","symmetry"],"status":"published"},{"id":"math-normal-009","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":2,"stem":"For $X ~ N(20, 25)$, what is the standardised value $z = (x − μ)/σ$ of $x = 20$?","stemZh":"已知 $X ~ N(20, 25)，x = 20$ 标准化后 $z = (x − μ)/σ$ 等于多少？","options":[{"id":"A","text":"0"},{"id":"B","text":"1"},{"id":"C","text":"−1"},{"id":"D","text":"5"}],"answer":"A","explanation":"Here $μ = 20$ and $σ = \\sqrt{25} = 5$, so $z = (20 − 20)/5 = 0$. The standardised value of the mean is always 0.","explanationZh":"$μ = 20，σ = \\sqrt{25} = 5，z = (20 − 20)/5 = 0。均值标准化后总是 0。$","tags":["easy","standardisation"],"status":"published"},{"id":"math-normal-010","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":1,"stem":"For a normal distribution, what happens to the curve if $σ$ is increased while $μ$ is kept fixed?","stemZh":"正态分布中保持 $μ$ 不变而增大 $σ，曲线会怎样变化？$","options":[{"id":"A","text":"It becomes taller and narrower"},{"id":"B","text":"It becomes flatter and wider"},{"id":"C","text":"It shifts to the right"},{"id":"D","text":"Its shape does not change"}],"answer":"B","explanation":"$σ$ controls the spread: a larger $σ$ spreads the same total area 1 over a wider interval, so the bell becomes flatter and wider. The centre stays at $μ$.","explanationZh":"$σ 决定离散程度：σ$ 越大，总面积 1 被摊到更宽的区间上，钟形曲线变得更矮更胖；对称轴仍在 $x = μ。$","tags":["easy","curve","standard-deviation"],"status":"published"},{"id":"math-normal-011","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(μ, σ^{2})$, evaluate $P(μ < X ≤ μ + σ)$.","stemZh":"$对 X ~ N(μ, σ^{2})，求 P(μ < X ≤ μ + σ)。$","options":[{"id":"A","text":"0.6826"},{"id":"B","text":"0.4772"},{"id":"C","text":"0.1587"},{"id":"D","text":"0.3413"}],"answer":"D","explanation":"The interval $(μ − σ, μ + σ]$ carries probability 0.6826 and is symmetric about $μ$, so each half carries $0.6826/2 = 0.3413$.","explanationZh":"区间 $(μ − σ, μ + σ]$ 的概率为 $0.6826，且关于 μ$ 对称，所以每半边为 $0.6826/2 = 0.3413。$","tags":["medium","empirical-rule","symmetry"],"status":"published"},{"id":"math-normal-012","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(μ, σ^{2})$, evaluate $P(X > μ + σ)$.","stemZh":"$对 X ~ N(μ, σ^{2})，求 P(X > μ + σ)。$","options":[{"id":"A","text":"0.3413"},{"id":"B","text":"0.1587"},{"id":"C","text":"0.3174"},{"id":"D","text":"0.8413"}],"answer":"B","explanation":"Outside $(μ − σ, μ + σ]$ the remaining probability is $1 − 0.6826 = 0.3174$, split equally between the two tails, so each tail is $0.3174/2 = 0.1587$.","explanationZh":"区间 $(μ − σ, μ + σ]$ 之外的概率为 $1 − 0.6826 = 0.3174，两侧尾部各占一半，故 P(X > μ + σ) = 0.3174/2 = 0.1587。$","tags":["medium","empirical-rule","tail"],"status":"published"},{"id":"math-normal-013","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(μ, σ^{2})$, evaluate $P(μ − σ < X ≤ μ + 2σ)$.","stemZh":"$对 X ~ N(μ, σ^{2})，求 P(μ − σ < X ≤ μ + 2σ)。$","options":[{"id":"A","text":"0.9544"},{"id":"B","text":"0.4772"},{"id":"C","text":"0.8185"},{"id":"D","text":"0.3413"}],"answer":"C","explanation":"Split the interval at $μ: P(μ − σ < X ≤ μ) = 0.6826/2 = 0.3413$ and $P(μ < X ≤ μ + 2σ) = 0.9544/2 = 0.4772$. Their sum is 0.8185.","explanationZh":"$以 μ 为界分段：P(μ − σ < X ≤ μ) = 0.6826/2 = 0.3413，P(μ < X ≤ μ + 2σ) = 0.9544/2 = 0.4772，相加得 0.8185。$","tags":["medium","empirical-rule","symmetry"],"status":"published"},{"id":"math-normal-014","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(50, 100)$, find the standardised value z of $x = 70$.","stemZh":"已知 $X ~ N(50, 100)，求 x = 70$ 的标准化值 z。","options":[{"id":"A","text":"2"},{"id":"B","text":"0.2"},{"id":"C","text":"20"},{"id":"D","text":"−2"}],"answer":"A","explanation":"$μ = 50$ and $σ = \\sqrt{100} = 10$, so $z = (70 − 50)/10 = 2$. Option B uses $σ^{2} = 100$ as the divisor.","explanationZh":"$μ = 50，σ = \\sqrt{100} = 10，z = (70 − 50)/10 = 2。B$ 是误用 $σ^{2} = 100$ 去除。","tags":["medium","standardisation"],"status":"published"},{"id":"math-normal-015","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(8, 9)$, evaluate $P(5 < X ≤ 11)$.","stemZh":"已知 $X ~ N(8, 9)，求 P(5 < X ≤ 11)。$","options":[{"id":"A","text":"0.3413"},{"id":"B","text":"0.6826"},{"id":"C","text":"0.9544"},{"id":"D","text":"0.9974"}],"answer":"B","explanation":"$μ = 8$ and $σ = \\sqrt{9} = 3$, so $5 = μ − σ$ and $11 = μ + σ$. Hence $P(5 < X ≤ 11) = 0.6826$.","explanationZh":"$μ = 8，σ = \\sqrt{9} = 3，5 = μ − σ，11 = μ + σ，所以概率为 0.6826。$","tags":["medium","empirical-rule"],"status":"published"},{"id":"math-normal-016","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(8, 9)$, evaluate $P(X > 11)$.","stemZh":"已知 $X ~ N(8, 9)，求 P(X > 11)。$","options":[{"id":"A","text":"0.8413"},{"id":"B","text":"0.3174"},{"id":"C","text":"0.3413"},{"id":"D","text":"0.1587"}],"answer":"D","explanation":"Here $11 = μ + σ$, so $P(X > 11) = (1 − 0.6826)/2 = 0.1587$. The value 0.3174 is the probability of lying outside $(μ − σ, μ + σ]$ altogether.","explanationZh":"$11 = μ + σ，所以 P(X > 11) = (1 − 0.6826)/2 = 0.1587。0.3174$ 是落在 $(μ − σ, μ + σ]$ 之外的总概率。","tags":["medium","empirical-rule","tail"],"status":"published"},{"id":"math-normal-017","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(100, 64)$, evaluate $P(84 < X ≤ 116)$.","stemZh":"已知 $X ~ N(100, 64)，求 P(84 < X ≤ 116)。$","options":[{"id":"A","text":"0.9544"},{"id":"B","text":"0.6826"},{"id":"C","text":"0.4772"},{"id":"D","text":"0.9974"}],"answer":"A","explanation":"$μ = 100$ and $σ = \\sqrt{64} = 8$, so $84 = μ − 2σ$ and $116 = μ + 2σ$. Hence the probability is 0.9544.","explanationZh":"$μ = 100，σ = \\sqrt{64} = 8，84 = μ − 2σ，116 = μ + 2σ，故概率为 0.9544。$","tags":["medium","empirical-rule"],"status":"published"},{"id":"math-normal-018","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(100, 64)$, evaluate $P(X ≤ 92)$.","stemZh":"已知 $X ~ N(100, 64)，求 P(X ≤ 92)。$","options":[{"id":"A","text":"0.8413"},{"id":"B","text":"0.1587"},{"id":"C","text":"0.3413"},{"id":"D","text":"0.3174"}],"answer":"B","explanation":"$92 = μ − σ$, and by symmetry the left tail equals the right $tail: P(X ≤ 92) = (1 − 0.6826)/2 = 0.1587$.","explanationZh":"$92 = μ − σ，由对称性左尾等于右尾：P(X ≤ 92) = (1 − 0.6826)/2 = 0.1587。$","tags":["medium","empirical-rule","symmetry"],"status":"published"},{"id":"math-normal-019","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"Compare the density curves of $X ~ N(0, 1)$ and $Y ~ N(0, 4)$. Which statement is true?","stemZh":"比较 $X ~ N(0, 1) 与 Y ~ N(0, 4)$ 的密度曲线，哪个说法正确？","options":[{"id":"A","text":"The curve of X is flatter and wider than that of Y"},{"id":"B","text":"The two curves are identical"},{"id":"C","text":"The curve of Y is flatter and wider than that of X"},{"id":"D","text":"The curve of Y is shifted to the right of that of X"}],"answer":"C","explanation":"Both have mean 0, so both are centred at $x = 0$. But $σ_Y = \\sqrt{4} = 2$ while $σ_X = 1$, so Y spreads more and its bell is flatter and wider.","explanationZh":"两者均值都为 $0，对称轴都是 x = 0；但 σ_Y = \\sqrt{4} = 2，σ_X = 1，Y$ 离散程度更大，曲线更矮更胖。","tags":["medium","curve","standard-deviation"],"status":"published"},{"id":"math-normal-020","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(70, 25)$, evaluate $P(X > 80)$.","stemZh":"已知 $X ~ N(70, 25)，求 P(X > 80)。$","options":[{"id":"A","text":"0.0456"},{"id":"B","text":"0.9772"},{"id":"C","text":"0.1587"},{"id":"D","text":"0.0228"}],"answer":"D","explanation":"$μ = 70$ and $σ = \\sqrt{25} = 5$, so $80 = μ + 2σ$ and $P(X > 80) = (1 − 0.9544)/2 = 0.0228$. The value 0.0456 is the total probability in both tails beyond $2σ$.","explanationZh":"$μ = 70，σ = \\sqrt{25} = 5，80 = μ + 2σ，P(X > 80) = (1 − 0.9544)/2 = 0.0228。0.0456$ 是两侧 $2σ$ 之外的总概率。","tags":["medium","empirical-rule","tail"],"status":"published"},{"id":"math-normal-021","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":4,"stem":"For $X ~ N(μ, σ^{2})$, evaluate $P(μ − 2σ < X ≤ μ − σ)$.","stemZh":"$对 X ~ N(μ, σ^{2})，求 P(μ − 2σ < X ≤ μ − σ)。$","options":[{"id":"A","text":"0.2718"},{"id":"B","text":"0.4772"},{"id":"C","text":"0.1359"},{"id":"D","text":"0.3413"}],"answer":"C","explanation":"Take the difference of the two empirical rules and halve $it: (0.9544 − 0.6826)/2 = 0.2718/2 = 0.1359$. The value 0.2718 counts the same band on both sides of $μ$.","explanationZh":"$用两条经验法则相减再取一半：(0.9544 − 0.6826)/2 = 0.2718/2 = 0.1359。0.2718 是 μ$ 两侧同一段的总和。","tags":["hard","empirical-rule","symmetry"],"status":"published"},{"id":"math-normal-022","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":4,"stem":"For $X ~ N(20, 16)$, evaluate $P(12 < X ≤ 24)$.","stemZh":"已知 $X ~ N(20, 16)，求 P(12 < X ≤ 24)。$","options":[{"id":"A","text":"0.4772"},{"id":"B","text":"0.3413"},{"id":"C","text":"0.9544"},{"id":"D","text":"0.8185"}],"answer":"D","explanation":"$μ = 20$ and $σ = \\sqrt{16} = 4$, so $12 = μ − 2σ$ and $24 = μ + σ$. Split at $μ: 0.9544/2 + 0.6826/2 = 0.4772 + 0.3413 = 0.8185$.","explanationZh":"$μ = 20，σ = \\sqrt{16} = 4，12 = μ − 2σ，24 = μ + σ。以 μ 分段：0.9544/2 + 0.6826/2 = 0.4772 + 0.3413 = 0.8185。$","tags":["hard","empirical-rule","symmetry"],"status":"published"},{"id":"math-normal-023","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":4,"stem":"For $X ~ N(20, 16)$, evaluate $P(X ≤ 12) + P(X > 28)$.","stemZh":"已知 $X ~ N(20, 16)，求 P(X ≤ 12) + P(X > 28)。$","options":[{"id":"A","text":"0.0456"},{"id":"B","text":"0.0228"},{"id":"C","text":"0.9544"},{"id":"D","text":"0.1359"}],"answer":"A","explanation":"Here $12 = μ − 2σ$ and $28 = μ + 2σ$, so the two tails together are the complement of $0.9544: 1 − 0.9544 = 0.0456. A$ single tail would be 0.0228.","explanationZh":"$12 = μ − 2σ，28 = μ + 2σ，两侧尾部之和为 0.9544 的补：1 − 0.9544 = 0.0456。单侧尾部才是 0.0228。$","tags":["hard","empirical-rule","tail"],"status":"published"},{"id":"math-normal-024","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":4,"stem":"For $X ~ N(μ, σ^{2})$, evaluate $P(X ≤ μ + σ)$.","stemZh":"$对 X ~ N(μ, σ^{2})，求 P(X ≤ μ + σ)。$","options":[{"id":"A","text":"0.6826"},{"id":"B","text":"0.8413"},{"id":"C","text":"0.3413"},{"id":"D","text":"0.1587"}],"answer":"B","explanation":"$P(X ≤ μ) = 0.5$ by symmetry, and $P(μ < X ≤ μ + σ) = 0.6826/2 = 0.3413$, so the total is $0.5 + 0.3413 = 0.8413$.","explanationZh":"由对称性 $P(X ≤ μ) = 0.5，且 P(μ < X ≤ μ + σ) = 0.6826/2 = 0.3413，合计 0.5 + 0.3413 = 0.8413。$","tags":["hard","empirical-rule","symmetry"],"status":"published"},{"id":"math-normal-025","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":5,"stem":"For $X ~ N(30, 4)$, evaluate $P(24 < X ≤ 30)$.","stemZh":"已知 $X ~ N(30, 4)，求 P(24 < X ≤ 30)。$","options":[{"id":"A","text":"0.9974"},{"id":"B","text":"0.4772"},{"id":"C","text":"0.4987"},{"id":"D","text":"0.3413"}],"answer":"C","explanation":"$μ = 30$ and $σ = \\sqrt{4} = 2$, so $24 = μ − 3σ$. By symmetry the left half of the $3σ$ band is $0.9974/2 = 0.4987 (not 0.5: a$ little probability lies beyond $3σ)$.","explanationZh":"$μ = 30，σ = \\sqrt{4} = 2，24 = μ − 3σ。由对称性，3σ$ 区间左半概率为 $0.9974/2 = 0.4987（不等于 0.5，因为 3σ$ 之外仍有少量概率）。","tags":["hard","empirical-rule","symmetry"],"status":"published"},{"id":"math-normal-026","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":5,"stem":"For $X ~ N(μ, σ^{2})$, which of the following statements is NOT correct?","stemZh":"$对 X ~ N(μ, σ^{2})，下列说法中不正确的是？$","options":[{"id":"A","text":"$P(μ − σ < X ≤ μ + σ) ≈ 0.6826$"},{"id":"B","text":"$P(X > μ) = 0.5$"},{"id":"C","text":"$P(μ − 3σ < X ≤ μ + 3σ) ≈ 0.9974$"},{"id":"D","text":"$P(μ − σ < X ≤ μ + 2σ) ≈ 0.9544$"}],"answer":"D","explanation":"The first three are the standard facts. For D, splitting at $μ$ gives $0.6826/2 + 0.9544/2 = 0.3413 + 0.4772 = 0.8185$, not 0.9544; 0.9544 belongs to the symmetric interval $(μ − 2σ, μ + 2σ]$.","explanationZh":"前三项都是标准结论。D 中以 $μ$ 分段得 $0.6826/2 + 0.9544/2 = 0.3413 + 0.4772 = 0.8185，而不是 0.9544；0.9544$ 对应对称区间 $(μ − 2σ, μ + 2σ]。$","tags":["hard","empirical-rule","error-distinction"],"status":"published"},{"id":"math-normal-027","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":4,"stem":"For $X ~ N(50, 25)$, evaluate $P(X ≤ 35)$.","stemZh":"已知 $X ~ N(50, 25)，求 P(X ≤ 35)。$","options":[{"id":"A","text":"0.0026"},{"id":"B","text":"0.0013"},{"id":"C","text":"0.9974"},{"id":"D","text":"0.0228"}],"answer":"B","explanation":"$μ = 50$ and $σ = \\sqrt{25} = 5$, so $35 = μ − 3σ$. The two tails beyond $3σ$ hold $1 − 0.9974 = 0.0026$ in total, so one tail is $0.0026/2 = 0.0013$.","explanationZh":"$μ = 50，σ = \\sqrt{25} = 5，35 = μ − 3σ。3σ$ 之外两侧共 $1 − 0.9974 = 0.0026，单侧为 0.0026/2 = 0.0013。$","tags":["hard","empirical-rule","tail"],"status":"published"},{"id":"math-normal-028","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":5,"stem":"For $X ~ N(12, 25)$, find the standardised value z of $x = 4.5$.","stemZh":"已知 $X ~ N(12, 25)，求 x = 4.5$ 的标准化值 z。","options":[{"id":"A","text":"−1.5"},{"id":"B","text":"1.5"},{"id":"C","text":"−0.3"},{"id":"D","text":"−7.5"}],"answer":"A","explanation":"$μ = 12$ and $σ = \\sqrt{25} = 5$, so $z = (4.5 − 12)/5 = −7.5/5 = −1.5$. Dividing by $σ^{2} = 25$ instead of $σ$ gives −0.3, and −7.5 is only the numerator.","explanationZh":"$μ = 12，σ = \\sqrt{25} = 5，z = (4.5 − 12)/5 = −7.5/5 = −1.5。若误除以 σ^{2} = 25 得 −0.3，−7.5$ 只是分子。","tags":["hard","standardisation"],"status":"published"},{"id":"math-normal-029","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":5,"stem":"Scores are distributed as $X ~ N(80, 36)$. Out of 1000 students, about how many score between 74 and 86?","stemZh":"成绩 $X ~ N(80, 36)。1000$ 名学生中，大约有多少人成绩在 74 到 86 之间？","options":[{"id":"A","text":"341"},{"id":"B","text":"954"},{"id":"C","text":"683"},{"id":"D","text":"318"}],"answer":"C","explanation":"$μ = 80$ and $σ = \\sqrt{36} = 6$, so $74 = μ − σ$ and $86 = μ + σ$. The probability is 0.6826, giving $1000·0.6826 = 682.6 ≈ 683$ students.","explanationZh":"$μ = 80，σ = \\sqrt{36} = 6，74 = μ − σ，86 = μ + σ，概率为 0.6826，人数约 1000 \\times 0.6826 = 682.6 ≈ 683$ 人。","tags":["hard","empirical-rule","expected-count"],"status":"published"},{"id":"math-normal-030","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":5,"stem":"Scores are distributed as $X ~ N(80, 36)$. Out of 1000 students, about how many score below 68 or above 92?","stemZh":"成绩 $X ~ N(80, 36)。1000$ 名学生中，大约有多少人成绩低于 68 或高于 92？","options":[{"id":"A","text":"23"},{"id":"B","text":"456"},{"id":"C","text":"954"},{"id":"D","text":"46"}],"answer":"D","explanation":"$μ = 80$ and $σ = 6$, so $68 = μ − 2σ$ and $92 = μ + 2σ$. The two tails together carry $1 − 0.9544 = 0.0456$, giving $1000·0.0456 = 45.6 ≈ 46$ students. One tail alone would be about 23.","explanationZh":"$μ = 80，σ = 6，68 = μ − 2σ，92 = μ + 2σ。两侧尾部共 1 − 0.9544 = 0.0456，人数约 1000 \\times 0.0456 = 45.6 ≈ 46$ 人。只算单侧约 23 人。","tags":["hard","empirical-rule","expected-count"],"status":"published"},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-uniform-acceleration-001","unitId":"kinematics","topicId":"uniform-acceleration","difficulty":1,"stem":"A car starts from rest and accelerates uniformly at $2\\ \\mathrm{m/s^{2}}$. What is its speed after $5\\ \\mathrm{s}$?","stemZh":"汽车由静止开始以 $2\\ \\mathrm{m/s^{2}} 匀加速直线运动，5\\ \\mathrm{s}$ 后的速度多大？","options":[{"id":"A","text":"$2.5\\ \\mathrm{m/s}$"},{"id":"B","text":"$5\\ \\mathrm{m/s}$"},{"id":"C","text":"$10\\ \\mathrm{m/s}$"},{"id":"D","text":"$20\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"With $v_0 = 0: v = v_0 +$ at $= 0 + 2 \\times 5 = 10\\ \\mathrm{m/s}$.","explanationZh":"$由 v_0=0，v=v_0+at=0+2\\times5=10\\ \\mathrm{m/s}。$","tags":["basic","kinematics","v-t-relation"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-uniform-acceleration-002","unitId":"kinematics","topicId":"uniform-acceleration","difficulty":1,"stem":"Starting from rest, an object accelerates uniformly at $4\\ \\mathrm{m/s^{2}}$. How far does it travel in $3\\ \\mathrm{s}$?","stemZh":"物体由静止开始以 $4\\ \\mathrm{m/s^{2}} 匀加速，3\\ \\mathrm{s}$ 内的位移是多少？","options":[{"id":"A","text":"$18\\ \\mathrm{m}$"},{"id":"B","text":"$12\\ \\mathrm{m}$"},{"id":"C","text":"$24\\ \\mathrm{m}$"},{"id":"D","text":"$36\\ \\mathrm{m}$"}],"answer":"A","explanation":"$s = \\dfrac{1}{2}at^{2} = \\dfrac{1}{2} \\times 4 \\times 3^{2} = 18\\ \\mathrm{m}$.","explanationZh":"$s=\\dfrac{1}{2}at^{2}=\\dfrac{1}{2}\\times4\\times3^{2}=18\\ \\mathrm{m}。$","tags":["basic","kinematics","s-t-relation"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-uniform-acceleration-003","unitId":"kinematics","topicId":"uniform-acceleration","difficulty":1,"stem":"An object accelerates from rest at $3\\ \\mathrm{m/s^{2}}$ over a distance of $24\\ \\mathrm{m}$. What is its final speed?","stemZh":"物体由静止以 $3\\ \\mathrm{m/s^{2}}$ 匀加速，运动 $24\\ \\mathrm{m}$ 后的末速度多大？","options":[{"id":"A","text":"$6\\ \\mathrm{m/s}$"},{"id":"B","text":"$8\\ \\mathrm{m/s}$"},{"id":"C","text":"$12\\ \\mathrm{m/s}$"},{"id":"D","text":"$24\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"$v^{2} = 2as = 2 \\times 3 \\times 24 = 144$, so $v = 12\\ \\mathrm{m/s}$.","explanationZh":"$v^{2}=2as=2\\times3\\times24=144，故 v=12\\ \\mathrm{m/s}。$","tags":["basic","kinematics","v-s-relation"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-uniform-acceleration-004","unitId":"kinematics","topicId":"uniform-acceleration","difficulty":2,"stem":"A car accelerates uniformly from rest to $20\\ \\mathrm{m/s}$ over $40\\ \\mathrm{m}$. What is its acceleration?","stemZh":"汽车由静止匀加速到 $20\\ \\mathrm{m/s}，位移 40\\ \\mathrm{m}，加速度多大？$","options":[{"id":"A","text":"$2.5\\ \\mathrm{m/s^{2}}$"},{"id":"B","text":"$10\\ \\mathrm{m/s^{2}}$"},{"id":"C","text":"$20\\ \\mathrm{m/s^{2}}$"},{"id":"D","text":"$5\\ \\mathrm{m/s^{2}}$"}],"answer":"D","explanation":"From $v^{2} = 2as: a = \\dfrac{v^{2}}{2s} = \\dfrac{400}{80} = 5\\ \\mathrm{m/s^{2}}$.","explanationZh":"$由 v^{2}=2as 得 a=\\dfrac{v^{2}}{2s}=\\dfrac{400}{80}=5\\ \\mathrm{m/s^{2}}。$","tags":["basic","kinematics","acceleration"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-uniform-acceleration-005","unitId":"kinematics","topicId":"uniform-acceleration","difficulty":2,"stem":"An object moving at $20\\ \\mathrm{m/s}$ decelerates uniformly at $4\\ \\mathrm{m/s^{2}}$. How long does it take to stop?","stemZh":"物体以 $20\\ \\mathrm{m/s}$ 运动，以 $4\\ \\mathrm{m/s^{2}}$ 匀减速，多久停下？","options":[{"id":"A","text":"$4\\ \\mathrm{s}$"},{"id":"B","text":"$5\\ \\mathrm{s}$"},{"id":"C","text":"$16\\ \\mathrm{s}$"},{"id":"D","text":"$80\\ \\mathrm{s}$"}],"answer":"B","explanation":"Coming to rest means $v = 0$, so $t = \\dfrac{v_0}{a} = \\dfrac{20}{4} = 5\\ \\mathrm{s}$.","explanationZh":"停下时 $v=0，t=\\dfrac{v_0}{a}=\\dfrac{20}{4}=5\\ \\mathrm{s}。$","tags":["basic","kinematics","deceleration"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-uniform-acceleration-006","unitId":"kinematics","topicId":"uniform-acceleration","difficulty":3,"stem":"$A 2\\ \\mathrm{kg}$ block on a smooth horizontal floor starts from rest and moves $8\\ \\mathrm{m}$ in $2\\ \\mathrm{s}$ under a constant horizontal force F. Find F.","stemZh":"质量 $2\\ \\mathrm{kg}$ 的物块在光滑水平面上由静止出发，在水平恒力 F 作用下 $2\\ \\mathrm{s}$ 内位移 $8\\ \\mathrm{m}，求$ F。","options":[{"id":"A","text":"$4\\ \\mathrm{N}$"},{"id":"B","text":"$16\\ \\mathrm{N}$"},{"id":"C","text":"$8\\ \\mathrm{N}$"},{"id":"D","text":"$2\\ \\mathrm{N}$"}],"answer":"C","explanation":"First $a = \\dfrac{2s}{t^{2}} = \\dfrac{16}{4} = 4\\ \\mathrm{m/s^{2}}$; then $F =$ ma $= 2 \\times 4 = 8\\ \\mathrm{N}$.","explanationZh":"先求 $a=\\dfrac{2s}{t^{2}}=\\dfrac{16}{4}=4\\ \\mathrm{m/s^{2}}，再由 F=ma=2\\times4=8\\ \\mathrm{N}。$","tags":["applied","kinematics","newton","fromExam"],"source":{"exam":"CSCA Physics","year":"2025-12","note":"Same structure as the recalled item: find a from s and t, then F = ma."}},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-uniform-acceleration-007","unitId":"kinematics","topicId":"uniform-acceleration","difficulty":3,"stem":"A car travelling at $20\\ \\mathrm{m/s}$ brakes with a uniform deceleration of $5\\ \\mathrm{m/s^{2}}$. What is its braking distance?","stemZh":"汽车以 $20\\ \\mathrm{m/s}$ 行驶，以 $5\\ \\mathrm{m/s^{2}}$ 匀减速刹车，刹车距离是多少？","options":[{"id":"A","text":"$40\\ \\mathrm{m}$"},{"id":"B","text":"$20\\ \\mathrm{m}$"},{"id":"C","text":"$80\\ \\mathrm{m}$"},{"id":"D","text":"$100\\ \\mathrm{m}$"}],"answer":"A","explanation":"$s = \\dfrac{v_0^{2}}{2a} = \\dfrac{400}{10} = 40\\ \\mathrm{m}$.","explanationZh":"$s=\\dfrac{v_0^{2}}{2a}=\\dfrac{400}{10}=40\\ \\mathrm{m}。$","tags":["applied","kinematics","braking"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-uniform-acceleration-008","unitId":"kinematics","topicId":"uniform-acceleration","difficulty":3,"stem":"An object accelerates from rest at $2\\ \\mathrm{m/s^{2}}$. What is its displacement during the 3rd second?","stemZh":"物体由静止以 $2\\ \\mathrm{m/s^{2}}$ 匀加速，第 3 秒内的位移是多少？","options":[{"id":"A","text":"$5\\ \\mathrm{m}$"},{"id":"B","text":"$3\\ \\mathrm{m}$"},{"id":"C","text":"$9\\ \\mathrm{m}$"},{"id":"D","text":"$4\\ \\mathrm{m}$"}],"answer":"A","explanation":"$\\Delta s = \\dfrac{1}{2}a(3^{2} - 2^{2}) = \\dfrac{1}{2} \\times 2 \\times 5 = 5\\ \\mathrm{m}$.","explanationZh":"$\\Delta s=\\dfrac{1}{2}a(3^{2}-2^{2})=\\dfrac{1}{2}\\times2\\times5=5\\ \\mathrm{m}。$","tags":["applied","kinematics","nth-second"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-uniform-acceleration-009","unitId":"kinematics","topicId":"uniform-acceleration","difficulty":3,"stem":"An object accelerates from rest at $3\\ \\mathrm{m/s^{2}}$ for $4\\ \\mathrm{s}$, then continues at constant speed for $6\\ \\mathrm{s}$. Find the total distance.","stemZh":"物体由静止以 $3\\ \\mathrm{m/s^{2}}$ 加速 $4\\ \\mathrm{s}，再匀速运动 6\\ \\mathrm{s}，求总路程。$","options":[{"id":"A","text":"$48\\ \\mathrm{m}$"},{"id":"B","text":"$96\\ \\mathrm{m}$"},{"id":"C","text":"$72\\ \\mathrm{m}$"},{"id":"D","text":"$120\\ \\mathrm{m}$"}],"answer":"B","explanation":"Stage $1: s_1 = \\dfrac{1}{2} \\times 3 \\times 16 = 24\\ \\mathrm{m}$, reaching $v = 12\\ \\mathrm{m/s}$. Stage $2: s_2 = 12 \\times 6 = 72\\ \\mathrm{m}$. Total $96\\ \\mathrm{m}$.","explanationZh":"第一段 $s_1=\\dfrac{1}{2}\\times3\\times16=24\\ \\mathrm{m}，末速 v=12\\ \\mathrm{m/s}；第二段 s_2=12\\times6=72\\ \\mathrm{m}，共 96\\ \\mathrm{m}。$","tags":["applied","kinematics","two-stage"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-uniform-acceleration-010","unitId":"kinematics","topicId":"uniform-acceleration","difficulty":3,"stem":"At $t = 0$ car A starts from rest with $a = 2\\ \\mathrm{m/s^{2}}$ while car B passes the same point at a constant $10\\ \\mathrm{m/s}$. When does A catch B?","stemZh":"$t=0 时 A$ 车由静止以 $2\\ \\mathrm{m/s^{2}}$ 启动，B 车同时同地以 $10\\ \\mathrm{m/s}$ 匀速通过，A 何时追上 B？","options":[{"id":"A","text":"$5\\ \\mathrm{s}$"},{"id":"B","text":"$20\\ \\mathrm{s}$"},{"id":"C","text":"$2.5\\ \\mathrm{s}$"},{"id":"D","text":"$10\\ \\mathrm{s}$"}],"answer":"D","explanation":"$A: s = \\dfrac{1}{2} \\times 2 \\times t^{2} = t^{2}. B: s = 10t$. Equal when $t^{2} = 10t$, so $t = 10\\ \\mathrm{s}$.","explanationZh":"$A 车 s=\\dfrac{1}{2}\\times2\\times t^{2}=t^{2}，B 车 s=10t。t^{2}=10t 得 t=10\\ \\mathrm{s}。$","tags":["applied","kinematics","catch-up"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-uniform-acceleration-011","unitId":"kinematics","topicId":"uniform-acceleration","difficulty":4,"stem":"A car moving at $20\\ \\mathrm{m/s}$ brakes at $5\\ \\mathrm{m/s^{2}}$. How far does it travel in the $5\\ \\mathrm{s}$ after braking begins?","stemZh":"汽车以 $20\\ \\mathrm{m/s}$ 行驶，刹车后以 $5\\ \\mathrm{m/s^{2}}$ 匀减速，刹车后 $5\\ \\mathrm{s}$ 内的位移是多少？","options":[{"id":"A","text":"$37.5\\ \\mathrm{m}$"},{"id":"B","text":"$40\\ \\mathrm{m}$"},{"id":"C","text":"$50\\ \\mathrm{m}$"},{"id":"D","text":"$62.5\\ \\mathrm{m}$"}],"answer":"B","explanation":"The car stops at $t = \\dfrac{20}{5} = 4\\ \\mathrm{s}$, before 5 s is up. Substituting $t = 5$ blindly gives $37.5\\ \\mathrm{m}$, but the car is already stationary. Real distance $= \\dfrac{v_0^{2}}{2a} = 40\\ \\mathrm{m}$.","explanationZh":"汽车在 $t=\\dfrac{20}{5}=4\\ \\mathrm{s}$ 时已停下，早于 5 s。若直接代入 $t=5 得 37.5\\ \\mathrm{m}$ 是错误的；实际位移 $=\\dfrac{v_0^{2}}{2a}=40\\ \\mathrm{m}。$","tags":["trap","kinematics","braking","stop-time"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-uniform-acceleration-012","unitId":"kinematics","topicId":"uniform-acceleration","difficulty":4,"stem":"A ball is thrown vertically upward at $30\\ \\mathrm{m/s}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its height above the launch point after $4\\ \\mathrm{s}$?","stemZh":"小球以 $30\\ \\mathrm{m/s}$ 竖直上抛，取 $g=10\\ \\mathrm{m/s^{2}}，4\\ \\mathrm{s}$ 时相对抛出点的高度是多少？","options":[{"id":"A","text":"$40\\ \\mathrm{m}$"},{"id":"B","text":"$80\\ \\mathrm{m}$"},{"id":"C","text":"$120\\ \\mathrm{m}$"},{"id":"D","text":"$20\\ \\mathrm{m}$"}],"answer":"A","explanation":"Take upward as $positive: s = v_0 t - \\dfrac{1}{2}gt^{2} = 30 \\times 4 - \\dfrac{1}{2} \\times 10 \\times 16 = 120 - 80 = 40\\ \\mathrm{m}$, still above the launch point.","explanationZh":"$取向上为正：s=v_0t-\\dfrac{1}{2}gt^{2}=30\\times4-\\dfrac{1}{2}\\times10\\times16=120-80=40\\ \\mathrm{m}，仍在抛出点上方。$","tags":["trap","kinematics","vertical-throw","sign"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-uniform-acceleration-013","images":[{"src":"/physics/diagrams/vt-graph.svg","alt":"Velocity-time graph: speed rises uniformly to 12 m/s in 3 s, then stays at 12 m/s for 2 s.","altZh":"v–t 图像：前 3 s 匀加速到 12 m/s，之后保持 12 m/s 匀速 2 s。"}],"unitId":"kinematics","topicId":"uniform-acceleration","difficulty":4,"stem":"A velocity–time graph shows an object accelerating uniformly from 0 to $12\\ \\mathrm{m/s}$ in $3\\ \\mathrm{s}$, then holding $12\\ \\mathrm{m/s}$ for $2\\ \\mathrm{s}$. Find the total displacement.","stemZh":"某物体的 v-t $图像：3\\ \\mathrm{s}$ 内由 0 匀加速到 $12\\ \\mathrm{m/s}，再以 12\\ \\mathrm{m/s}$ 匀速 $2\\ \\mathrm{s}。求总位移。$","options":[{"id":"A","text":"$30\\ \\mathrm{m}$"},{"id":"B","text":"$36\\ \\mathrm{m}$"},{"id":"C","text":"$54\\ \\mathrm{m}$"},{"id":"D","text":"$42\\ \\mathrm{m}$"}],"answer":"D","explanation":"Displacement is the area under the graph: triangle $\\dfrac{1}{2} \\times 3 \\times 12 = 18\\ \\mathrm{m}$ plus rectangle $12 \\times 2 = 24\\ \\mathrm{m}$, total $42\\ \\mathrm{m}$.","explanationZh":"位移等于图线与时间轴围成的面积：三角形 $\\dfrac{1}{2}\\times3\\times12=18\\ \\mathrm{m}，矩形 12\\times2=24\\ \\mathrm{m}，共 42\\ \\mathrm{m}。$","tags":["trap","kinematics","v-t-graph","area"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-uniform-acceleration-014","unitId":"kinematics","topicId":"uniform-acceleration","difficulty":5,"stem":"Object A starts from rest at $a = 1\\ \\mathrm{m/s^{2}}$. Two seconds later B starts from the same point at $a = 4\\ \\mathrm{m/s^{2}}$. How long after A started does B catch up?","stemZh":"A 物体由静止以 $1\\ \\mathrm{m/s^{2}} 出发，2\\ \\mathrm{s} 后 B$ 从同地以 $4\\ \\mathrm{m/s^{2}}$ 出发。A 出发后多久 B 追上 A？","options":[{"id":"A","text":"$2\\ \\mathrm{s}$"},{"id":"B","text":"$3\\ \\mathrm{s}$"},{"id":"C","text":"$4\\ \\mathrm{s}$"},{"id":"D","text":"$6\\ \\mathrm{s}$"}],"answer":"C","explanation":"Let t be A's time. $A: s = \\dfrac{1}{2}t^{2}. B: s = \\dfrac{1}{2} \\times 4(t-2)^{2} = 2(t-2)^{2}. Equating: \\dfrac{1}{2}t^{2} = 2(t-2)^{2} \\Rightarrow 3t^{2} - 16t + 16 = 0$, giving $t = 4\\ \\mathrm{s} (the\\text{ root }t = \\dfrac{4}{3}\\ \\mathrm{s}\\text{ is before }B starts)$.","explanationZh":"设 A 运动时间为 $t。A：s=\\dfrac{1}{2}t^{2}；B：s=2(t-2)^{2}。联立 \\dfrac{1}{2}t^{2}=2(t-2)^{2} 得 3t^{2}-16t+16=0，解得 t=4\\ \\mathrm{s}（另一根 t=\\dfrac{4}{3}\\ \\mathrm{s} 在 B$ 出发前，舍去）。","tags":["trap","kinematics","catch-up","quadratic"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-uniform-acceleration-015","unitId":"kinematics","topicId":"uniform-acceleration","difficulty":5,"stem":"A block slides from rest down a smooth incline and reaches $8\\ \\mathrm{m/s}$ after $4\\ \\mathrm{s}$. It then travels on a rough horizontal surface and stops after a further $4\\ \\mathrm{s}$. Find the total distance.","stemZh":"$物块沿光滑斜面由静止下滑，4\\ \\mathrm{s}$ 后速度达 $8\\ \\mathrm{m/s}，之后在粗糙水平面上再经 4\\ \\mathrm{s}$ 停下。求总路程。","options":[{"id":"A","text":"$16\\ \\mathrm{m}$"},{"id":"B","text":"$24\\ \\mathrm{m}$"},{"id":"C","text":"$32\\ \\mathrm{m}$"},{"id":"D","text":"$48\\ \\mathrm{m}$"}],"answer":"C","explanation":"$Incline: a = \\dfrac{8}{4} = 2\\ \\mathrm{m/s^{2}}, s_1 = \\dfrac{1}{2} \\times 2 \\times 16 = 16\\ \\mathrm{m}. Horizontal$: average speed $\\dfrac{8+0}{2} = 4\\ \\mathrm{m/s}$, so $s_2 = 4 \\times 4 = 16\\ \\mathrm{m}$. Total $32\\ \\mathrm{m}$.","explanationZh":"$斜面：a=\\dfrac{8}{4}=2\\ \\mathrm{m/s^{2}}，s_1=\\dfrac{1}{2}\\times2\\times16=16\\ \\mathrm{m}。水平：平均速度 \\dfrac{8+0}{2}=4\\ \\mathrm{m/s}，s_2=4\\times4=16\\ \\mathrm{m}。共 32\\ \\mathrm{m}。$","tags":["trap","kinematics","two-stage","average-speed"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-common-forces-001","unitId":"forces-newton","topicId":"common-forces","difficulty":1,"stem":"Gravity on the Moon is about $\\dfrac{1}{6}$ of that on Earth. For a mass $m = 6\\ \\mathrm{kg}$ with $g = 10\\ \\mathrm{m/s^{2}}$ on Earth, what is its weight on the Moon?","stemZh":"月球上的重力约为地球的 $\\dfrac{1}{6}。质量 m=6\\ \\mathrm{kg}$ 的物体，地球上 $g=10\\ \\mathrm{m/s^{2}}，它在月球上的重力是多少？$","options":[{"id":"A","text":"$36\\ \\mathrm{N}$"},{"id":"B","text":"$10\\ \\mathrm{N}$"},{"id":"C","text":"$60\\ \\mathrm{N}$"},{"id":"D","text":"$6\\ \\mathrm{N}$"}],"answer":"B","explanation":"On Earth $G =$ mg $= 60\\ \\mathrm{N}$; on the Moon G' $= \\dfrac{60}{6} = 10\\ \\mathrm{N}$. Mass itself does not change.","explanationZh":"地球上 $G=mg=60\\ \\mathrm{N}，月球上 G'=\\dfrac{60}{6}=10\\ \\mathrm{N}。质量不变。$","tags":["basic","forces","gravity","fromExam"],"source":{"exam":"CSCA Physics","year":"2026-04","note":"Recalled item uses Mars (1/3); here the Moon variant with fresh numbers."}},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-common-forces-002","unitId":"forces-newton","topicId":"common-forces","difficulty":1,"stem":"A spring of stiffness $k = 50\\ \\mathrm{N/m}$ and natural length $10\\ \\mathrm{cm}$ is stretched to $14\\ \\mathrm{cm}$. What pull is required?","stemZh":"劲度系数 $k=50\\ \\mathrm{N/m}$ 的弹簧原长 $10\\ \\mathrm{cm}，拉长到 14\\ \\mathrm{cm}，所需拉力多大？$","options":[{"id":"A","text":"$6\\ \\mathrm{N}$"},{"id":"B","text":"$4\\ \\mathrm{N}$"},{"id":"C","text":"$2\\ \\mathrm{N}$"},{"id":"D","text":"$8\\ \\mathrm{N}$"}],"answer":"C","explanation":"Extension $\\Delta x = 14 - 10 = 4\\ \\mathrm{cm} = 0.04\\ \\mathrm{m}$, so $F = k\\Delta x = 50 \\times 0.04 = 2\\ \\mathrm{N}$.","explanationZh":"伸长量 $\\Delta x=14-10=4\\ \\mathrm{cm}=0.04\\ \\mathrm{m}，F=k\\Delta x=50\\times0.04=2\\ \\mathrm{N}。$","tags":["basic","forces","spring","fromExam"],"source":{"exam":"CSCA Physics","year":"2026-04","note":"Same spring item structure as the recalled paper."}},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-common-forces-003","unitId":"forces-newton","topicId":"common-forces","difficulty":1,"stem":"A spring has natural length $0.7\\ \\mathrm{m}$ and stiffness $k = 10\\ \\mathrm{N/m}$. When pulled with $3\\ \\mathrm{N}$, what is its new length?","stemZh":"弹簧原长 $0.7\\ \\mathrm{m}，劲度系数 k=10\\ \\mathrm{N/m}。施加 3\\ \\mathrm{N}$ 拉力后长度是多少？","options":[{"id":"A","text":"$0.73\\ \\mathrm{m}$"},{"id":"B","text":"$1.0\\ \\mathrm{m}$"},{"id":"C","text":"$0.4\\ \\mathrm{m}$"},{"id":"D","text":"$0.3\\ \\mathrm{m}$"}],"answer":"B","explanation":"$\\Delta x = \\dfrac{F}{k} = \\dfrac{3}{10} = 0.3\\ \\mathrm{m}$, so the length is $0.7 + 0.3 = 1.0\\ \\mathrm{m}$.","explanationZh":"$\\Delta x=\\dfrac{F}{k}=\\dfrac{3}{10}=0.3\\ \\mathrm{m}，长度为 0.7+0.3=1.0\\ \\mathrm{m}。$","tags":["basic","forces","spring","fromExam"],"source":{"exam":"CSCA Physics","year":"2026-01","note":"Recalled item asks for the new length, not the extension."}},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-common-forces-004","unitId":"forces-newton","topicId":"common-forces","difficulty":2,"stem":"$A 5\\ \\mathrm{kg}$ block slides on a horizontal surface with coefficient of kinetic friction $\\mu = 0.2$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the friction force?","stemZh":"质量 $5\\ \\mathrm{kg}$ 的物块在水平面上滑动，动摩擦因数 $\\mu=0.2，取 g=10\\ \\mathrm{m/s^{2}}，摩擦力多大？$","options":[{"id":"A","text":"$10\\ \\mathrm{N}$"},{"id":"B","text":"$5\\ \\mathrm{N}$"},{"id":"C","text":"$2\\ \\mathrm{N}$"},{"id":"D","text":"$50\\ \\mathrm{N}$"}],"answer":"A","explanation":"On a horizontal surface $N =$ mg $= 50\\ \\mathrm{N}$, so $f = \\mu N = 0.2 \\times 50 = 10\\ \\mathrm{N}$.","explanationZh":"水平面上 $N=mg=50\\ \\mathrm{N}，f=\\mu N=0.2\\times50=10\\ \\mathrm{N}。$","tags":["basic","forces","friction"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-common-forces-005","images":[{"src":"/physics/diagrams/fbd-push.svg","alt":"Free-body diagram of a block on a horizontal floor: a push to the right and static friction to the left (block does not move).","altZh":"物块在水平地面上的受力图：向右的推力与向左的静摩擦力（物块未动）。"}],"unitId":"forces-newton","topicId":"common-forces","difficulty":2,"stem":"A block rests on a horizontal floor. A horizontal force of $5\\ \\mathrm{N}$ pushes it to the right but it does not move. What is the direction of the static friction on the block?","stemZh":"物块静止在水平地面上，用 $5\\ \\mathrm{N}$ 的水平力向右推但未推动，物块受到的静摩擦力方向如何？","options":[{"id":"A","text":"To the left"},{"id":"B","text":"To the right"},{"id":"C","text":"Vertically upward"},{"id":"D","text":"Zero — there is no friction"}],"optionsZh":[{"id":"A","text":"向左"},{"id":"B","text":"向右"},{"id":"C","text":"竖直向上"},{"id":"D","text":"为零，无摩擦力"}],"answer":"A","explanation":"Static friction opposes the tendency to slip. Since the applied force tends to move the block right, friction acts to the left and equals $5\\ \\mathrm{N}$.","explanationZh":"静摩擦力阻碍相对运动趋势。外力使物体有向右运动趋势，故静摩擦力向左，大小等于 $5\\ \\mathrm{N}。$","tags":["basic","forces","friction","direction"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-common-forces-006","images":[{"src":"/physics/diagrams/fbd-push.svg","alt":"Free-body diagram of a block pulled on a horizontal floor; static friction balances the pull while the block stays at rest.","altZh":"水平面上被拉动的物块受力图；物块静止时静摩擦力与拉力平衡。"}],"unitId":"forces-newton","topicId":"common-forces","difficulty":3,"stem":"$A 5\\ \\mathrm{kg}$ block sits on a horizontal $floor, \\mu = 0.2$, and is pulled horizontally with $F = 8\\ \\mathrm{N}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the friction force on the block?","stemZh":"质量 $5\\ \\mathrm{kg} 的物块在水平面上，\\mu=0.2，用 F=8\\ \\mathrm{N}$ 的水平力拉，取 $g=10\\ \\mathrm{m/s^{2}}，物块受到的摩擦力多大？$","options":[{"id":"A","text":"$8\\ \\mathrm{N}$"},{"id":"B","text":"$4\\ \\mathrm{N}$"},{"id":"C","text":"$0\\ \\mathrm{N}$"},{"id":"D","text":"$10\\ \\mathrm{N}$"}],"answer":"A","explanation":"Maximum static friction is about $\\mu$ mg $= 0.2 \\times 5 \\times 10 = 10\\ \\mathrm{N}$. Since $8\\ \\mathrm{N} < 10\\ \\mathrm{N}$ the block stays at rest, so static friction simply balances the $pull: f = 8\\ \\mathrm{N} (not 10\\ \\mathrm{N})$.","explanationZh":"最大静摩擦力约为 $\\mu mg=0.2\\times5\\times10=10\\ \\mathrm{N}。因 8\\ \\mathrm{N}<10\\ \\mathrm{N}，物块仍静止，静摩擦力与外力平衡，f=8\\ \\mathrm{N}（不是 10\\ \\mathrm{N}）。$","tags":["applied","forces","friction","static","fromExam"],"source":{"exam":"CSCA Physics","year":"2026-04","note":"Recurring recalled-exam trap: the pull is below the static limit."}},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-common-forces-007","unitId":"forces-newton","topicId":"common-forces","difficulty":3,"stem":"$A 2\\ \\mathrm{kg}$ mass hangs at rest from a vertical spring of stiffness $k = 200\\ \\mathrm{N/m}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the extension?","stemZh":"质量 $2\\ \\mathrm{kg}$ 的物体竖直悬挂在劲度系数 $k=200\\ \\mathrm{N/m}$ 的弹簧下端静止，取 $g=10\\ \\mathrm{m/s^{2}}，弹簧伸长多少？$","options":[{"id":"A","text":"$0.01\\ \\mathrm{m}$"},{"id":"B","text":"$0.2\\ \\mathrm{m}$"},{"id":"C","text":"$1\\ \\mathrm{m}$"},{"id":"D","text":"$0.1\\ \\mathrm{m}$"}],"answer":"D","explanation":"At equilibrium $k\\Delta x = mg$, so $\\Delta x = \\dfrac{mg}{k} = \\dfrac{20}{200} = 0.1\\ \\mathrm{m}$.","explanationZh":"平衡时 $k\\Delta x=mg，故 \\Delta x=\\dfrac{mg}{k}=\\dfrac{20}{200}=0.1\\ \\mathrm{m}。$","tags":["applied","forces","spring","equilibrium"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-common-forces-008","images":[{"src":"/physics/diagrams/incline-smooth-30.svg","alt":"Block on a smooth 30-degree incline; the forces are weight mg and the normal reaction N (no friction).","altZh":"光滑 30° 斜面上的物块；受力为重力 mg 与支持力 N（无摩擦）。"}],"unitId":"forces-newton","topicId":"common-forces","difficulty":3,"stem":"$A 2\\ \\mathrm{kg}$ block rests on a smooth incline of angle $\\theta = 30^{\\circ}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the normal force on the block?","stemZh":"质量 $2\\ \\mathrm{kg}$ 的物块静止在倾角 $\\theta=30^{\\circ}$ 的光滑斜面上，取 $g=10\\ \\mathrm{m/s^{2}}，支持力多大？$","options":[{"id":"A","text":"$10\\ \\mathrm{N}$"},{"id":"B","text":"$20\\ \\mathrm{N}$"},{"id":"C","text":"$17.3\\ \\mathrm{N}$"},{"id":"D","text":"$8.7\\ \\mathrm{N}$"}],"answer":"C","explanation":"Perpendicular to the plane, forces $balance: N = mg\\cos\\theta = 2 \\times 10 \\times \\cos 30^{\\circ} \\approx 17.3\\ \\mathrm{N}$. Note $N \\neq$ mg.","explanationZh":"$垂直斜面方向受力平衡：N=mg\\cos\\theta=2\\times10\\times\\cos30^{\\circ}\\approx17.3\\ \\mathrm{N}。注意 N\\neq$ mg。","tags":["applied","forces","incline","normal-force"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-common-forces-009","unitId":"forces-newton","topicId":"common-forces","difficulty":3,"stem":"You hold $a 0.5\\ \\mathrm{kg}$ bottle stationary in your hand. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the static friction exerted by your hand on the bottle?","stemZh":"用手握住质量 $0.5\\ \\mathrm{kg}$ 的瓶子保持静止，取 $g=10\\ \\mathrm{m/s^{2}}，手对瓶子的静摩擦力多大？$","options":[{"id":"A","text":"$0\\ \\mathrm{N}$"},{"id":"B","text":"$10\\ \\mathrm{N}$"},{"id":"C","text":"$5\\ \\mathrm{N}$ downward"},{"id":"D","text":"$5\\ \\mathrm{N}$ upward"}],"optionsZh":[{"id":"A","text":"$0\\ \\mathrm{N}$"},{"id":"B","text":"$10\\ \\mathrm{N}$"},{"id":"C","text":"$5\\ \\mathrm{N}，向下$"},{"id":"D","text":"$5\\ \\mathrm{N}，向上$"}],"answer":"D","explanation":"Vertically the bottle is in equilibrium, so friction must balance its $weight: f =$ mg $= 5\\ \\mathrm{N}$, directed upward to oppose the tendency to slip down.","explanationZh":"$瓶子竖直方向平衡，静摩擦力与重力平衡：f=mg=5\\ \\mathrm{N}，方向向上以阻碍下滑趋势。$","tags":["applied","forces","friction","static","fromExam"],"source":{"exam":"CSCA Physics","note":"Holding/gripping context appears in the recalled papers."}},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-common-forces-010","images":[{"src":"/physics/diagrams/vectors-resultant.svg","alt":"Two forces of 3 N and 5 N acting at a point; their resultant magnitude lies between 2 N and 8 N.","altZh":"大小为 3 N 与 5 N 的两个共点力；合力大小介于 2 N 与 8 N 之间。"}],"unitId":"forces-newton","topicId":"common-forces","difficulty":3,"stem":"An object is acted on by two forces of magnitude $3\\ \\mathrm{N}$ and $5\\ \\mathrm{N}$. Which of the following cannot be the magnitude of their resultant?","stemZh":"物体受两个大小分别为 $3\\ \\mathrm{N} 和 5\\ \\mathrm{N}$ 的力作用，下列哪个不可能是合力的大小？","options":[{"id":"A","text":"$2\\ \\mathrm{N}$"},{"id":"B","text":"$4\\ \\mathrm{N}$"},{"id":"C","text":"$8\\ \\mathrm{N}$"},{"id":"D","text":"$1\\ \\mathrm{N}$"}],"answer":"D","explanation":"The resultant lies between $|F_1 - F_2| = 2\\ \\mathrm{N}$ and $F_1 + F_2 = 8\\ \\mathrm{N}$. So $1\\ \\mathrm{N}$ is outside the range and impossible.","explanationZh":"合力范围在 $|F_1-F_2|=2\\ \\mathrm{N} 与 F_1+F_2=8\\ \\mathrm{N} 之间，1\\ \\mathrm{N}$ 超出范围，不可能。","tags":["applied","forces","resultant","range","fromExam"],"source":{"exam":"CSCA Physics","year":"2026-01","note":"Recalled item asks which resultant is impossible."}},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-common-forces-011","images":[{"src":"/physics/diagrams/fbd-push.svg","alt":"Free-body diagram of a block pulled on a horizontal floor; the pull is below the static-friction limit so the block does not move.","altZh":"水平面上被拉动的物块受力图；拉力小于最大静摩擦力，物块不动。"}],"unitId":"forces-newton","topicId":"common-forces","difficulty":4,"stem":"$A 4\\ \\mathrm{kg}$ block on a horizontal floor has $\\mu = 0.2$ and is pulled horizontally with $F = 6\\ \\mathrm{N}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the friction force once it is moving?","stemZh":"质量 $4\\ \\mathrm{kg} 的物块在水平面上，\\mu=0.2，用 F=6\\ \\mathrm{N}$ 的水平力拉，取 $g=10\\ \\mathrm{m/s^{2}}，物块运动后受到的摩擦力多大？$","options":[{"id":"A","text":"$6\\ \\mathrm{N}$"},{"id":"B","text":"$10\\ \\mathrm{N}$"},{"id":"C","text":"$8\\ \\mathrm{N}$"},{"id":"D","text":"$0\\ \\mathrm{N}$"}],"answer":"C","explanation":"The static limit is about $\\mu$ mg $= 0.2 \\times 4 \\times 10 = 8\\ \\mathrm{N}$. Here $6\\ \\mathrm{N} < 8\\ \\mathrm{N}$, so the block does not start moving at all — it stays at rest and static friction is $6\\ \\mathrm{N}. Careful$: if it were already sliding, kinetic friction would be $8\\ \\mathrm{N}$.","explanationZh":"最大静摩擦力约 $\\mu mg=0.2\\times4\\times10=8\\ \\mathrm{N}。因 6\\ \\mathrm{N}<8\\ \\mathrm{N}，物块根本不会动，静摩擦力为 6\\ \\mathrm{N}。注意：若已在滑动，滑动摩擦力才是 8\\ \\mathrm{N}。$","tags":["trap","forces","friction","static-vs-kinetic"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-common-forces-012","unitId":"forces-newton","topicId":"common-forces","difficulty":4,"stem":"A uniform spring of stiffness k is cut into two equal halves. What is the stiffness of each half?","stemZh":"一根劲度系数为 k 的均匀弹簧被剪成等长的两段，每段的劲度系数是多少？","options":[{"id":"A","text":"$\\dfrac{k}{2}$"},{"id":"B","text":"k"},{"id":"C","text":"2k"},{"id":"D","text":"4k"}],"answer":"C","explanation":"For a uniform spring, stiffness is inversely proportional to length $(k \\propto \\dfrac{1}{L})$. Halving the length doubles the stiffness, giving 2k.","explanationZh":"$均匀弹簧的劲度系数与长度成反比（k\\propto\\dfrac{1}{L}）。长度减半，劲度系数加倍，为 2k。$","tags":["trap","forces","spring","stiffness"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-common-forces-013","unitId":"forces-newton","topicId":"common-forces","difficulty":4,"stem":"A block sits at rest relative to a conveyor belt that runs to the right at constant speed. What friction acts on the block?","stemZh":"物块相对传送带静止，传送带向右匀速运行，物块受到的摩擦力如何？","options":[{"id":"A","text":"Kinetic friction to the right"},{"id":"B","text":"Static friction to the left"},{"id":"C","text":"Zero"},{"id":"D","text":"Static friction upward"}],"optionsZh":[{"id":"A","text":"滑动摩擦力向右"},{"id":"B","text":"静摩擦力向左"},{"id":"C","text":"为零"},{"id":"D","text":"静摩擦力向上"}],"answer":"C","explanation":"Moving at constant speed with the belt, the block has no acceleration and no tendency to slip, so the net horizontal force — and hence friction — is zero.","explanationZh":"物块随传送带匀速运动，无加速度、无相对滑动趋势，水平方向合力为零，故摩擦力为零。","tags":["trap","forces","friction","conveyor"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-common-forces-014","images":[{"src":"/physics/diagrams/stacked-blocks.svg","alt":"A 1 kg block resting on a 4 kg block on a smooth floor, pulled by a horizontal force on the lower block; friction drives the upper block.","altZh":"光滑地面上 1 kg 物块叠放在 4 kg 物块上，对下层施加水平力；摩擦力带动上层。"}],"unitId":"forces-newton","topicId":"common-forces","difficulty":5,"stem":"$A 1\\ \\mathrm{kg}$ block rests on top of $a 4\\ \\mathrm{kg}$ block on a smooth floor. A horizontal force $F = 10\\ \\mathrm{N}$ pulls the lower block and the two move together. What friction acts on the upper block?","stemZh":"质量 $1\\ \\mathrm{kg}$ 的物块叠放在 $4\\ \\mathrm{kg}$ 物块上，地面光滑。用 $F=10\\ \\mathrm{N}$ 的水平力拉下面物块，两者一起运动。上面物块受到的摩擦力多大？","options":[{"id":"A","text":"$2\\ \\mathrm{N}$"},{"id":"B","text":"$8\\ \\mathrm{N}$"},{"id":"C","text":"$10\\ \\mathrm{N}$"},{"id":"D","text":"$0\\ \\mathrm{N}$"}],"answer":"A","explanation":"Treat the pair as one $body: a = \\dfrac{F}{m+M} = \\dfrac{10}{5} = 2\\ \\mathrm{m/s^{2}}$. The only horizontal force on the upper block is friction, so $f =$ ma $= 1 \\times 2 = 2\\ \\mathrm{N}$.","explanationZh":"$整体法：a=\\dfrac{F}{m+M}=\\dfrac{10}{5}=2\\ \\mathrm{m/s^{2}}。上面物块水平方向只受摩擦力，f=ma=1\\times2=2\\ \\mathrm{N}。$","tags":["trap","forces","friction","connected","whole-body"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-common-forces-015","images":[{"src":"/physics/diagrams/incline-rough.svg","alt":"Block on a rough 30-degree incline; weight mg, the normal reaction N, and kinetic friction up the slope.","altZh":"粗糙 30° 斜面上的物块；受力为重力 mg、支持力 N 与沿斜面向上的滑动摩擦力。"}],"unitId":"forces-newton","topicId":"common-forces","difficulty":5,"stem":"$A 2\\ \\mathrm{kg}$ block is released on an incline of $\\theta = 30^{\\circ}$ with $\\mu = 0.2$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what friction acts on it?","stemZh":"质量 $2\\ \\mathrm{kg}$ 的物块放在倾角 $\\theta=30^{\\circ}、\\mu=0.2$ 的斜面上由静止释放，取 $g=10\\ \\mathrm{m/s^{2}}，它受到的摩擦力多大？$","options":[{"id":"A","text":"About $3.5\\ \\mathrm{N}$ up the slope"},{"id":"B","text":"$10\\ \\mathrm{N}$ down the slope"},{"id":"C","text":"About $17.3\\ \\mathrm{N}$ perpendicular to the slope"},{"id":"D","text":"$0\\ \\mathrm{N}$"}],"optionsZh":[{"id":"A","text":"$约 3.5\\ \\mathrm{N}，沿斜面向上$"},{"id":"B","text":"$10\\ \\mathrm{N}，沿斜面向下$"},{"id":"C","text":"$约 17.3\\ \\mathrm{N}，垂直斜面$"},{"id":"D","text":"$0\\ \\mathrm{N}$"}],"answer":"A","explanation":"Down-slope component $mg\\sin\\theta = 10\\ \\mathrm{N}$ exceeds the static limit $\\mu mg\\cos\\theta = 0.2 \\times 17.3 \\approx 3.5\\ \\mathrm{N}$, so it slides. Kinetic friction is $\\mu N \\approx 3.5\\ \\mathrm{N}$, directed up the slope opposing motion.","explanationZh":"沿斜面向下的分力 $mg\\sin\\theta=10\\ \\mathrm{N}$ 大于最大静摩擦力 $\\mu mg\\cos\\theta=0.2\\times17.3\\approx3.5\\ \\mathrm{N}，故物块下滑。滑动摩擦力 \\mu N\\approx3.5\\ \\mathrm{N}，方向沿斜面向上阻碍运动。$","tags":["trap","forces","friction","incline"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-newton-incline-001","unitId":"forces-newton","topicId":"newton-incline","difficulty":1,"stem":"A horizontal force of $6\\ \\mathrm{N}$ pulls $a 2\\ \\mathrm{kg}$ block across a smooth horizontal floor. What is its acceleration?","stemZh":"$用 6\\ \\mathrm{N}$ 的水平力拉光滑水平面上质量 $2\\ \\mathrm{kg}$ 的物块，加速度多大？","options":[{"id":"A","text":"$3\\ \\mathrm{m/s^{2}}$"},{"id":"B","text":"$12\\ \\mathrm{m/s^{2}}$"},{"id":"C","text":"$0.33\\ \\mathrm{m/s^{2}}$"},{"id":"D","text":"$6\\ \\mathrm{m/s^{2}}$"}],"answer":"A","explanation":"$a = \\dfrac{F}{m} = \\dfrac{6}{2} = 3\\ \\mathrm{m/s^{2}}$.","explanationZh":"$a=\\dfrac{F}{m}=\\dfrac{6}{2}=3\\ \\mathrm{m/s^{2}}。$","tags":["basic","newton","second-law"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-newton-incline-002","unitId":"forces-newton","topicId":"newton-incline","difficulty":1,"stem":"$A 5\\ \\mathrm{kg}$ object on a smooth horizontal floor accelerates at $2\\ \\mathrm{m/s^{2}}$. What is the net horizontal force on it?","stemZh":"质量 $5\\ \\mathrm{kg}$ 的物体在光滑水平面上以 $2\\ \\mathrm{m/s^{2}}$ 加速，所受水平合力多大？","options":[{"id":"A","text":"$2.5\\ \\mathrm{N}$"},{"id":"B","text":"$10\\ \\mathrm{N}$"},{"id":"C","text":"$20\\ \\mathrm{N}$"},{"id":"D","text":"$7\\ \\mathrm{N}$"}],"answer":"B","explanation":"$F_{net} =$ ma $= 5 \\times 2 = 10\\ \\mathrm{N}$.","explanationZh":"$F_{\\text{合}}=ma=5\\times2=10\\ \\mathrm{N}。$","tags":["basic","newton","second-law"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-newton-incline-003","unitId":"forces-newton","topicId":"newton-incline","difficulty":1,"stem":"A lift accelerates upward at $2\\ \\mathrm{m/s^{2}}$. For $a 50\\ \\mathrm{kg}$ person inside, taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the normal force from the floor?","stemZh":"电梯以 $2\\ \\mathrm{m/s^{2}}$ 竖直向上加速，其中质量 $50\\ \\mathrm{kg}$ 的人，取 $g=10\\ \\mathrm{m/s^{2}}，地板对他的支持力多大？$","options":[{"id":"A","text":"$500\\ \\mathrm{N}$"},{"id":"B","text":"$400\\ \\mathrm{N}$"},{"id":"C","text":"$600\\ \\mathrm{N}$"},{"id":"D","text":"$100\\ \\mathrm{N}$"}],"answer":"C","explanation":"Taking upward as positive: N - mg $= ma$, so $N = m(g + a) = 50 \\times 12 = 600\\ \\mathrm{N}$. This is the 'overweight' (超重) case.","explanationZh":"$取向上为正：N-mg=ma，故 N=m(g+a)=50\\times12=600\\ \\mathrm{N}。这是超重情形。$","tags":["basic","newton","lift","overweight","fromExam"],"source":{"exam":"CSCA Physics","year":"2026-01","note":"Recalled item uses the same lift-acceleration numbers."}},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-newton-incline-004","images":[{"src":"/physics/diagrams/incline-smooth-30.svg","alt":"Block released on a smooth 30-degree incline; only weight and the normal reaction act.","altZh":"光滑 30° 斜面上由静止释放的物块；仅受重力与支持力。"}],"unitId":"forces-newton","topicId":"newton-incline","difficulty":2,"stem":"A block is released from rest on a smooth incline of angle $\\theta = 30^{\\circ}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its acceleration?","stemZh":"物块由静止在倾角 $\\theta=30^{\\circ}$ 的光滑斜面上释放，取 $g=10\\ \\mathrm{m/s^{2}}，加速度多大？$","options":[{"id":"A","text":"$10\\ \\mathrm{m/s^{2}}$"},{"id":"B","text":"$5\\ \\mathrm{m/s^{2}}$"},{"id":"C","text":"$8.66\\ \\mathrm{m/s^{2}}$"},{"id":"D","text":"$2.5\\ \\mathrm{m/s^{2}}$"}],"answer":"B","explanation":"Along the slope the only force component is $mg\\sin\\theta$, so $a = g\\sin\\theta = 10 \\times 0.5 = 5\\ \\mathrm{m/s^{2}}$.","explanationZh":"沿斜面方向只有 $mg\\sin\\theta，故 a=g\\sin\\theta=10\\times0.5=5\\ \\mathrm{m/s^{2}}。$","tags":["basic","newton","incline","fromExam"],"source":{"exam":"CSCA Physics","year":"2025-12","note":"Smooth 30-degree incline is a recalled-exam staple."}},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-newton-incline-005","unitId":"forces-newton","topicId":"newton-incline","difficulty":2,"stem":"A lift descends with a downward acceleration of $2\\ \\mathrm{m/s^{2}}$. For $a 50\\ \\mathrm{kg} person$, taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the normal force?","stemZh":"电梯以 $2\\ \\mathrm{m/s^{2}}$ 的加速度竖直下降，质量 $50\\ \\mathrm{kg}$ 的人，取 $g=10\\ \\mathrm{m/s^{2}}，支持力多大？$","options":[{"id":"A","text":"$600\\ \\mathrm{N}$"},{"id":"B","text":"$500\\ \\mathrm{N}$"},{"id":"C","text":"$400\\ \\mathrm{N}$"},{"id":"D","text":"$300\\ \\mathrm{N}$"}],"answer":"C","explanation":"Taking downward as positive: mg $- N = ma$, so $N = m(g - a) = 50 \\times 8 = 400\\ \\mathrm{N}$. This is the 'underweight' (失重) case.","explanationZh":"$取向下为正：mg-N=ma，故 N=m(g-a)=50\\times8=400\\ \\mathrm{N}。这是失重情形。$","tags":["basic","newton","lift","underweight"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-newton-incline-006","unitId":"forces-newton","topicId":"newton-incline","difficulty":3,"stem":"$A 5\\ \\mathrm{kg}$ block slides from rest down a smooth $30^{\\circ}$ incline. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its displacement in $2\\ \\mathrm{s}$?","stemZh":"质量 $5\\ \\mathrm{kg}$ 的物块沿倾角 $30^{\\circ}$ 的光滑斜面由静止下滑，取 $g=10\\ \\mathrm{m/s^{2}}，2\\ \\mathrm{s}$ 内位移多大？","options":[{"id":"A","text":"$5\\ \\mathrm{m}$"},{"id":"B","text":"$20\\ \\mathrm{m}$"},{"id":"C","text":"$15\\ \\mathrm{m}$"},{"id":"D","text":"$10\\ \\mathrm{m}$"}],"answer":"D","explanation":"First $a = g\\sin 30^{\\circ} = 5\\ \\mathrm{m/s^{2}}$; then $s = \\dfrac{1}{2}at^{2} = \\dfrac{1}{2} \\times 5 \\times 4 = 10\\ \\mathrm{m}$.","explanationZh":"先求 $a=g\\sin30^{\\circ}=5\\ \\mathrm{m/s^{2}}，再由 s=\\dfrac{1}{2}at^{2}=\\dfrac{1}{2}\\times5\\times4=10\\ \\mathrm{m}。$","tags":["applied","newton","incline","fromExam"],"source":{"exam":"CSCA Physics","year":"2025-12","note":"Same two-step structure as the recalled incline item."}},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-newton-incline-007","unitId":"forces-newton","topicId":"newton-incline","difficulty":3,"stem":"$A 4\\ \\mathrm{kg}$ block on a horizontal $floor, \\mu = 0.2$, is pulled horizontally with $F = 20\\ \\mathrm{N}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its acceleration?","stemZh":"质量 $4\\ \\mathrm{kg} 的物块在水平面上，\\mu=0.2，用 F=20\\ \\mathrm{N}$ 水平拉，取 $g=10\\ \\mathrm{m/s^{2}}，加速度多大？$","options":[{"id":"A","text":"$5\\ \\mathrm{m/s^{2}}$"},{"id":"B","text":"$3\\ \\mathrm{m/s^{2}}$"},{"id":"C","text":"$2\\ \\mathrm{m/s^{2}}$"},{"id":"D","text":"$4\\ \\mathrm{m/s^{2}}$"}],"answer":"B","explanation":"Friction $f = \\mu$ mg $= 0.2 \\times 4 \\times 10 = 8\\ \\mathrm{N}$. Net force $20 - 8 = 12\\ \\mathrm{N}$, so $a = \\dfrac{12}{4} = 3\\ \\mathrm{m/s^{2}}$.","explanationZh":"摩擦力 $f=\\mu mg=0.2\\times4\\times10=8\\ \\mathrm{N}，合力 20-8=12\\ \\mathrm{N}，a=\\dfrac{12}{4}=3\\ \\mathrm{m/s^{2}}。$","tags":["applied","newton","friction","horizontal"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-newton-incline-008","images":[{"src":"/physics/diagrams/incline-rough.svg","alt":"Block sliding down a rough 30-degree incline; friction acts up the slope.","altZh":"沿粗糙 30° 斜面下滑的物块；摩擦力沿斜面向上。"}],"unitId":"forces-newton","topicId":"newton-incline","difficulty":3,"stem":"A block slides down a rough incline of $\\theta = 30^{\\circ}$ with $\\mu = 0.2$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its acceleration?","stemZh":"物块沿倾角 $\\theta=30^{\\circ}、\\mu=0.2$ 的粗糙斜面下滑，取 $g=10\\ \\mathrm{m/s^{2}}，加速度多大？$","options":[{"id":"A","text":"$5\\ \\mathrm{m/s^{2}}$"},{"id":"B","text":"$4.3\\ \\mathrm{m/s^{2}}$"},{"id":"C","text":"$3.3\\ \\mathrm{m/s^{2}}$"},{"id":"D","text":"$1.7\\ \\mathrm{m/s^{2}}$"}],"answer":"C","explanation":"$a = g(\\sin\\theta - \\mu\\cos\\theta) = 10(0.5 - 0.2 \\times 0.866) = 10 \\times 0.327 \\approx 3.3\\ \\mathrm{m/s^{2}}$.","explanationZh":"$a=g(\\sin\\theta-\\mu\\cos\\theta)=10(0.5-0.2\\times0.866)=10\\times0.327\\approx3.3\\ \\mathrm{m/s^{2}}。$","tags":["applied","newton","incline","friction"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-newton-incline-009","images":[{"src":"/physics/diagrams/connected-floor.svg","alt":"Two blocks of 2 kg and 4 kg joined by a string on a smooth floor, pulled by a force on the 4 kg block.","altZh":"光滑地面上 2 kg 与 4 kg 两物块用绳连接，对 4 kg 物块施加拉力。"}],"unitId":"forces-newton","topicId":"newton-incline","difficulty":3,"stem":"Two blocks of mass $2\\ \\mathrm{kg}$ and $4\\ \\mathrm{kg}$ are connected by a light string on a smooth horizontal floor. A force $F = 12\\ \\mathrm{N}$ pulls the $4\\ \\mathrm{kg}$ block. What is the acceleration of the system?","stemZh":"质量 $2\\ \\mathrm{kg} 和 4\\ \\mathrm{kg}$ 的两物块用轻绳连接放在光滑水平面上，用 $F=12\\ \\mathrm{N} 拉 4\\ \\mathrm{kg}$ 物块，系统加速度多大？","options":[{"id":"A","text":"$6\\ \\mathrm{m/s^{2}}$"},{"id":"B","text":"$3\\ \\mathrm{m/s^{2}}$"},{"id":"C","text":"$4\\ \\mathrm{m/s^{2}}$"},{"id":"D","text":"$2\\ \\mathrm{m/s^{2}}$"}],"answer":"D","explanation":"Treat the two blocks as one $body: a = \\dfrac{F}{m_1 + m_2} = \\dfrac{12}{6} = 2\\ \\mathrm{m/s^{2}}$.","explanationZh":"$整体法：a=\\dfrac{F}{m_1+m_2}=\\dfrac{12}{6}=2\\ \\mathrm{m/s^{2}}。$","tags":["applied","newton","connected","whole-body"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-newton-incline-010","images":[{"src":"/physics/diagrams/connected-floor.svg","alt":"Two blocks of 2 kg and 4 kg joined by a string on a smooth floor; the string tension is found by isolating one block.","altZh":"光滑地面上 2 kg 与 4 kg 两物块用绳连接；隔离一物块求绳的张力。"}],"unitId":"forces-newton","topicId":"newton-incline","difficulty":3,"stem":"Two blocks of mass $2\\ \\mathrm{kg}$ and $4\\ \\mathrm{kg}$ are joined by a light string on a smooth floor, and a force pulls the $4\\ \\mathrm{kg}$ block so that the pair accelerates at $2\\ \\mathrm{m/s^{2}}$. What is the string tension?","stemZh":"质量 $2\\ \\mathrm{kg} 和 4\\ \\mathrm{kg}$ 的两物块用轻绳连接放在光滑水平面上，拉 $4\\ \\mathrm{kg}$ 物块使系统以 $2\\ \\mathrm{m/s^{2}}$ 加速，绳的张力多大？","options":[{"id":"A","text":"$4\\ \\mathrm{N}$"},{"id":"B","text":"$8\\ \\mathrm{N}$"},{"id":"C","text":"$12\\ \\mathrm{N}$"},{"id":"D","text":"$6\\ \\mathrm{N}$"}],"answer":"A","explanation":"Isolate the $2\\ \\mathrm{kg} block$: its only horizontal force is the tension, so $T = m_1 a = 2 \\times 2 = 4\\ \\mathrm{N}$.","explanationZh":"隔离 $2\\ \\mathrm{kg} 物块：水平方向只受绳的张力，T=m_1a=2\\times2=4\\ \\mathrm{N}。$","tags":["applied","newton","connected","isolation"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-newton-incline-011","unitId":"forces-newton","topicId":"newton-incline","difficulty":4,"stem":"A block is held at rest on a smooth $30^{\\circ}$ incline by a light string parallel to the slope. The string is cut. What is the block's acceleration immediately afterwards?","stemZh":"物块被平行于斜面的轻绳拉住静止在倾角 $30^{\\circ}$ 的光滑斜面上，剪断轻绳后瞬间，物块的加速度多大？","options":[{"id":"A","text":"0"},{"id":"B","text":"$10\\ \\mathrm{m/s^{2}}$ vertically downward"},{"id":"C","text":"$5\\ \\mathrm{m/s^{2}}$ down the slope"},{"id":"D","text":"$8.66\\ \\mathrm{m/s^{2}}$ down the slope"}],"optionsZh":[{"id":"A","text":"0"},{"id":"B","text":"$10\\ \\mathrm{m/s^{2}}，竖直向下$"},{"id":"C","text":"$5\\ \\mathrm{m/s^{2}}，沿斜面向下$"},{"id":"D","text":"$8.66\\ \\mathrm{m/s^{2}}，沿斜面向下$"}],"answer":"C","explanation":"The string tension vanishes instantly, but gravity and the normal force are unchanged. Along the slope only $mg\\sin\\theta remains$, so $a = g\\sin 30^{\\circ} = 5\\ \\mathrm{m/s^{2}}$ down the slope — not g.","explanationZh":"绳的张力瞬间消失，重力和支持力不变。沿斜面只剩 $mg\\sin\\theta，故 a=g\\sin30^{\\circ}=5\\ \\mathrm{m/s^{2}}$ 沿斜面向下，而不是 g。","tags":["trap","newton","incline","instantaneous"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-newton-incline-012","images":[{"src":"/physics/diagrams/atwood.svg","alt":"Atwood machine with 3 kg and 5 kg masses over a pulley; the heavier 5 kg mass descends.","altZh":"阿特伍德机：3 kg 与 5 kg 两物体跨滑轮悬挂，较重的 5 kg 物体下降。"}],"unitId":"forces-newton","topicId":"newton-incline","difficulty":4,"stem":"In an Atwood machine, masses $3\\ \\mathrm{kg}$ and $5\\ \\mathrm{kg}$ hang over a light frictionless pulley. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the acceleration?","stemZh":"$阿特伍德机中，3\\ \\mathrm{kg} 和 5\\ \\mathrm{kg}$ 的两物体跨过轻质光滑滑轮悬挂，取 $g=10\\ \\mathrm{m/s^{2}}，加速度多大？$","options":[{"id":"A","text":"$2.5\\ \\mathrm{m/s^{2}}$"},{"id":"B","text":"$5\\ \\mathrm{m/s^{2}}$"},{"id":"C","text":"$10\\ \\mathrm{m/s^{2}}$"},{"id":"D","text":"$4\\ \\mathrm{m/s^{2}}$"}],"answer":"A","explanation":"For the $system: a = \\dfrac{(m_2 - m_1)g}{m_1 + m_2} = \\dfrac{2 \\times 10}{8} = 2.5\\ \\mathrm{m/s^{2}}$. The heavier mass descends.","explanationZh":"$对系统：a=\\dfrac{(m_2-m_1)g}{m_1+m_2}=\\dfrac{2\\times10}{8}=2.5\\ \\mathrm{m/s^{2}}，重物下降。$","tags":["trap","newton","pulley","atwood"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-newton-incline-013","unitId":"forces-newton","topicId":"newton-incline","difficulty":4,"stem":"$A 60\\ \\mathrm{kg}$ person stands on a scale in a lift. The scale reads $720\\ \\mathrm{N}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the lift's acceleration?","stemZh":"质量 $60\\ \\mathrm{kg}$ 的人站在电梯中的体重计上，示数为 $720\\ \\mathrm{N}，取 g=10\\ \\mathrm{m/s^{2}}，电梯加速度如何？$","options":[{"id":"A","text":"$2\\ \\mathrm{m/s^{2}}$ upward"},{"id":"B","text":"$2\\ \\mathrm{m/s^{2}}$ downward"},{"id":"C","text":"$12\\ \\mathrm{m/s^{2}}$ upward"},{"id":"D","text":"0 — moving at constant speed"}],"optionsZh":[{"id":"A","text":"$2\\ \\mathrm{m/s^{2}}，向上$"},{"id":"B","text":"$2\\ \\mathrm{m/s^{2}}，向下$"},{"id":"C","text":"$12\\ \\mathrm{m/s^{2}}，向上$"},{"id":"D","text":"0，匀速运动"}],"answer":"A","explanation":"The reading is the normal $force: N = m(g + a)$, so $a = \\dfrac{N}{m} - g = \\dfrac{720}{60} - 10 = 2\\ \\mathrm{m/s^{2}}$, directed upward since $N >$ mg.","explanationZh":"$示数即支持力：N=m(g+a)，故 a=\\dfrac{N}{m}-g=\\dfrac{720}{60}-10=2\\ \\mathrm{m/s^{2}}；因 N>mg，方向向上。$","tags":["trap","newton","lift","scale-reading"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-newton-incline-014","images":[{"src":"/physics/diagrams/pulley-table.svg","alt":"A 4 kg block on a table connected over a pulley to a hanging 2 kg mass on a smooth setup.","altZh":"桌面上 4 kg 物块经滑轮与悬挂的 2 kg 物块相连（光滑情形）。"}],"unitId":"forces-newton","topicId":"newton-incline","difficulty":5,"stem":"$A 4\\ \\mathrm{kg}$ block on a smooth horizontal table is connected by a light string over a pulley to a hanging $2\\ \\mathrm{kg}$ mass. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the acceleration?","stemZh":"光滑水平桌面上质量 $4\\ \\mathrm{kg}$ 的物块，通过轻绳跨过滑轮与悬挂的 $2\\ \\mathrm{kg}$ 物体相连，取 $g=10\\ \\mathrm{m/s^{2}}，加速度多大？$","options":[{"id":"A","text":"$3.33\\ \\mathrm{m/s^{2}}$"},{"id":"B","text":"$5\\ \\mathrm{m/s^{2}}$"},{"id":"C","text":"$6.67\\ \\mathrm{m/s^{2}}$"},{"id":"D","text":"$2\\ \\mathrm{m/s^{2}}$"}],"answer":"A","explanation":"Treat the two as one system: the only external driving force along the motion is $m_2 g$, so $a = \\dfrac{m_2 g}{m_1 + m_2} = \\dfrac{20}{6} \\approx 3.33\\ \\mathrm{m/s^{2}}$.","explanationZh":"整体法：沿运动方向的外力只有 $m_2g，a=\\dfrac{m_2g}{m_1+m_2}=\\dfrac{20}{6}\\approx3.33\\ \\mathrm{m/s^{2}}。$","tags":["trap","newton","pulley","connected"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-newton-incline-015","images":[{"src":"/physics/diagrams/incline-rough.svg","alt":"Block on a rough 30-degree incline with mu $= 0.5$; first decide whether it slides, then find acceleration.","altZh":"μ=0.5 的粗糙 30° 斜面上的物块；先判断是否下滑，再求加速度。"}],"unitId":"forces-newton","topicId":"newton-incline","difficulty":5,"stem":"$A 2\\ \\mathrm{kg}$ block is released on an incline of $\\theta = 30^{\\circ}$ with $\\mu = 0.5$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what happens?","stemZh":"质量 $2\\ \\mathrm{kg}$ 的物块在倾角 $\\theta=30^{\\circ}、\\mu=0.5$ 的斜面上由静止释放，取 $g=10\\ \\mathrm{m/s^{2}}，结果如何？$","options":[{"id":"A","text":"It stays at rest"},{"id":"B","text":"It accelerates down at $5.0\\ \\mathrm{m/s^{2}}$"},{"id":"C","text":"It accelerates down at about $0.67\\ \\mathrm{m/s^{2}}$"},{"id":"D","text":"It accelerates down at $3.3\\ \\mathrm{m/s^{2}}$"}],"optionsZh":[{"id":"A","text":"保持静止"},{"id":"B","text":"$以 5.0\\ \\mathrm{m/s^{2}}$ 沿斜面下加速"},{"id":"C","text":"以约 $0.67\\ \\mathrm{m/s^{2}}$ 沿斜面下加速"},{"id":"D","text":"$以 3.3\\ \\mathrm{m/s^{2}}$ 沿斜面下加速"}],"answer":"C","explanation":"Compare first: down-slope component $mg\\sin\\theta = 10\\ \\mathrm{N}$; maximum static friction $\\mu mg\\cos\\theta = 0.5 \\times 17.3 \\approx 8.66\\ \\mathrm{N}$. Since $10 > 8.66$ it slides, and then $a = g(\\sin\\theta - \\mu\\cos\\theta) = 10(0.5 - 0.433) \\approx 0.67\\ \\mathrm{m/s^{2}} —$ much less than the smooth-incline value.","explanationZh":"先判断：沿斜面向下分力 $mg\\sin\\theta=10\\ \\mathrm{N}；最大静摩擦力 \\mu mg\\cos\\theta=0.5\\times17.3\\approx8.66\\ \\mathrm{N}。因 10>8.66 故下滑，a=g(\\sin\\theta-\\mu\\cos\\theta)=10(0.5-0.433)\\approx0.67\\ \\mathrm{m/s^{2}}，远小于光滑斜面的值。$","tags":["trap","newton","incline","friction","decide-then-solve"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-charge-coulomb-001","unitId":"electrostatics","topicId":"charge-coulomb","difficulty":1,"stem":"A small object carries a net charge of $-3.2\\times10^{-19}\\ \\mathrm{C}$. Taking the elementary charge as $e = 1.6\\times10^{-19}\\ \\mathrm{C}$, how many excess electrons does it carry?","stemZh":"某物体带净电荷 $-3.2\\times10^{-19}\\ \\mathrm{C}，取元电荷 e=1.6\\times10^{-19}\\ \\mathrm{C}，它多余多少个电子？$","options":[{"id":"A","text":"2"},{"id":"B","text":"3"},{"id":"C","text":"5"},{"id":"D","text":"0.5"}],"answer":"A","explanation":"Charge is $quantised: q = ne$, so $n = \\dfrac{|q|}{e} = \\dfrac{3.2\\times10^{-19}}{1.6\\times10^{-19}} = 2$. The excess of electrons makes the charge negative.","explanationZh":"电荷量子化 $q=ne，故 n=\\dfrac{|q|}{e}=\\dfrac{3.2\\times10^{-19}}{1.6\\times10^{-19}}=2；多出电子使电荷带负电。$","tags":["basic","electrostatics","charge-quantisation"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-charge-coulomb-002","unitId":"electrostatics","topicId":"charge-coulomb","difficulty":1,"stem":"Two point charges are separated by a distance d and repel each other with a force F. If the magnitude of one of the charges is doubled while the distance is unchanged, the repulsive force becomes","stemZh":"两个点电荷相距 d 时相互排斥，斥力为 F。若其中一个电荷量加倍而距离不变，则斥力变为","options":[{"id":"A","text":"$\\dfrac{F}{2}$"},{"id":"B","text":"F"},{"id":"C","text":"4F"},{"id":"D","text":"2F"}],"answer":"D","explanation":"By Coulomb's law $F = k\\dfrac{q_1q_2}{r^{2}}$, so $F \\propto q_1q_2$. Doubling one charge doubles the force, giving 2F.","explanationZh":"由库仑定律 $F=k\\dfrac{q_1q_2}{r^{2}}，F\\propto q_1q_2。一个电荷量加倍，力加倍，为 2F。$","tags":["basic","electrostatics","coulomb-scaling"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-charge-coulomb-003","unitId":"electrostatics","topicId":"charge-coulomb","difficulty":1,"stem":"Two point charges attract each other with a force F. If the distance between them is doubled and the charges are unchanged, the force becomes","stemZh":"两个点电荷之间的引力为 F。若它们之间的距离加倍而电荷量不变，则作用力变为","options":[{"id":"A","text":"2F"},{"id":"B","text":"$\\dfrac{F}{2}$"},{"id":"C","text":"$\\dfrac{F}{4}$"},{"id":"D","text":"$\\dfrac{F}{8}$"}],"answer":"C","explanation":"$F \\propto \\dfrac{1}{r^{2}}$. Doubling the distance makes the force $\\dfrac{1}{2^{2}} = \\dfrac{1}{4}$ of its original value.","explanationZh":"$F\\propto\\dfrac{1}{r^{2}}。距离加倍，力变为原来的 \\dfrac{1}{2^{2}}=\\dfrac{1}{4}。$","tags":["basic","electrostatics","coulomb-scaling"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-charge-coulomb-004","unitId":"electrostatics","topicId":"charge-coulomb","difficulty":2,"stem":"Two identical metal spheres carry charges $+4\\ \\mathrm{C}$ and $-6\\ \\mathrm{C}$. They are brought into contact and then separated. What is the charge on each sphere afterwards?","stemZh":"两个相同的金属球分别带 $+4\\ \\mathrm{C} 和 -6\\ \\mathrm{C}$ 的电荷，接触后再分开，每个球的电荷量是多少？","options":[{"id":"A","text":"$+1\\ \\mathrm{C}$"},{"id":"B","text":"$-1\\ \\mathrm{C}$"},{"id":"C","text":"$-2\\ \\mathrm{C}$"},{"id":"D","text":"$-5\\ \\mathrm{C}$"}],"answer":"B","explanation":"Charge is conserved: the total is $(+4) + (-6) = -2\\ \\mathrm{C}$. Identical spheres share the total equally, so each carries $\\dfrac{-2}{2} = -1\\ \\mathrm{C}$.","explanationZh":"电荷守恒：总量为 $(+4)+(-6)=-2\\ \\mathrm{C}。相同金属球平分总电荷，每个带 \\dfrac{-2}{2}=-1\\ \\mathrm{C}。$","tags":["basic","electrostatics","charge-sharing"],"images":[{"src":"/physics/diagrams/coulomb-spheres-contact.svg","alt":"Two identical metal spheres carrying +4 C and -6 C are brought into contact, after which each carries -1 C.","altZh":"两个相同金属球分别带 +4 C 与 −6 C，接触后各带 −1 C。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-charge-coulomb-005","unitId":"electrostatics","topicId":"charge-coulomb","difficulty":2,"stem":"Two point charges of $+2.0\\times10^{-6}\\ \\mathrm{C}$ each are placed $0.30\\ \\mathrm{m}$ apart in vacuum. What is the magnitude of the electric force between them? $(k = 9.0\\times10^{9}\\ \\mathrm{N\\cdot m^{2}/C^{2}})$","stemZh":"真空中两个各带 $+2.0\\times10^{-6}\\ \\mathrm{C}$ 的点电荷相距 $0.30\\ \\mathrm{m}，它们之间的静电力大小为多少？（k=9.0\\times10^{9}\\ \\mathrm{N\\cdot m^{2}/C^{2}}）$","options":[{"id":"A","text":"$0.20\\ \\mathrm{N}$"},{"id":"B","text":"$0.40\\ \\mathrm{N}$"},{"id":"C","text":"$0.90\\ \\mathrm{N}$"},{"id":"D","text":"$1.2\\ \\mathrm{N}$"}],"answer":"B","explanation":"$F = k\\dfrac{q_1q_2}{r^{2}} = 9.0\\times10^{9}\\times\\dfrac{(2.0\\times10^{-6})^{2}}{0.30^{2}} = 0.40\\ \\mathrm{N}$. Both charges are positive, so the force is repulsive.","explanationZh":"$F=k\\dfrac{q_1q_2}{r^{2}}=9.0\\times10^{9}\\times\\dfrac{(2.0\\times10^{-6})^{2}}{0.30^{2}}=0.40\\ \\mathrm{N}。两电荷同号，为斥力。$","tags":["basic","electrostatics","coulomb-calculation"],"images":[{"src":"/physics/diagrams/coulomb-two-charges.svg","alt":"Two positive point charges q1 and q2 separated by a distance r, repelling each other with force F.","altZh":"两个正点电荷 q1、q2 相距 r，相互排斥，斥力为 F。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-charge-coulomb-006","unitId":"electrostatics","topicId":"charge-coulomb","difficulty":3,"stem":"Two point charges $+1.0\\times10^{-6}\\ \\mathrm{C}$ and $+3.0\\times10^{-6}\\ \\mathrm{C}$ are $0.30\\ \\mathrm{m}$ apart in vacuum. What is the magnitude of the force between them? $(k = 9.0\\times10^{9}\\ \\mathrm{N\\cdot m^{2}/C^{2}})$","stemZh":"真空中两个点电荷 $+1.0\\times10^{-6}\\ \\mathrm{C} 与 +3.0\\times10^{-6}\\ \\mathrm{C}$ 相距 $0.30\\ \\mathrm{m}，它们之间的作用力大小为多少？（k=9.0\\times10^{9}\\ \\mathrm{N\\cdot m^{2}/C^{2}}）$","options":[{"id":"A","text":"$0.10\\ \\mathrm{N}$"},{"id":"B","text":"$0.90\\ \\mathrm{N}$"},{"id":"C","text":"$0.30\\ \\mathrm{N}$"},{"id":"D","text":"$3.0\\ \\mathrm{N}$"}],"answer":"C","explanation":"$F = k\\dfrac{q_1q_2}{r^{2}} = 9.0\\times10^{9}\\times\\dfrac{1.0\\times10^{-6}\\times3.0\\times10^{-6}}{0.30^{2}} = 0.30\\ \\mathrm{N}$.","explanationZh":"$F=k\\dfrac{q_1q_2}{r^{2}}=9.0\\times10^{9}\\times\\dfrac{1.0\\times10^{-6}\\times3.0\\times10^{-6}}{0.30^{2}}=0.30\\ \\mathrm{N}。$","tags":["applied","electrostatics","coulomb-calculation"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-charge-coulomb-007","unitId":"electrostatics","topicId":"charge-coulomb","difficulty":3,"stem":"Two point charges exert a force F on each other when their separation is r. If one charge is doubled and the separation is halved, the new force is","stemZh":"两个点电荷相距 r 时相互作用力为 F。若一个电荷量加倍、距离减半，则新的作用力为","options":[{"id":"A","text":"F"},{"id":"B","text":"2F"},{"id":"C","text":"4F"},{"id":"D","text":"8F"}],"answer":"D","explanation":"$F \\propto \\dfrac{q_1q_2}{r^{2}}$. Doubling one charge multiplies the force by 2; halving r multiplies it by 4. The total factor is $2\\times4 = 8$, so the force is 8F.","explanationZh":"$F\\propto\\dfrac{q_1q_2}{r^{2}}。一个电荷量加倍使力变 2$ 倍，距离减半使力变 4 倍，共 $2\\times4=8$ 倍，即 8F。","tags":["applied","electrostatics","coulomb-scaling"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-charge-coulomb-008","unitId":"electrostatics","topicId":"charge-coulomb","difficulty":3,"stem":"Two positive point charges +4q and +q are fixed on a straight $line, 0.60\\ \\mathrm{m}$ apart. A third charge is placed on the line between them. At what distance from +4q is the net electric force on the third charge zero?","stemZh":"两个正点电荷 +4q 与 +q 固定在一条直线上，相距 $0.60\\ \\mathrm{m}。在二者连线上放第三个电荷，它距 +4q$ 多远时受到的合力为零？","options":[{"id":"A","text":"$0.20\\ \\mathrm{m}$"},{"id":"B","text":"$0.30\\ \\mathrm{m}$"},{"id":"C","text":"$0.40\\ \\mathrm{m}$"},{"id":"D","text":"$0.48\\ \\mathrm{m}$"}],"answer":"C","explanation":"Let the point be x from +4q, so it is $(0.60 - x)$ from +q. Setting the two forces $equal: \\dfrac{4q}{x^{2}} = \\dfrac{q}{(0.60-x)^{2}}$, hence $\\dfrac{2}{x} = \\dfrac{1}{0.60-x}$ and $x = 0.40\\ \\mathrm{m}$. The balance point is always nearer the smaller charge.","explanationZh":"设该点距 +4q 为 x，则距 $+q 为 (0.60-x)。两力相等：\\dfrac{4q}{x^{2}}=\\dfrac{q}{(0.60-x)^{2}}，得 \\dfrac{2}{x}=\\dfrac{1}{0.60-x}，x=0.40\\ \\mathrm{m}。平衡点总是靠近电荷量较小的一方。$","tags":["applied","electrostatics","zero-force-point"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-charge-coulomb-009","unitId":"electrostatics","topicId":"charge-coulomb","difficulty":3,"stem":"A point charge at P experiences a force of $3.0\\ \\mathrm{N}$ directed along the +x axis because of charge $Q_1$, and a force of $4.0\\ \\mathrm{N}$ directed along the +y axis because of charge $Q_2$. What is the magnitude of the resultant force on the charge at P?","stemZh":"位于 P 的点电荷受电荷 $Q_1$ 的作用力为 $3.0\\ \\mathrm{N}，方向沿 +x$ 轴；受电荷 $Q_2$ 的作用力为 $4.0\\ \\mathrm{N}，方向沿 +y$ 轴。P 处电荷所受合力大小为多少？","options":[{"id":"A","text":"$5.0\\ \\mathrm{N}$"},{"id":"B","text":"$1.0\\ \\mathrm{N}$"},{"id":"C","text":"$7.0\\ \\mathrm{N}$"},{"id":"D","text":"$25\\ \\mathrm{N}$"}],"answer":"A","explanation":"The two forces are perpendicular, so they combine by $Pythagoras: F = \\sqrt{3.0^{2} + 4.0^{2}} = 5.0\\ \\mathrm{N}$. Simply adding them to get $7.0\\ \\mathrm{N}$ ignores their directions.","explanationZh":"$两力互相垂直，用勾股定理合成：F=\\sqrt{3.0^{2}+4.0^{2}}=5.0\\ \\mathrm{N}。直接相加得 7.0\\ \\mathrm{N}$ 忽略了方向。","tags":["applied","electrostatics","vector-addition"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-charge-coulomb-010","unitId":"electrostatics","topicId":"charge-coulomb","difficulty":3,"stem":"Two identical conducting spheres carry charges $+6.0\\times10^{-6}\\ \\mathrm{C}$ and $+2.0\\times10^{-6}\\ \\mathrm{C}$ and repel each other with a force F when separated by r. They are touched together and then separated back to the same distance r. The new force between them is","stemZh":"两个相同的导体球分别带 $+6.0\\times10^{-6}\\ \\mathrm{C} 与 +2.0\\times10^{-6}\\ \\mathrm{C}，相距 r$ 时斥力为 F。接触后再分开回到距离 r，它们之间的作用力变为","options":[{"id":"A","text":"$\\dfrac{4F}{3}$"},{"id":"B","text":"$\\dfrac{F}{3}$"},{"id":"C","text":"2F"},{"id":"D","text":"3F"}],"answer":"A","explanation":"After contact each sphere carries $\\dfrac{6.0 + 2.0}{2} = 4.0\\times10^{-6}\\ \\mathrm{C}$. Since $F \\propto q_1q_2, \\dfrac{F'}{F} = \\dfrac{4\\times4}{6\\times2} = \\dfrac{4}{3}$, so the force grows to $\\dfrac{4F}{3}$.","explanationZh":"接触后每球带 $\\dfrac{6.0+2.0}{2}=4.0\\times10^{-6}\\ \\mathrm{C}。由 F\\propto q_1q_2，\\dfrac{F'}{F}=\\dfrac{4\\times4}{6\\times2}=\\dfrac{4}{3}，力变为 \\dfrac{4F}{3}。$","tags":["applied","electrostatics","charge-sharing"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-charge-coulomb-011","unitId":"electrostatics","topicId":"charge-coulomb","difficulty":5,"stem":"Three point charges A, B and C are fixed at the vertices of an equilateral triangle. A and B are both positive. The resultant electric force on A makes an angle of less than $60^{\\circ}$ with the line BA produced, as shown. Which statement about charge C is correct?","stemZh":"三个点电荷 A、B、C 固定在等边三角形的三个顶点，A、B 均带正电。A 所受静电力的合力与 BA 延长线的夹角小于 $60^{\\circ}（如图）。关于电荷 C$ 的判断正确的是","options":[{"id":"A","text":"C is positive and $|q_C| < q_B$"},{"id":"B","text":"C is negative and $|q_C| > q_B$"},{"id":"C","text":"C is positive and $|q_C| > q_B$"},{"id":"D","text":"C is negative and $|q_C| < q_B$"}],"answer":"D","explanation":"The repulsion from B alone would push A exactly along BA produced. To swing the resultant towards AC by less than $60^{\\circ}, C$ must pull A — that is, C is negative — and its force must be smaller than B's, so $|q_C| < q_B$. If $|q_C| \\geq q_B$ the resultant would be deflected by $60^{\\circ}$ or more.","explanationZh":"B 单独对 A 的斥力沿 BA 延长线方向。要使合力向 AC 一侧偏转小于 $60^{\\circ}，C$ 必须吸引 A（即带负电），且其力小于 B 的力，故 $|q_C| E_b$"},{"id":"B","text":"$E_a < E_b$"},{"id":"C","text":"$E_a = E_b$"},{"id":"D","text":"It cannot be decided from the figure."}],"answer":"A","explanation":"The field is stronger where the field lines are more closely packed. Near the charge the lines are bunched together and a is the closer point, so $E_a > E_b$.","explanationZh":"电场线越密处场强越大。靠近电荷处电场线密集，而 a 点离电荷更近，故 $E_a>E_b。$","tags":["basic","electrostatics","field-lines"],"images":[{"src":"/physics/diagrams/field-lines-point-positive.svg","alt":"Radial field lines pointing outwards from a positive point charge, with point a close to the charge and point b further away.","altZh":"正点电荷向外辐射的电场线，a 点靠近电荷，b 点较远。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-field-003","unitId":"electrostatics","topicId":"electric-field","difficulty":1,"stem":"The direction of the electric field at a point is defined as","stemZh":"电场中某点电场强度的方向规定为","options":[{"id":"A","text":"the direction of the force on a negative charge placed at the point."},{"id":"B","text":"the direction of motion of a positive charge released from rest there."},{"id":"C","text":"the direction of the force on a positive test charge placed at the point."},{"id":"D","text":"the direction in which the electric potential increases most rapidly."}],"answer":"C","explanation":"By definition $\\vec{E}$ points along the force acting on a small positive test charge. Option A is the opposite direction, and option D is reversed: the field points towards decreasing potential.","explanationZh":"规定 $\\vec{E}$ 的方向与放在该点的正试探电荷所受电场力方向相同。A 方向相反；D 也反了——电场指向电势降低的方向。","tags":["basic","electrostatics","field-direction"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-field-004","unitId":"electrostatics","topicId":"electric-field","difficulty":2,"stem":"A point charge $Q = +2.0\\times10^{-8}\\ \\mathrm{C}$ is isolated in vacuum. What is the electric field strength at a point $0.30\\ \\mathrm{m}$ from it? $(k = 9.0\\times10^{9}\\ \\mathrm{N\\cdot m^{2}/C^{2}})$","stemZh":"真空中孤立点电荷 $Q=+2.0\\times10^{-8}\\ \\mathrm{C}，距它 0.30\\ \\mathrm{m} 处的电场强度为多少？（k=9.0\\times10^{9}\\ \\mathrm{N\\cdot m^{2}/C^{2}}）$","options":[{"id":"A","text":"$6.0\\times10^{2}\\ \\mathrm{N/C}$"},{"id":"B","text":"$2.0\\times10^{3}\\ \\mathrm{N/C}$"},{"id":"C","text":"$6.0\\times10^{3}\\ \\mathrm{N/C}$"},{"id":"D","text":"$1.8\\times10^{4}\\ \\mathrm{N/C}$"}],"answer":"B","explanation":"$E = k\\dfrac{Q}{r^{2}} = 9.0\\times10^{9}\\times\\dfrac{2.0\\times10^{-8}}{0.30^{2}} = 2.0\\times10^{3}\\ \\mathrm{N/C}$, directed radially away from the charge.","explanationZh":"$E=k\\dfrac{Q}{r^{2}}=9.0\\times10^{9}\\times\\dfrac{2.0\\times10^{-8}}{0.30^{2}}=2.0\\times10^{3}\\ \\mathrm{N/C}，方向沿径向向外。$","tags":["basic","electrostatics","point-charge-field"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-field-005","unitId":"electrostatics","topicId":"electric-field","difficulty":2,"stem":"In a uniform electric field, the field lines are","stemZh":"匀强电场的电场线是","options":[{"id":"A","text":"curved and converging towards a point."},{"id":"B","text":"always circles."},{"id":"C","text":"parallel but getting farther apart."},{"id":"D","text":"parallel straight lines that are equally spaced."}],"answer":"D","explanation":"A uniform field has the same magnitude and direction everywhere, so its field lines are parallel, straight and equally spaced.","explanationZh":"匀强电场各处场强大小和方向都相同，故电场线是等间距的平行直线。","tags":["basic","electrostatics","uniform-field"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-field-006","unitId":"electrostatics","topicId":"electric-field","difficulty":3,"stem":"A point charge Q produces an electric field of strength E at a distance r from it. What is the field strength at a distance 3r from the same charge?","stemZh":"点电荷 Q 在距它 r 处产生的电场强度为 E。在距它 3r 处，电场强度为多少？","options":[{"id":"A","text":"3E"},{"id":"B","text":"$\\dfrac{E}{3}$"},{"id":"C","text":"$\\dfrac{E}{9}$"},{"id":"D","text":"$\\dfrac{E}{6}$"}],"answer":"C","explanation":"$E = k\\dfrac{Q}{r^{2}} \\propto \\dfrac{1}{r^{2}}$. Tripling the distance gives $\\dfrac{1}{3^{2}} = \\dfrac{1}{9}$ of the original field, so E' $= \\dfrac{E}{9}$.","explanationZh":"$E=k\\dfrac{Q}{r^{2}}\\propto\\dfrac{1}{r^{2}}。距离变为 3$ 倍，场强变为 $\\dfrac{1}{3^{2}}=\\dfrac{1}{9}，即 E'=\\dfrac{E}{9}。$","tags":["applied","electrostatics","point-charge-field"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-field-007","unitId":"electrostatics","topicId":"electric-field","difficulty":3,"stem":"A charge of $+3.0\\times10^{-9}\\ \\mathrm{C}$ is placed in a uniform electric field of strength $6.0\\times10^{4}\\ \\mathrm{N/C}$. What is the magnitude of the electric force on the charge?","stemZh":"把电荷量 $+3.0\\times10^{-9}\\ \\mathrm{C}$ 的电荷放入场强 $6.0\\times10^{4}\\ \\mathrm{N/C}$ 的匀强电场中，它所受电场力的大小为多少？","options":[{"id":"A","text":"$1.8\\times10^{-4}\\ \\mathrm{N}$"},{"id":"B","text":"$2.0\\times10^{-5}\\ \\mathrm{N}$"},{"id":"C","text":"$5.0\\times10^{-5}\\ \\mathrm{N}$"},{"id":"D","text":"$1.8\\times10^{-13}\\ \\mathrm{N}$"}],"answer":"A","explanation":"$F =$ qE $= 3.0\\times10^{-9}\\times6.0\\times10^{4} = 1.8\\times10^{-4}\\ \\mathrm{N}$. The force is along the field because the charge is positive.","explanationZh":"$F=qE=3.0\\times10^{-9}\\times6.0\\times10^{4}=1.8\\times10^{-4}\\ \\mathrm{N}。电荷为正，受力方向与场强方向相同。$","tags":["applied","electrostatics","force-in-field"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-field-008","unitId":"electrostatics","topicId":"electric-field","difficulty":3,"stem":"Two point charges $+2.0\\times10^{-9}\\ \\mathrm{C}$ and $-2.0\\times10^{-9}\\ \\mathrm{C}$ are $4.0\\ \\mathrm{cm}$ apart. What is the electric field strength at the midpoint of the line joining them? $(k = 9.0\\times10^{9}\\ \\mathrm{N\\cdot m^{2}/C^{2}})$","stemZh":"两个点电荷 $+2.0\\times10^{-9}\\ \\mathrm{C} 与 -2.0\\times10^{-9}\\ \\mathrm{C}$ 相距 $4.0\\ \\mathrm{cm}。连线中点处的电场强度为多少？（k=9.0\\times10^{9}\\ \\mathrm{N\\cdot m^{2}/C^{2}}）$","options":[{"id":"A","text":"0"},{"id":"B","text":"$4.5\\times10^{4}\\ \\mathrm{N/C}$"},{"id":"C","text":"$9.0\\times10^{4}\\ \\mathrm{N/C}$"},{"id":"D","text":"$1.8\\times10^{5}\\ \\mathrm{N/C}$"}],"answer":"C","explanation":"The midpoint is $0.020\\ \\mathrm{m}$ from each charge, so each produces $E = k\\dfrac{q}{r^{2}} = 9.0\\times10^{9}\\times\\dfrac{2.0\\times10^{-9}}{0.020^{2}} = 4.5\\times10^{4}\\ \\mathrm{N/C}$. Both fields point from the positive towards the negative charge — the same way — so they add to $9.0\\times10^{4}\\ \\mathrm{N/C}$.","explanationZh":"中点到两电荷均为 $0.020\\ \\mathrm{m}，各自产生 E=k\\dfrac{q}{r^{2}}=9.0\\times10^{9}\\times\\dfrac{2.0\\times10^{-9}}{0.020^{2}}=4.5\\times10^{4}\\ \\mathrm{N/C}。两场方向都由正电荷指向负电荷，方向相同，故叠加为 9.0\\times10^{4}\\ \\mathrm{N/C}。$","tags":["applied","electrostatics","superposition"],"images":[{"src":"/physics/diagrams/efield-two-charges.svg","alt":"Two point charges of equal magnitude and opposite sign 4.0 cm apart, with the electric field at the midpoint M drawn to the right.","altZh":"两个等量异号点电荷相距 4.0 cm，中点 M 处的场强方向向右。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-field-009","unitId":"electrostatics","topicId":"electric-field","difficulty":3,"stem":"Two equal positive point charges +Q are held a distance 2d apart. What is the electric field strength at the midpoint of the line joining them?","stemZh":"两个等量正点电荷 +Q 相距 2d，连线中点处的电场强度为多少？","options":[{"id":"A","text":"0"},{"id":"B","text":"$k\\dfrac{Q}{d^{2}}$"},{"id":"C","text":"$2k\\dfrac{Q}{d^{2}}$"},{"id":"D","text":"$4k\\dfrac{Q}{d^{2}}$"}],"answer":"A","explanation":"Each charge produces a field of magnitude $k\\dfrac{Q}{d^{2}}$ at the midpoint, but the two fields point in opposite directions, so they cancel and the resultant field is zero. (Note that the potential there is not zero.)","explanationZh":"两电荷在中点各自产生 $k\\dfrac{Q}{d^{2}}$ 的场强，但方向相反，相互抵消，合场强为零。（注意该点电势不为零。）","tags":["applied","electrostatics","superposition"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-field-010","unitId":"electrostatics","topicId":"electric-field","difficulty":3,"stem":"The electric field at a point $0.20\\ \\mathrm{m}$ from a small charged sphere has magnitude $4.5\\times10^{4}\\ \\mathrm{N/C}$ and points away from the sphere. What is the charge on the sphere? $(k = 9.0\\times10^{9}\\ \\mathrm{N\\cdot m^{2}/C^{2}})$","stemZh":"距一个带电小球 $0.20\\ \\mathrm{m}$ 处电场强度大小为 $4.5\\times10^{4}\\ \\mathrm{N/C}，方向背离小球。小球的电荷量为多少？（k=9.0\\times10^{9}\\ \\mathrm{N\\cdot m^{2}/C^{2}}）$","options":[{"id":"A","text":"$1.0\\times10^{-7}\\ \\mathrm{C}$"},{"id":"B","text":"$2.0\\times10^{-7}\\ \\mathrm{C}$"},{"id":"C","text":"$4.0\\times10^{-7}\\ \\mathrm{C}$"},{"id":"D","text":"$2.0\\times10^{-6}\\ \\mathrm{C}$"}],"answer":"B","explanation":"From $E = k\\dfrac{Q}{r^{2}}: Q = \\dfrac{Er^{2}}{k} = \\dfrac{4.5\\times10^{4}\\times0.20^{2}}{9.0\\times10^{9}} = 2.0\\times10^{-7}\\ \\mathrm{C}$. The field points away from the sphere, so the charge is positive.","explanationZh":"$由 E=k\\dfrac{Q}{r^{2}} 得 Q=\\dfrac{Er^{2}}{k}=\\dfrac{4.5\\times10^{4}\\times0.20^{2}}{9.0\\times10^{9}}=2.0\\times10^{-7}\\ \\mathrm{C}。场强背离小球，故带正电。$","tags":["applied","electrostatics","point-charge-field"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-field-011","unitId":"electrostatics","topicId":"electric-field","difficulty":4,"stem":"The figure shows the electric field between two parallel charged plates, with points a and b marked. Which statement is correct?","stemZh":"图为两块平行带电板之间的电场，标出了 a、b 两点。下列说法正确的是","options":[{"id":"A","text":"$E_a > E_b$"},{"id":"B","text":"$E_a < E_b$"},{"id":"C","text":"It is impossible to compare $E_a$ and $E_b$ from the figure."},{"id":"D","text":"$E_a = E_b$, because the field is uniform."}],"answer":"D","explanation":"Between parallel plates the field lines are parallel, straight and equally spaced, so the field is $uniform: E_a = E_b$.","explanationZh":"平行板间的电场线是等间距的平行直线，属匀强电场，故 $E_a=E_b。$","tags":["trap","electrostatics","uniform-field"],"images":[{"src":"/physics/diagrams/field-lines-uniform.svg","alt":"Uniform electric field between two parallel plates drawn as equally spaced parallel field lines, with points a and b.","altZh":"两平行板间的匀强电场，电场线等间距平行，标有 a、b 两点。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-field-012","unitId":"electrostatics","topicId":"electric-field","difficulty":4,"stem":"The figure shows the field lines of an electric field, with points a and b marked. Which statement is correct?","stemZh":"图为某电场的电场线分布，标出了 a、b 两点。下列说法正确的是","options":[{"id":"A","text":"$E_a > E_b$, and the field is uniform."},{"id":"B","text":"$E_a > E_b$, and the field is not uniform."},{"id":"C","text":"$E_a < E_b$"},{"id":"D","text":"$E_a = E_b$"}],"answer":"B","explanation":"The lines are packed more closely near a than near b, so $E_a > E_b$. Because the spacing of the lines changes from place to place, the field is not uniform.","explanationZh":"a 点附近电场线比 b 点密集，故 $E_a>E_b；由于电场线间距处处不同，该电场不是匀强电场。$","tags":["trap","electrostatics","field-lines"],"images":[{"src":"/physics/diagrams/field-lines-nonuniform.svg","alt":"A non-uniform electric field whose field lines are close together near point a and far apart near point b.","altZh":"非匀强电场的电场线，靠近 a 点密集、靠近 b 点稀疏。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-field-013","unitId":"electrostatics","topicId":"electric-field","difficulty":4,"stem":"Which statement about electric field lines is correct?","stemZh":"关于电场线，下列说法正确的是","options":[{"id":"A","text":"They start on positive charges and end on negative charges, and they never cross."},{"id":"B","text":"Two field lines may cross at a point where the field is zero."},{"id":"C","text":"In an electrostatic field they always form closed loops."},{"id":"D","text":"They are always straight lines."}],"answer":"A","explanation":"Field lines begin on positive charge and end on negative charge. If two lines crossed, the field at that point would have two different directions at once, which is impossible. Closed loops occur for magnetic field lines, not for electrostatic ones.","explanationZh":"电场线始于正电荷、止于负电荷。若两条电场线相交，交点处场强将有两个方向，这是不可能的。闭合的场线是磁场线的特征，不是静电场的。","tags":["trap","electrostatics","field-lines"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-field-014","unitId":"electrostatics","topicId":"electric-field","difficulty":5,"stem":"The electric field at a certain point in space is zero. Which statement must be true?","stemZh":"空间中某点的电场强度为零，则下列哪项一定成立？","options":[{"id":"A","text":"The electric potential at that point must be zero."},{"id":"B","text":"No electric field lines pass anywhere near that point."},{"id":"C","text":"The point must be equidistant from all the charges present."},{"id":"D","text":"A charge placed at that point experiences no net electric force."}],"answer":"D","explanation":"Since $\\vec{F} = q\\vec{E}$, if $E = 0$ the net electric force on any charge placed there is zero. The potential need not be zero — for example, at the centre of a uniformly charged ring $E = 0$ while the potential is non-zero.","explanationZh":"$由 \\vec{F}=q\\vec{E}，E=0$ 时放在该点的电荷所受合力为零。但电势不一定为零——例如均匀带电圆环圆心处 $E=0，电势不为零。$","tags":["trap","electrostatics","field-and-force"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-field-015","unitId":"electrostatics","topicId":"electric-field","difficulty":5,"stem":"A negative point charge is released from rest in a uniform electric field. Ignoring gravity, the charge","stemZh":"把一个负点电荷从静止释放于匀强电场中，不计重力，该电荷将","options":[{"id":"A","text":"move in the direction of the field with constant acceleration."},{"id":"B","text":"move opposite to the field with constant acceleration."},{"id":"C","text":"move in the direction of the field with steadily increasing acceleration."},{"id":"D","text":"stay at rest, because the field is uniform."}],"answer":"B","explanation":"For a negative charge $\\vec{F} = q\\vec{E}$ with $q < 0$, so the force — and therefore the acceleration — is opposite to the field. The field is uniform, so the force is constant and the acceleration is constant.","explanationZh":"负电荷满足 $\\vec{F}=q\\vec{E} 且 q<0，故受力方向与场强相反，加速度方向也与场强相反。匀强电场中受力恒定，加速度恒定。$","tags":["trap","electrostatics","field-and-force"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-potential-001","unitId":"electrostatics","topicId":"electric-potential","difficulty":1,"stem":"A charge of $+4.0\\times10^{-6}\\ \\mathrm{C}$ is moved from A to B by the electric field, which does $+8.0\\times10^{-4}\\ \\mathrm{J}$ of work. What is the potential difference $U_{AB} = V_A - V_B$?","stemZh":"电荷量 $+4.0\\times10^{-6}\\ \\mathrm{C}$ 的点电荷在电场力作用下从 A 移到 B，电场力做功 $+8.0\\times10^{-4}\\ \\mathrm{J}。求电势差 U_{AB}=V_A-V_B。$","options":[{"id":"A","text":"$3.2\\times10^{-9}\\ \\mathrm{V}$"},{"id":"B","text":"$-200\\ \\mathrm{V}$"},{"id":"C","text":"$200\\ \\mathrm{V}$"},{"id":"D","text":"$500\\ \\mathrm{V}$"}],"answer":"C","explanation":"$W_{AB} = qU_{AB}$, so $U_{AB} = \\dfrac{W}{q} = \\dfrac{8.0\\times10^{-4}}{4.0\\times10^{-6}} = 200\\ \\mathrm{V}$.","explanationZh":"$由 W_{AB}=qU_{AB}，U_{AB}=\\dfrac{W}{q}=\\dfrac{8.0\\times10^{-4}}{4.0\\times10^{-6}}=200\\ \\mathrm{V}。$","tags":["basic","electrostatics","potential-difference"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-potential-002","unitId":"electrostatics","topicId":"electric-potential","difficulty":1,"stem":"A charge $q = +2.0\\times10^{-6}\\ \\mathrm{C}$ moves between two points whose potential difference is $U = 100\\ \\mathrm{V}$. How much work does the electric field do on the charge?","stemZh":"电荷量 $q=+2.0\\times10^{-6}\\ \\mathrm{C}$ 在两电势差 $U=100\\ \\mathrm{V}$ 的点之间移动，电场力做功多少？","options":[{"id":"A","text":"$2.0\\times10^{-4}\\ \\mathrm{J}$"},{"id":"B","text":"$5.0\\times10^{-5}\\ \\mathrm{J}$"},{"id":"C","text":"$5.0\\times10^{-3}\\ \\mathrm{J}$"},{"id":"D","text":"$2.0\\times10^{-8}\\ \\mathrm{J}$"}],"answer":"A","explanation":"$W =$ qU $= 2.0\\times10^{-6}\\times100 = 2.0\\times10^{-4}\\ \\mathrm{J}$.","explanationZh":"$W=qU=2.0\\times10^{-6}\\times100=2.0\\times10^{-4}\\ \\mathrm{J}。$","tags":["basic","electrostatics","work-by-field"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-potential-003","unitId":"electrostatics","topicId":"electric-potential","difficulty":1,"stem":"Moving along an electric field line in the direction of the field, the electric potential","stemZh":"沿电场线方向移动，电势","options":[{"id":"A","text":"increases."},{"id":"B","text":"stays constant."},{"id":"C","text":"is always zero."},{"id":"D","text":"decreases."}],"answer":"D","explanation":"The electric field always points from higher to lower potential, so moving along the field direction the potential falls.","explanationZh":"电场方向总是由高电势指向低电势，故沿电场线方向电势降低。","tags":["basic","electrostatics","potential-and-field"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-potential-004","unitId":"electrostatics","topicId":"electric-potential","difficulty":2,"stem":"A positive charge is moved slowly from a point of lower potential to a point of higher potential, with no change in its kinetic energy. Its electric potential energy","stemZh":"把正电荷缓慢地从低电势点移到高电势点，动能不变，它的电势能将","options":[{"id":"A","text":"increases."},{"id":"B","text":"decreases."},{"id":"C","text":"stays the same."},{"id":"D","text":"becomes zero."}],"answer":"A","explanation":"For a positive charge $E_p = qV$, so going to a higher potential raises the potential energy. An external force must do positive work against the field.","explanationZh":"正电荷 $E_p=qV，电势升高则电势能增大。此过程外力必须克服电场力做正功。$","tags":["basic","electrostatics","potential-energy"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-potential-005","unitId":"electrostatics","topicId":"electric-potential","difficulty":2,"stem":"In a uniform electric field of strength $5.0\\times10^{2}\\ \\mathrm{N/C}$, two points are separated by $0.040\\ \\mathrm{m}$ measured along the direction of the field. What is the potential difference between them?","stemZh":"场强为 $5.0\\times10^{2}\\ \\mathrm{N/C}$ 的匀强电场中，两点沿电场方向相距 $0.040\\ \\mathrm{m}，它们的电势差为多少？$","options":[{"id":"A","text":"$1.25\\times10^{4}\\ \\mathrm{V}$"},{"id":"B","text":"$2.0\\ \\mathrm{V}$"},{"id":"C","text":"$0.08\\ \\mathrm{V}$"},{"id":"D","text":"$20\\ \\mathrm{V}$"}],"answer":"D","explanation":"$U =$ Ed $= 5.0\\times10^{2}\\times0.040 = 20\\ \\mathrm{V}$.","explanationZh":"$U=Ed=5.0\\times10^{2}\\times0.040=20\\ \\mathrm{V}。$","tags":["basic","electrostatics","uniform-field-potential"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-potential-006","unitId":"electrostatics","topicId":"electric-potential","difficulty":3,"stem":"A thin rod ab of length L is placed in a uniform electric field of strength E, making an angle $\\theta$ with the direction of the field, as shown. What is the potential difference between the ends a and b?","stemZh":"长为 L 的细杆 ab 放在场强为 E 的匀强电场中，与电场方向成 $\\theta$ 角（如图）。a、b 两端的电势差为多少？","options":[{"id":"A","text":"EL"},{"id":"B","text":"$EL\\sin\\theta$"},{"id":"C","text":"$EL\\cos\\theta$"},{"id":"D","text":"$EL\\tan\\theta$"}],"answer":"C","explanation":"In a uniform field $U = E\\,d$, where d is the distance measured along the field direction. Here $d = L\\cos\\theta$, so $U = EL\\cos\\theta$.","explanationZh":"匀强电场中 $U=Ed，d$ 为沿电场方向的距离。此处 $d=L\\cos\\theta，故 U=EL\\cos\\theta。$","tags":["applied","electrostatics","uniform-field-potential"],"images":[{"src":"/physics/diagrams/potential-rod-angle.svg","alt":"A rod a b of length L lying at an angle theta to a uniform electric field E.","altZh":"长为 L 的细杆 ab 与匀强电场 E 成 θ 角。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-potential-007","unitId":"electrostatics","topicId":"electric-potential","difficulty":3,"stem":"A charge of $-2.0\\times10^{-6}\\ \\mathrm{C}$ is moved from A to B, where the potential difference $V_A - V_B = +50\\ \\mathrm{V}$. How much work does the electric field do on the charge?","stemZh":"电荷量 $-2.0\\times10^{-6}\\ \\mathrm{C} 从 A$ 移到 B，已知 $V_A-V_B=+50\\ \\mathrm{V}，电场力做功多少？$","options":[{"id":"A","text":"$+1.0\\times10^{-4}\\ \\mathrm{J}$"},{"id":"B","text":"$-1.0\\times10^{-4}\\ \\mathrm{J}$"},{"id":"C","text":"$+2.5\\times10^{-4}\\ \\mathrm{J}$"},{"id":"D","text":"$-2.5\\times10^{-4}\\ \\mathrm{J}$"}],"answer":"B","explanation":"$W = q(V_A - V_B) = (-2.0\\times10^{-6})\\times50 = -1.0\\times10^{-4}\\ \\mathrm{J}. A$ negative charge moving to a lower potential has its potential energy increased, so the field does negative work on it.","explanationZh":"$W=q(V_A-V_B)=(-2.0\\times10^{-6})\\times50=-1.0\\times10^{-4}\\ \\mathrm{J}。负电荷移向低电势处，电势能增大，故电场力做负功。$","tags":["applied","electrostatics","work-by-field"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-potential-008","unitId":"electrostatics","topicId":"electric-potential","difficulty":3,"stem":"An electron is accelerated from rest through a potential difference of $100\\ \\mathrm{V}$. What is the kinetic energy it gains? $(e = 1.6\\times10^{-19}\\ \\mathrm{C})$","stemZh":"电子从静止经 $100\\ \\mathrm{V} 的电势差加速，它获得的动能为多少？（e=1.6\\times10^{-19}\\ \\mathrm{C}）$","options":[{"id":"A","text":"$1.6\\times10^{-17}\\ \\mathrm{J}$"},{"id":"B","text":"$1.6\\times10^{-19}\\ \\mathrm{J}$"},{"id":"C","text":"$6.25\\times10^{-18}\\ \\mathrm{J}$"},{"id":"D","text":"$1.0\\times10^{-17}\\ \\mathrm{J}$"}],"answer":"A","explanation":"The field does work $|q|U$, and starting from rest all of it becomes kinetic $energy: E_k = 1.6\\times10^{-19}\\times100 = 1.6\\times10^{-17}\\ \\mathrm{J}$.","explanationZh":"电场力做功 $|q|U，由静止开始全部转化为动能：E_k=1.6\\times10^{-19}\\times100=1.6\\times10^{-17}\\ \\mathrm{J}。$","tags":["applied","electrostatics","work-energy"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-potential-009","unitId":"electrostatics","topicId":"electric-potential","difficulty":3,"stem":"A charge of $+5.0\\times10^{-6}\\ \\mathrm{C}$ is moved between two points and the electric field does $1.0\\times10^{-3}\\ \\mathrm{J}$ of work. What is the potential difference between the two points?","stemZh":"电荷量 $+5.0\\times10^{-6}\\ \\mathrm{C}$ 在两点间移动时电场力做功 $1.0\\times10^{-3}\\ \\mathrm{J}，这两点间的电势差为多少？$","options":[{"id":"A","text":"$5.0\\times10^{-9}\\ \\mathrm{V}$"},{"id":"B","text":"$5.0\\times10^{3}\\ \\mathrm{V}$"},{"id":"C","text":"$20\\ \\mathrm{V}$"},{"id":"D","text":"$200\\ \\mathrm{V}$"}],"answer":"D","explanation":"$U = \\dfrac{W}{q} = \\dfrac{1.0\\times10^{-3}}{5.0\\times10^{-6}} = 200\\ \\mathrm{V}$.","explanationZh":"$U=\\dfrac{W}{q}=\\dfrac{1.0\\times10^{-3}}{5.0\\times10^{-6}}=200\\ \\mathrm{V}。$","tags":["applied","electrostatics","potential-difference"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-potential-010","unitId":"electrostatics","topicId":"electric-potential","difficulty":3,"stem":"Two points lie on the same equipotential surface. How much work does the electric field do when a charge is moved from one point to the other along the surface?","stemZh":"两点位于同一等势面上。沿该等势面把电荷从一点移到另一点，电场力做功多少？","options":[{"id":"A","text":"qU, where U is the potential of the surface."},{"id":"B","text":"Zero."},{"id":"C","text":"A negative amount equal to qU."},{"id":"D","text":"It depends on the path taken along the surface."}],"answer":"B","explanation":"The two points are at the same potential, so $U = 0$ and $W =$ qU $= 0$. The electrostatic force is conservative, so the work is zero whatever path is taken.","explanationZh":"两点等势，故 $U=0，W=qU=0。静电力是保守力，无论沿什么路径做功都为零。$","tags":["applied","electrostatics","equipotential"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-potential-011","unitId":"electrostatics","topicId":"electric-potential","difficulty":4,"stem":"In the field of a positive point charge Q, a small positive charge +q is moved slowly from B to A, where A is nearer to Q. The potential difference between the two points is U. Which statement is correct?","stemZh":"在正点电荷 Q 的电场中，把正电荷 +q 从 B 缓慢移到离 Q 更近的 A 点，两点间电势差为 U。下列说法正确的是","options":[{"id":"A","text":"The external force does work qU against the field, and the potential energy increases by qU."},{"id":"B","text":"The electric field does work qU, and the potential energy increases by qU."},{"id":"C","text":"The external force does work qU against the field, and the potential energy decreases by qU."},{"id":"D","text":"The electric field does work qU, and the potential energy decreases by qU."}],"answer":"A","explanation":"Moving a positive charge towards a positive charge is against the repulsive field force, so an external force must do positive work qU and the potential energy rises by qU. The field itself does work -qU.","explanationZh":"把正电荷移向正电荷是克服斥力做功，外力做正功 qU，电势能增加 qU；电场力本身做 -qU 的功。","tags":["trap","electrostatics","work-sign"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-potential-012","unitId":"electrostatics","topicId":"electric-potential","difficulty":4,"stem":"A student claims: \"Wherever the electric field is zero, the electric potential must also be zero.\" Is this claim correct?","stemZh":"有同学认为：“电场强度为零的地方，电势一定为零。”这一说法是否正确？","options":[{"id":"A","text":"Yes, because $E = -\\dfrac{\\Delta V}{\\Delta x}$."},{"id":"B","text":"Yes, because the potential and the field are proportional to each other."},{"id":"C","text":"$No: E = 0$ only means the potential is constant in that neighbourhood, not that it is zero."},{"id":"D","text":"No: the potential at such a point must be negative."}],"answer":"C","explanation":"E measures how fast the potential changes with position. $E = 0$ means the potential is momentarily constant there, but it may have any value — for example, at the centre of a uniformly charged ring the field is zero while the potential is not.","explanationZh":"$E 反映电势随位置变化的快慢。E=0$ 只说明该处附近电势不变，其数值可任意——例如均匀带电圆环圆心处场强为零而电势不为零。","tags":["trap","electrostatics","potential-vs-field"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-potential-013","unitId":"electrostatics","topicId":"electric-potential","difficulty":4,"stem":"In an electrostatic field, the equipotential surfaces are","stemZh":"静电场中的等势面","options":[{"id":"A","text":"always parallel to the field lines."},{"id":"B","text":"always perpendicular to the field lines, and no work is done moving a charge along them."},{"id":"C","text":"always spheres centred on each charge."},{"id":"D","text":"surfaces on which the field strength is constant."}],"answer":"B","explanation":"If an equipotential were not perpendicular to the field, the field would have a component along it and the potential would change — so equipotentials cross field lines at right angles. Moving along one therefore does no work.","explanationZh":"若等势面不与电场线垂直，电场沿面方向存在分量，电势就会变化，故等势面必与电场线垂直。沿等势面移动电荷电场力不做功。","tags":["trap","electrostatics","equipotential"],"images":[{"src":"/physics/diagrams/equipotential-lines.svg","alt":"Vertical electric field lines crossed at right angles by dashed equipotential lines.","altZh":"竖直的电场线被虚线等势面垂直穿过。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-potential-014","unitId":"electrostatics","topicId":"electric-potential","difficulty":5,"stem":"Two large parallel plates are $0.020\\ \\mathrm{m}$ apart and the potential difference between them is $120\\ \\mathrm{V}$. What is the electric field strength between the plates?","stemZh":"两块大平行板相距 $0.020\\ \\mathrm{m}，板间电势差为 120\\ \\mathrm{V}。板间电场强度为多少？$","options":[{"id":"A","text":"$2.4\\ \\mathrm{N/C}$"},{"id":"B","text":"$6000\\ \\mathrm{N/C}$"},{"id":"C","text":"$240\\ \\mathrm{N/C}$"},{"id":"D","text":"$0.17\\ \\mathrm{N/C}$"}],"answer":"B","explanation":"For parallel plates the field between them is uniform, and $E = \\dfrac{U}{d} = \\dfrac{120}{0.020} = 6000\\ \\mathrm{N/C}$.","explanationZh":"$平行板间为匀强电场，E=\\dfrac{U}{d}=\\dfrac{120}{0.020}=6000\\ \\mathrm{N/C}。$","tags":["trap","electrostatics","uniform-field-potential"],"images":[{"src":"/physics/diagrams/parallel-plate-capacitor.svg","alt":"Two parallel charged plates separated by a distance d with a potential difference U, and a uniform electric field between them.","altZh":"两块平行带电板相距 d，板间电势差为 U，其间为匀强电场。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-electric-potential-015","unitId":"electrostatics","topicId":"electric-potential","difficulty":5,"stem":"A positive charge is carried around a closed path in an electrostatic field and returned to its starting point. What is the total work done by the electric field during the trip?","stemZh":"把正电荷沿闭合路径在静电场中移动一周回到出发点，整个过程电场力做功为多少？","options":[{"id":"A","text":"qU, where U is the largest potential difference crossed."},{"id":"B","text":"Always positive."},{"id":"C","text":"Zero."},{"id":"D","text":"Always negative."}],"answer":"C","explanation":"The electrostatic force is conservative: the potential at the start and the end is the same, so the total change in potential is zero and the work done around any closed path is zero.","explanationZh":"静电力是保守力：起点与终点电势相同，总电势差为零，故沿任意闭合路径电场力做功均为零。","tags":["trap","electrostatics","conservative-field"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-dc-circuits-001","unitId":"dc-circuits","topicId":"dc-circuits","difficulty":1,"stem":"A steady current of $2\\ \\mathrm{A}$ flows through a wire. How much charge passes through its cross-section in $5\\ \\mathrm{s}$?","stemZh":"导线中通过 $2\\ \\mathrm{A} 的恒定电流，5\\ \\mathrm{s}$ 内通过横截面的电荷量是多少？","options":[{"id":"A","text":"$0.4\\ \\mathrm{C}$"},{"id":"B","text":"$2.5\\ \\mathrm{C}$"},{"id":"C","text":"$10\\ \\mathrm{C}$"},{"id":"D","text":"$7\\ \\mathrm{C}$"}],"answer":"C","explanation":"By the definition of $current, q =$ It $= 2 \\times 5 = 10\\ \\mathrm{C}$.","explanationZh":"由电流定义式 $q=It=2\\times5=10\\ \\mathrm{C}。$","tags":["basic","circuits","current-definition"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-dc-circuits-002","unitId":"dc-circuits","topicId":"dc-circuits","difficulty":1,"stem":"Two resistors of $4\\ \\Omega$ and $6\\ \\Omega$ are connected in series. What is their total resistance?","stemZh":"两个 $4\\ \\Omega 和 6\\ \\Omega$ 的电阻串联，总电阻是多少？","options":[{"id":"A","text":"$10\\ \\Omega$"},{"id":"B","text":"$2.4\\ \\Omega$"},{"id":"C","text":"$5\\ \\Omega$"},{"id":"D","text":"$24\\ \\Omega$"}],"answer":"A","explanation":"Resistors in series $add: R = R_1 + R_2 = 4 + 6 = 10\\ \\Omega$.","explanationZh":"$串联电阻相加：R=R_1+R_2=4+6=10\\ \\Omega。$","tags":["basic","circuits","series-resistance"],"images":[{"src":"/physics/diagrams/circuit-series-2r.svg","alt":"Two resistors R1 and R2 connected in series in a loop with a battery.","altZh":"电阻 R1 与 R2 串联，与电源构成回路。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-dc-circuits-003","unitId":"dc-circuits","topicId":"dc-circuits","difficulty":2,"stem":"A voltage of $12\\ \\mathrm{V}$ is applied across $a 4\\ \\Omega$ resistor. What current flows through it?","stemZh":"$在 4\\ \\Omega$ 的电阻两端加 $12\\ \\mathrm{V}$ 的电压，通过它的电流是多大？","options":[{"id":"A","text":"$0.33\\ \\mathrm{A}$"},{"id":"B","text":"$3\\ \\mathrm{A}$"},{"id":"C","text":"$48\\ \\mathrm{A}$"},{"id":"D","text":"$8\\ \\mathrm{A}$"}],"answer":"B","explanation":"Ohm's $law: I = \\dfrac{U}{R} = \\dfrac{12}{4} = 3\\ \\mathrm{A}$.","explanationZh":"由欧姆定律 $I=\\dfrac{U}{R}=\\dfrac{12}{4}=3\\ \\mathrm{A}。$","tags":["basic","circuits","ohms-law"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-dc-circuits-004","unitId":"dc-circuits","topicId":"dc-circuits","difficulty":2,"stem":"Two resistors of $6\\ \\Omega$ and $3\\ \\Omega$ are connected in parallel. What is their combined resistance?","stemZh":"两个 $6\\ \\Omega 和 3\\ \\Omega$ 的电阻并联，总电阻是多少？","options":[{"id":"A","text":"$9\\ \\Omega$"},{"id":"B","text":"$4.5\\ \\Omega$"},{"id":"C","text":"$18\\ \\Omega$"},{"id":"D","text":"$2\\ \\Omega$"}],"answer":"D","explanation":"For parallel resistors $\\dfrac{1}{R} = \\dfrac{1}{R_1} + \\dfrac{1}{R_2} = \\dfrac{1}{6} + \\dfrac{1}{3} = \\dfrac{1}{2}$, so $R = 2\\ \\Omega$.","explanationZh":"并联时 $\\dfrac{1}{R}=\\dfrac{1}{R_1}+\\dfrac{1}{R_2}=\\dfrac{1}{6}+\\dfrac{1}{3}=\\dfrac{1}{2}，故 R=2\\ \\Omega。$","tags":["basic","circuits","parallel-resistance"],"images":[{"src":"/physics/diagrams/circuit-parallel-2r.svg","alt":"Two resistors R1 and R2 connected in parallel across a battery.","altZh":"电阻 R1 与 R2 并联在电源两端。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-dc-circuits-005","unitId":"dc-circuits","topicId":"dc-circuits","difficulty":2,"stem":"An electric lamp carries a current of $0.5\\ \\mathrm{A}$ when connected to $a 220\\ \\mathrm{V}$ supply. What is its power?","stemZh":"一盏电灯接在 $220\\ \\mathrm{V}$ 的电源上，通过它的电流为 $0.5\\ \\mathrm{A}，它的电功率是多少？$","options":[{"id":"A","text":"$55\\ \\mathrm{W}$"},{"id":"B","text":"$220\\ \\mathrm{W}$"},{"id":"C","text":"$110\\ \\mathrm{W}$"},{"id":"D","text":"$440\\ \\mathrm{W}$"}],"answer":"C","explanation":"$P =$ UI $= 220 \\times 0.5 = 110\\ \\mathrm{W}$.","explanationZh":"$P=UI=220\\times0.5=110\\ \\mathrm{W}。$","tags":["basic","circuits","electric-power"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-dc-circuits-006","unitId":"dc-circuits","topicId":"dc-circuits","difficulty":3,"stem":"Resistors $R_1 = 10\\ \\Omega$ and $R_2 = 30\\ \\Omega$ are connected in parallel across a battery. What is the ratio $I_1 : I_2$ of the currents through them?","stemZh":"电阻 $R_1 = 10\\ \\Omega 和 R_2 = 30\\ \\Omega$ 并联在电源两端，通过它们的电流之比 $I_1 : I_2$ 是多少？","options":[{"id":"A","text":"1 : 3"},{"id":"B","text":"1 : 1"},{"id":"C","text":"9 : 1"},{"id":"D","text":"3 : 1"}],"answer":"D","explanation":"Parallel branches share the same voltage, so $I \\propto \\dfrac{1}{R}$ and $I_1 : I_2 = R_2 : R_1 = 30 : 10 = 3 : 1$.","explanationZh":"$并联支路电压相等，I\\propto\\dfrac{1}{R}，故 I_1:I_2=R_2:R_1=30:10=3:1。$","tags":["applied","circuits","parallel-current-ratio"],"images":[{"src":"/physics/diagrams/circuit-parallel-2r.svg","alt":"Two resistors R1 and R2 connected in parallel across a battery.","altZh":"电阻 R1 与 R2 并联在电源两端。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-dc-circuits-007","unitId":"dc-circuits","topicId":"dc-circuits","difficulty":3,"stem":"Resistors $R_1 = 20\\ \\Omega$ and $R_2 = 40\\ \\Omega$ are connected in series across $a 12\\ \\mathrm{V}$ supply. What is the voltage across $R_2$?","stemZh":"电阻 $R_1 = 20\\ \\Omega 和 R_2 = 40\\ \\Omega$ 串联接在 $12\\ \\mathrm{V} 的电源上，R_2$ 两端的电压是多少？","options":[{"id":"A","text":"$8\\ \\mathrm{V}$"},{"id":"B","text":"$4\\ \\mathrm{V}$"},{"id":"C","text":"$6\\ \\mathrm{V}$"},{"id":"D","text":"$12\\ \\mathrm{V}$"}],"answer":"A","explanation":"Total resistance $= 20 + 40 = 60\\ \\Omega$, so $I = \\dfrac{12}{60} = 0.2\\ \\mathrm{A}$ and $U_2 = I R_2 = 0.2 \\times 40 = 8\\ \\mathrm{V}$.","explanationZh":"总电阻 $=20+40=60\\ \\Omega，故 I=\\dfrac{12}{60}=0.2\\ \\mathrm{A}，U_2=IR_2=0.2\\times40=8\\ \\mathrm{V}。$","tags":["applied","circuits","series-voltage-division"],"images":[{"src":"/physics/diagrams/circuit-series-2r.svg","alt":"Two resistors R1 and R2 connected in series in a loop with a battery.","altZh":"电阻 R1 与 R2 串联，与电源构成回路。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-dc-circuits-008","unitId":"dc-circuits","topicId":"dc-circuits","difficulty":3,"stem":"A current of $2\\ \\mathrm{A}$ flows through $a 5\\ \\Omega$ resistor. How much power is dissipated as heat?","stemZh":"通过 $5\\ \\Omega$ 电阻的电流为 $2\\ \\mathrm{A}，它发热损耗的功率是多少？$","options":[{"id":"A","text":"$10\\ \\mathrm{W}$"},{"id":"B","text":"$40\\ \\mathrm{W}$"},{"id":"C","text":"$50\\ \\mathrm{W}$"},{"id":"D","text":"$20\\ \\mathrm{W}$"}],"answer":"D","explanation":"$P = I^{2}R = 2^{2} \\times 5 = 20\\ \\mathrm{W}$.","explanationZh":"$P=I^{2}R=2^{2}\\times5=20\\ \\mathrm{W}。$","tags":["applied","circuits","joule-heating"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-dc-circuits-009","unitId":"dc-circuits","topicId":"dc-circuits","difficulty":3,"stem":"$A 100\\ \\Omega$ resistor and $a 200\\ \\Omega$ resistor are connected in parallel. What is their combined resistance?","stemZh":"一个 $100\\ \\Omega$ 和一个 $200\\ \\Omega$ 的电阻并联，总电阻是多少？","options":[{"id":"A","text":"$300\\ \\Omega$"},{"id":"B","text":"$66.7\\ \\Omega$"},{"id":"C","text":"$150\\ \\Omega$"},{"id":"D","text":"$33.3\\ \\Omega$"}],"answer":"B","explanation":"$\\dfrac{1}{R} = \\dfrac{1}{100} + \\dfrac{1}{200} = \\dfrac{3}{200}$, so $R = \\dfrac{200}{3} \\approx 66.7\\ \\Omega$.","explanationZh":"$\\dfrac{1}{R}=\\dfrac{1}{100}+\\dfrac{1}{200}=\\dfrac{3}{200}，故 R=\\dfrac{200}{3}\\approx66.7\\ \\Omega。$","tags":["applied","circuits","parallel-resistance"],"images":[{"src":"/physics/diagrams/circuit-parallel-2r.svg","alt":"Two resistors R1 and R2 connected in parallel across a battery.","altZh":"电阻 R1 与 R2 并联在电源两端。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-dc-circuits-010","unitId":"dc-circuits","topicId":"dc-circuits","difficulty":3,"stem":"An electric heater carries a current of $4\\ \\mathrm{A}$. How much charge passes through it in $2\\ \\mathrm{minutes}$?","stemZh":"电热器中的电流为 $4\\ \\mathrm{A}，2\\ \\mathrm{min}$ 内通过它的电荷量是多少？","options":[{"id":"A","text":"$8\\ \\mathrm{C}$"},{"id":"B","text":"$120\\ \\mathrm{C}$"},{"id":"C","text":"$480\\ \\mathrm{C}$"},{"id":"D","text":"$240\\ \\mathrm{C}$"}],"answer":"C","explanation":"Convert the $time: t = 2 \\times 60 = 120\\ \\mathrm{s}$. Then $q =$ It $= 4 \\times 120 = 480\\ \\mathrm{C}$.","explanationZh":"先换算时间 $t=2\\times60=120\\ \\mathrm{s}，则 q=It=4\\times120=480\\ \\mathrm{C}。$","tags":["applied","circuits","current-definition"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-dc-circuits-011","unitId":"dc-circuits","topicId":"dc-circuits","difficulty":4,"stem":"An electric motor has a coil resistance of $2\\ \\Omega$ and operates at $100\\ \\mathrm{V}$ drawing a current of $3\\ \\mathrm{A}$. What is the power lost as heat in the coil?","stemZh":"电动机线圈电阻为 $2\\ \\Omega，在 100\\ \\mathrm{V}$ 电压下工作，电流为 $3\\ \\mathrm{A}，线圈发热损耗的功率是多少？$","options":[{"id":"A","text":"$300\\ \\mathrm{W}$"},{"id":"B","text":"$18\\ \\mathrm{W}$"},{"id":"C","text":"$6\\ \\mathrm{W}$"},{"id":"D","text":"$200\\ \\mathrm{W}$"}],"answer":"B","explanation":"A motor is not a pure resistor, so only the resistive part turns into $heat: P_{heat} = I^{2}R = 3^{2} \\times 2 = 18\\ \\mathrm{W}$. The input power UI $= 300\\ \\mathrm{W}$ mostly becomes mechanical output.","explanationZh":"$电动机不是纯电阻，只有电阻部分转化为热：P_{\\text{热}}=I^{2}R=3^{2}\\times2=18\\ \\mathrm{W}。输入功率 UI=300\\ \\mathrm{W}$ 大部分转化为机械能输出。","tags":["trap","circuits","non-ohmic-power"],"images":[{"src":"/physics/diagrams/circuit-motor-power.svg","alt":"A battery driving a motor M, with an ammeter A in the circuit.","altZh":"电源驱动电动机 M，电路中接有电流表 A。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-dc-circuits-012","unitId":"dc-circuits","topicId":"dc-circuits","difficulty":4,"stem":"A battery is joined to a resistor $R = 2\\ \\Omega$ through a switch, with a voltmeter across R. With the switch open the voltmeter reads $3\\ \\mathrm{V}$; with the switch closed it reads $2\\ \\mathrm{V}$. What is the internal resistance of the battery?","stemZh":"电源经开关与电阻 $R = 2\\ \\Omega$ 相连，电压表并联在 R 两端。开关断开时电压表示数为 $3\\ \\mathrm{V}，闭合时示数为 2\\ \\mathrm{V}。电源的内阻是多少？$","options":[{"id":"A","text":"$0.5\\ \\Omega$"},{"id":"B","text":"$3\\ \\Omega$"},{"id":"C","text":"$2\\ \\Omega$"},{"id":"D","text":"$1\\ \\Omega$"}],"answer":"D","explanation":"Open switch $→$ no current, so the voltmeter reads the $emf: E = 3\\ \\mathrm{V}$. Closed $→$ terminal voltage $U = 2\\ \\mathrm{V}$, giving $I = \\dfrac{U}{R} = \\dfrac{2}{2} = 1\\ \\mathrm{A}$. Then $E = U +$ Ir $\\Rightarrow 3 = 2 + 1 \\cdot r$, so $r = 1\\ \\Omega$.","explanationZh":"开关断开时无电流，电压表示数即电动势 $E=3\\ \\mathrm{V}；闭合时路端电压 U=2\\ \\mathrm{V}，I=\\dfrac{U}{R}=\\dfrac{2}{2}=1\\ \\mathrm{A}。由 E=U+Ir 得 3=2+1\\cdot$ r，故 $r=1\\ \\Omega。$","tags":["trap","circuits","closed-circuit-ohms-law"],"images":[{"src":"/physics/diagrams/circuit-emf-internal-r.svg","alt":"A battery with internal resistance r, a switch S, a resistor R and a voltmeter across R.","altZh":"含内阻 r 的电源、开关 S、电阻 R，电压表并联在 R 两端。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-dc-circuits-013","unitId":"dc-circuits","topicId":"dc-circuits","difficulty":5,"stem":"Three identical resistors are connected in various ways (all three always used). The smallest total resistance that can be obtained is $9\\ \\Omega$. What is the resistance of one resistor?","stemZh":"三个相同的电阻以各种方式连接（每次都用三个）。能得到的最小总电阻为 $9\\ \\Omega，则每个电阻的阻值是多少？$","options":[{"id":"A","text":"$3\\ \\Omega$"},{"id":"B","text":"$9\\ \\Omega$"},{"id":"C","text":"$27\\ \\Omega$"},{"id":"D","text":"$18\\ \\Omega$"}],"answer":"C","explanation":"The smallest total is obtained with all three in $parallel: R_{total} = \\dfrac{R}{3} = 9\\ \\Omega$, so $R = 27\\ \\Omega$.","explanationZh":"$三个电阻全部并联时总电阻最小：R_{\\text{总}}=\\dfrac{R}{3}=9\\ \\Omega，故 R=27\\ \\Omega。$","tags":["trap","circuits","combination-resistance"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-dc-circuits-014","unitId":"dc-circuits","topicId":"dc-circuits","difficulty":5,"stem":"Resistors $6\\ \\Omega$ and $3\\ \\Omega$ are connected in parallel, and this pair is then connected in series with $a 4\\ \\Omega$ resistor across $a 12\\ \\mathrm{V}$ supply. What is the current through the $4\\ \\Omega$ resistor?","stemZh":"电阻 $6\\ \\Omega 与 3\\ \\Omega$ 并联后，再与一个 $4\\ \\Omega$ 的电阻串联接在 $12\\ \\mathrm{V}$ 的电源上。通过 $4\\ \\Omega$ 电阻的电流是多少？","options":[{"id":"A","text":"$2\\ \\mathrm{A}$"},{"id":"B","text":"$1.33\\ \\mathrm{A}$"},{"id":"C","text":"$3\\ \\mathrm{A}$"},{"id":"D","text":"$0.5\\ \\mathrm{A}$"}],"answer":"A","explanation":"The parallel pair gives $\\dfrac{1}{R_p} = \\dfrac{1}{6} + \\dfrac{1}{3} = \\dfrac{1}{2}$, so $R_p = 2\\ \\Omega$. Total $R = 2 + 4 = 6\\ \\Omega$, hence $I = \\dfrac{12}{6} = 2\\ \\mathrm{A} —$ the whole current passes through the series $4\\ \\Omega$ resistor.","explanationZh":"并联部分 $\\dfrac{1}{R_p}=\\dfrac{1}{6}+\\dfrac{1}{3}=\\dfrac{1}{2}，R_p=2\\ \\Omega；总电阻 R=2+4=6\\ \\Omega，故 I=\\dfrac{12}{6}=2\\ \\mathrm{A}，即全部电流都通过串联的 4\\ \\Omega$ 电阻。","tags":["trap","circuits","series-parallel-combination"],"images":[{"src":"/physics/diagrams/circuit-series-parallel-combo.svg","alt":"R1 and R2 in parallel, connected in series with R3 and a battery.","altZh":"R1 与 R2 并联，再与 R3 及电源串联。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-dc-circuits-015","unitId":"dc-circuits","topicId":"dc-circuits","difficulty":5,"stem":"Two lamps rated $6\\ \\mathrm{V},\\ 3\\ \\mathrm{W}$ and $6\\ \\mathrm{V},\\ 6\\ \\mathrm{W}$ are connected in series across $a 6\\ \\mathrm{V}$ supply. Which lamp is brighter?","stemZh":"额定值为 $6\\ \\mathrm{V},\\ 3\\ \\mathrm{W} 和 6\\ \\mathrm{V},\\ 6\\ \\mathrm{W}$ 的两只灯泡串联接在 $6\\ \\mathrm{V}$ 的电源上，哪只灯泡更亮？","options":[{"id":"A","text":"The $3\\ \\mathrm{W}$ lamp"},{"id":"B","text":"The $6\\ \\mathrm{W}$ lamp"},{"id":"C","text":"Both are equally bright"},{"id":"D","text":"It cannot be determined"}],"answer":"A","explanation":"From $R = \\dfrac{U^{2}}{P}$: the $3\\ \\mathrm{W}$ lamp has $R = \\dfrac{36}{3} = 12\\ \\Omega$ and the $6\\ \\mathrm{W}$ lamp $R = \\dfrac{36}{6} = 6\\ \\Omega$. In series they carry the same current, and $P = I^{2}R$ is larger for the larger resistance, so the $3\\ \\mathrm{W}$ lamp is brighter.","explanationZh":"$由 R=\\dfrac{U^{2}}{P}，3\\ \\mathrm{W} 灯 R=\\dfrac{36}{3}=12\\ \\Omega，6\\ \\mathrm{W} 灯 R=\\dfrac{36}{6}=6\\ \\Omega。串联电流相同，P=I^{2}R$ 随电阻增大而增大，故 $3\\ \\mathrm{W}$ 灯更亮。","tags":["trap","circuits","power-comparison"],"images":[{"src":"/physics/diagrams/circuit-two-lamps.svg","alt":"Two lamps L1 and L2 connected in series with a battery.","altZh":"两只灯泡 L1 与 L2 串联，与电源构成回路。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-source-power-001","unitId":"dc-circuits","topicId":"source-power","difficulty":1,"stem":"With no current flowing, a voltmeter connected directly across the terminals of a battery reads $6.0\\ \\mathrm{V}$. What does this reading represent?","stemZh":"电路断开时，直接接在电池两端的电压表读数为 $6.0\\ \\mathrm{V}，该读数表示什么？$","options":[{"id":"A","text":"the internal resistance of the battery"},{"id":"B","text":"the electromotive force of the battery"},{"id":"C","text":"the terminal voltage while the battery is supplying current"},{"id":"D","text":"the charge stored in the battery"}],"answer":"B","explanation":"With no current the internal voltage drop Ir is zero, so the terminal voltage equals the emf E. The terminal voltage only differs from E once a current flows.","explanationZh":"断路时内阻上的电压降 $Ir=0，路端电压等于电动势$ E。只有在有电流时路端电压才小于 E。","tags":["basic","electromagnetism","emf"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-source-power-002","unitId":"dc-circuits","topicId":"source-power","difficulty":1,"stem":"A resistor of resistance $10\\ \\Omega$ carries a current of $2.0\\ \\mathrm{A}$. What is the power dissipated in it?","stemZh":"阻值 $10\\ \\Omega$ 的电阻中通过 $2.0\\ \\mathrm{A}$ 的电流，它消耗的功率为多少？","options":[{"id":"A","text":"$40\\ \\mathrm{W}$"},{"id":"B","text":"$20\\ \\mathrm{W}$"},{"id":"C","text":"$12\\ \\mathrm{W}$"},{"id":"D","text":"$5.0\\ \\mathrm{W}$"}],"answer":"A","explanation":"$P = I^{2}R = (2.0)^{2}\\times10 = 40\\ \\mathrm{W}$.","explanationZh":"$P=I^{2}R=(2.0)^{2}\\times10=40\\ \\mathrm{W}。$","tags":["basic","electromagnetism","electric-power"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-source-power-003","unitId":"dc-circuits","topicId":"source-power","difficulty":2,"stem":"A lamp is marked \"$220\\ \\mathrm{V}, 100\\ \\mathrm{W}$\". What is the resistance of its filament when it is lit?","stemZh":"$某灯泡标有“220\\ \\mathrm{V}，100\\ \\mathrm{W}”，正常发光时灯丝电阻为多少？$","options":[{"id":"A","text":"$2.2\\ \\Omega$"},{"id":"B","text":"$220\\ \\Omega$"},{"id":"C","text":"$100\\ \\Omega$"},{"id":"D","text":"$484\\ \\Omega$"}],"answer":"D","explanation":"$R = \\dfrac{U^{2}}{P} = \\dfrac{220^{2}}{100} = 484\\ \\Omega$.","explanationZh":"$R=\\dfrac{U^{2}}{P}=\\dfrac{220^{2}}{100}=484\\ \\Omega。$","tags":["basic","electromagnetism","electric-power"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-source-power-004","unitId":"dc-circuits","topicId":"source-power","difficulty":2,"stem":"An electric motor has a winding resistance of $2\\ \\Omega$ and draws a current of $3.0\\ \\mathrm{A}$ when running. What is the power lost as heat in the windings?","stemZh":"某电动机线圈电阻为 $2\\ \\Omega，正常工作电流 3.0\\ \\mathrm{A}，线圈发热损耗的功率为多少？$","options":[{"id":"A","text":"$6.0\\ \\mathrm{W}$"},{"id":"B","text":"$12\\ \\mathrm{W}$"},{"id":"C","text":"$18\\ \\mathrm{W}$"},{"id":"D","text":"$300\\ \\mathrm{W}$"}],"answer":"C","explanation":"The heating loss is the resistive loss $P = I^{2}R = (3.0)^{2}\\times2 = 18\\ \\mathrm{W}$. It is not UI — that is the total electrical power taken from the supply.","explanationZh":"发热损耗即电阻损耗 $P=I^{2}R=(3.0)^{2}\\times2=18\\ \\mathrm{W}。不是$ UI，UI 是从电源取用的总电功率。","tags":["basic","electromagnetism","non-ohmic-power"],"images":[{"src":"/physics/diagrams/circuit-motor-power.svg","alt":"An electric motor M connected to a supply, with an ammeter and a voltmeter.","altZh":"电动机 M 接在电源上，电路中有电流表和电压表。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-source-power-005","unitId":"dc-circuits","topicId":"source-power","difficulty":3,"stem":"In the circuit $shown, R = 2\\ \\Omega$. With the switch S open the voltmeter reads $3.0\\ \\mathrm{V}$; with S closed it reads $2.0\\ \\mathrm{V}$. What is the internal resistance r of the source?","stemZh":"$如图电路，R=2\\ \\Omega。开关 S$ 断开时电压表示数为 $3.0\\ \\mathrm{V}，闭合时示数为 2.0\\ \\mathrm{V}。电源内阻 r$ 为多少？","options":[{"id":"A","text":"$0.50\\ \\Omega$"},{"id":"B","text":"$2.0\\ \\Omega$"},{"id":"C","text":"$1.0\\ \\Omega$"},{"id":"D","text":"$1.5\\ \\Omega$"}],"answer":"C","explanation":"With S open no current flows, so the reading is the $emf: E = 3.0\\ \\mathrm{V}$. With $S closed, U = 2.0\\ \\mathrm{V}$ and $I = \\dfrac{U}{R} = \\dfrac{2.0}{2} = 1.0\\ \\mathrm{A}$. From $E = U + Ir: r = \\dfrac{E-U}{I} = \\dfrac{3.0-2.0}{1.0} = 1.0\\ \\Omega$.","explanationZh":"S 断开时无电流，读数即电动势 $E=3.0\\ \\mathrm{V}。S$ 闭合时路端电压 $U=2.0\\ \\mathrm{V}，I=\\dfrac{U}{R}=\\dfrac{2.0}{2}=1.0\\ \\mathrm{A}。由 E=U+Ir 得 r=\\dfrac{E-U}{I}=\\dfrac{3.0-2.0}{1.0}=1.0\\ \\Omega。$","tags":["applied","electromagnetism","internal-resistance"],"images":[{"src":"/physics/diagrams/circuit-emf-internal-r.svg","alt":"A source with internal resistance r feeding an external resistor R through a switch, with a voltmeter across the source terminals.","altZh":"内阻为 r 的电源经开关给外电阻 R 供电，电压表接在电源两端。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-source-power-006","unitId":"dc-circuits","topicId":"source-power","difficulty":3,"stem":"An electric motor runs from $a 100\\ \\mathrm{V}$ supply and draws $3.0\\ \\mathrm{A}$. Its winding resistance is $2\\ \\Omega$. What is its mechanical output power?","stemZh":"电动机接在 $100\\ \\mathrm{V}$ 电源上，电流 $3.0\\ \\mathrm{A}，线圈电阻 2\\ \\Omega，它的机械输出功率为多少？$","options":[{"id":"A","text":"$300\\ \\mathrm{W}$"},{"id":"B","text":"$282\\ \\mathrm{W}$"},{"id":"C","text":"$18\\ \\mathrm{W}$"},{"id":"D","text":"$318\\ \\mathrm{W}$"}],"answer":"B","explanation":"Total electrical power $in: P_{in} =$ UI $= 100\\times3.0 = 300\\ \\mathrm{W}$. Heating $loss: P_{heat} = I^{2}R = 18\\ \\mathrm{W}$. Mechanical $output: P_{out} = P_{in} - P_{heat} = 282\\ \\mathrm{W}$.","explanationZh":"输入总电功率 $P_{in}=UI=100\\times3.0=300\\ \\mathrm{W}；发热损耗 P_{heat}=I^{2}R=18\\ \\mathrm{W}；机械输出 P_{out}=P_{in}-P_{heat}=282\\ \\mathrm{W}。$","tags":["applied","electromagnetism","non-ohmic-power"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-source-power-007","unitId":"dc-circuits","topicId":"source-power","difficulty":3,"stem":"A source of emf $12\\ \\mathrm{V}$ and internal resistance $1.0\\ \\Omega$ is connected to $a 5.0\\ \\Omega$ resistor. What is the terminal voltage of the source?","stemZh":"电动势 $12\\ \\mathrm{V}、内阻 1.0\\ \\Omega$ 的电源接 $5.0\\ \\Omega$ 的电阻，电源的路端电压为多少？","options":[{"id":"A","text":"$10\\ \\mathrm{V}$"},{"id":"B","text":"$12\\ \\mathrm{V}$"},{"id":"C","text":"$2.0\\ \\mathrm{V}$"},{"id":"D","text":"$11\\ \\mathrm{V}$"}],"answer":"A","explanation":"$I = \\dfrac{E}{R+r} = \\dfrac{12}{5.0+1.0} = 2.0\\ \\mathrm{A}$. Terminal voltage $U =$ IR $= 2.0\\times5.0 = 10\\ \\mathrm{V}$, which also equals E - Ir $= 12 - 2.0 = 10\\ \\mathrm{V}$.","explanationZh":"$I=\\dfrac{E}{R+r}=\\dfrac{12}{5.0+1.0}=2.0\\ \\mathrm{A}。路端电压 U=IR=2.0\\times5.0=10\\ \\mathrm{V}，也等于 E-Ir=12-2.0=10\\ \\mathrm{V}。$","tags":["applied","electromagnetism","internal-resistance"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-source-power-008","unitId":"dc-circuits","topicId":"source-power","difficulty":3,"stem":"An electric motor takes in $300\\ \\mathrm{W}$ of electrical power and delivers $270\\ \\mathrm{W}$ of mechanical power. What is its efficiency?","stemZh":"某电动机输入电功率 $300\\ \\mathrm{W}，输出机械功率 270\\ \\mathrm{W}，它的效率为多少？$","options":[{"id":"A","text":"$11\\%$"},{"id":"B","text":"$30\\%$"},{"id":"C","text":"$111\\%$"},{"id":"D","text":"$90\\%$"}],"answer":"D","explanation":"$\\eta = \\dfrac{P_{out}}{P_{in}} = \\dfrac{270}{300} = 0.90 = 90\\%$. The missing $30\\ \\mathrm{W}$ is lost as heat in the windings and to friction.","explanationZh":"$\\eta=\\dfrac{P_{out}}{P_{in}}=\\dfrac{270}{300}=0.90=90\\%。缺少的 30\\ \\mathrm{W}$ 为线圈发热与摩擦损耗。","tags":["applied","electromagnetism","efficiency"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-source-power-009","unitId":"dc-circuits","topicId":"source-power","difficulty":4,"stem":"A motor of winding resistance R runs from a supply of voltage U and draws a current I. Which expression gives its mechanical output power?","stemZh":"线圈电阻为 R 的电动机接在电压 U 的电源上，电流为 I，它的机械输出功率表达式是","options":[{"id":"A","text":"UI"},{"id":"B","text":"$I^{2}R$"},{"id":"C","text":"UI $- I^{2}R$"},{"id":"D","text":"$\\dfrac{U^{2}}{R}$"}],"answer":"C","explanation":"UI is the total electrical power taken in and $I^{2}R$ is the heat lost in the windings, so the mechanical output is the difference UI $- I^{2}R$. The expressions UI and $\\dfrac{U^{2}}{R}$ describe a pure resistor, not a motor.","explanationZh":"UI $是输入总电功率，I^{2}R$ 是线圈发热损耗，故机械输出为二者之差 $UI-I^{2}R。UI 和 \\dfrac{U^{2}}{R}$ 只适用于纯电阻，不适用于电动机。","tags":["trap","electromagnetism","non-ohmic-power"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-source-power-010","unitId":"dc-circuits","topicId":"source-power","difficulty":4,"stem":"A student writes $P = \\dfrac{U^{2}}{R}$ and uses it to find the power of a running motor whose winding resistance is R. Why is this wrong?","stemZh":"有同学用 $P=\\dfrac{U^{2}}{R}$ 计算一台正在运转的电动机（线圈电阻为 R）的功率，为什么这样做是错的？","options":[{"id":"A","text":"It is not wrong — the formula always applies."},{"id":"B","text":"The formula assumes the device is a pure resistor, but a running motor has a back emf so $U \\neq$ IR."},{"id":"C","text":"The formula gives the heat loss rather than the power."},{"id":"D","text":"The formula only works for alternating current."}],"answer":"B","explanation":"$P = \\dfrac{U^{2}}{R}$ comes from combining $P =$ UI with Ohm's law $U =$ IR. A running motor is not a pure resistor — it generates a back emf, so $U \\neq$ IR and the substitution is invalid.","explanationZh":"$P=\\dfrac{U^{2}}{R} 由 P=UI$ 与欧姆定律 $U=IR$ 得出。运转中的电动机不是纯电阻，它产生反电动势，故 $U\\neq$ IR，该代换不成立。","tags":["trap","electromagnetism","non-ohmic-power"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-source-power-011","unitId":"dc-circuits","topicId":"source-power","difficulty":5,"stem":"A source of emf E and internal resistance r supplies a variable external resistor R. As R is increased, the terminal voltage of the source","stemZh":"电动势 E、内阻 r 的电源给可变外电阻 R 供电。当 R 增大时，电源的路端电压","options":[{"id":"A","text":"decreases, because the current increases."},{"id":"B","text":"stays equal to E."},{"id":"C","text":"falls to zero."},{"id":"D","text":"increases towards E, because the current and the drop Ir get smaller."}],"answer":"D","explanation":"$U = E -$ Ir with $I = \\dfrac{E}{R+r}$. Increasing R reduces I, so the internal drop Ir shrinks and U rises, approaching E as R becomes very large.","explanationZh":"$U=E-Ir，I=\\dfrac{E}{R+r}。R$ 增大时 I 减小，内电压降 Ir 减小，路端电压升高并趋近 E。","tags":["trap","electromagnetism","internal-resistance"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-source-power-012","unitId":"dc-circuits","topicId":"source-power","difficulty":5,"stem":"A source of emf $6.0\\ \\mathrm{V}$ and internal resistance $0.50\\ \\Omega$ is short-circuited by a wire of negligible resistance. What current flows?","stemZh":"电动势 $6.0\\ \\mathrm{V}、内阻 0.50\\ \\Omega$ 的电源被电阻可忽略的导线短路，电流为多大？","options":[{"id":"A","text":"$12\\ \\mathrm{A}$"},{"id":"B","text":"$3.0\\ \\mathrm{A}$"},{"id":"C","text":"$0\\ \\mathrm{A}$"},{"id":"D","text":"$6.0\\ \\mathrm{A}$"}],"answer":"A","explanation":"With $R \\approx 0, I = \\dfrac{E}{r} = \\dfrac{6.0}{0.50} = 12\\ \\mathrm{A}$. The terminal voltage is then zero, so all of the power EI is dissipated inside the source — which is why short circuits are dangerous.","explanationZh":"$R\\approx0 时 I=\\dfrac{E}{r}=\\dfrac{6.0}{0.50}=12\\ \\mathrm{A}。此时路端电压为零，全部功率$ EI 消耗在电源内部，这就是短路危险的原因。","tags":["trap","electromagnetism","internal-resistance"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-magnetic-field-force-001","unitId":"magnetic-field","topicId":"magnetic-field-force","difficulty":1,"stem":"A straight wire of length $0.50\\ \\mathrm{m}$ carries a current of $10\\ \\mathrm{A}$ and is placed perpendicular to a uniform magnetic field. It experiences a force of $1.5\\ \\mathrm{N}$. What is the magnetic flux density?","stemZh":"$长 0.50\\ \\mathrm{m}$ 的直导线通 $10\\ \\mathrm{A}$ 电流，垂直放入匀强磁场，所受安培力 $1.5\\ \\mathrm{N}。磁感应强度为多少？$","options":[{"id":"A","text":"$0.15\\ \\mathrm{T}$"},{"id":"B","text":"$0.30\\ \\mathrm{T}$"},{"id":"C","text":"$3.0\\ \\mathrm{T}$"},{"id":"D","text":"$7.5\\ \\mathrm{T}$"}],"answer":"B","explanation":"For a wire perpendicular to the $field, F = BIL$, so $B = \\dfrac{F}{IL} = \\dfrac{1.5}{10\\times0.50} = 0.30\\ \\mathrm{T}$.","explanationZh":"导线垂直磁场时 $F=BIL，故 B=\\dfrac{F}{IL}=\\dfrac{1.5}{10\\times0.50}=0.30\\ \\mathrm{T}。$","tags":["basic","electromagnetism","ampere-force"],"images":[{"src":"/physics/diagrams/magnetic-wire-ampere-force.svg","alt":"A straight current-carrying wire in a uniform magnetic field directed into the page, with the Ampere force F pointing upwards.","altZh":"通电直导线处在垂直纸面向里的匀强磁场中，安培力 F 竖直向上。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-magnetic-field-force-002","unitId":"magnetic-field","topicId":"magnetic-field-force","difficulty":1,"stem":"What are the SI units of magnetic flux density and of magnetic flux?","stemZh":"磁感应强度和磁通量的国际单位分别是什么？","options":[{"id":"A","text":"tesla and weber"},{"id":"B","text":"weber and tesla"},{"id":"C","text":"newton and joule"},{"id":"D","text":"henry and farad"}],"answer":"A","explanation":"Magnetic flux density B is measured in tesla (T) and magnetic flux $\\Phi$ in weber $(Wb)$, with $1\\ \\mathrm{Wb} = 1\\ \\mathrm{T\\cdot m^{2}}$.","explanationZh":"磁感应强度 B 的单位是特斯拉（T），磁通量 $\\Phi 的单位是韦伯（Wb），1\\ \\mathrm{Wb}=1\\ \\mathrm{T\\cdot m^{2}}。$","tags":["basic","electromagnetism","magnetic-units"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-magnetic-field-force-003","unitId":"magnetic-field","topicId":"magnetic-field-force","difficulty":1,"stem":"The direction of the force on a current-carrying conductor in a magnetic field is found using","stemZh":"判断磁场对通电导线作用力方向所用的定则是","options":[{"id":"A","text":"Fleming's right-hand rule."},{"id":"B","text":"Lenz's law."},{"id":"C","text":"Ampère's circuital law."},{"id":"D","text":"Fleming's left-hand rule."}],"answer":"D","explanation":"The motor effect uses Fleming's left-hand rule $(force, field, current)$. The right-hand rule is used for the induced current in a moving conductor, and Lenz's law gives the direction of an induced current.","explanationZh":"电动机效应（通电导线受力）用左手定则；右手定则用于判断导体切割磁感线时的感应电流方向；楞次定律判断感应电流方向。","tags":["basic","electromagnetism","ampere-force"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-magnetic-field-force-004","unitId":"magnetic-field","topicId":"magnetic-field-force","difficulty":2,"stem":"A wire of length $0.20\\ \\mathrm{m}$ is placed in a uniform field of $0.80\\ \\mathrm{T}$. The largest force it can experience is $0.080\\ \\mathrm{N}$. What current does it carry?","stemZh":"$长 0.20\\ \\mathrm{m}$ 的导线放在 $0.80\\ \\mathrm{T}$ 的匀强磁场中，它所受最大安培力为 $0.080\\ \\mathrm{N}，导线中的电流为多大？$","options":[{"id":"A","text":"$0.50\\ \\mathrm{A}$"},{"id":"B","text":"$2.0\\ \\mathrm{A}$"},{"id":"C","text":"$0.20\\ \\mathrm{A}$"},{"id":"D","text":"$5.0\\ \\mathrm{A}$"}],"answer":"A","explanation":"The force is largest when the wire is perpendicular to the field, so $F =$ BIL and $I = \\dfrac{F}{BL} = \\dfrac{0.080}{0.80\\times0.20} = 0.50\\ \\mathrm{A}$.","explanationZh":"$导线垂直磁场时受力最大，F=BIL，故 I=\\dfrac{F}{BL}=\\dfrac{0.080}{0.80\\times0.20}=0.50\\ \\mathrm{A}。$","tags":["basic","electromagnetism","ampere-force"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-magnetic-field-force-005","unitId":"magnetic-field","topicId":"magnetic-field-force","difficulty":3,"stem":"A straight wire of length $0.50\\ \\mathrm{m}$ carrying $2.0\\ \\mathrm{A}$ is placed in a uniform field of $0.40\\ \\mathrm{T}$. The field makes an angle of $30^{\\circ}$ with the direction of the current. What is the force on the wire?","stemZh":"$长 0.50\\ \\mathrm{m}、通 2.0\\ \\mathrm{A}$ 的直导线放在 $0.40\\ \\mathrm{T}$ 的匀强磁场中，磁场方向与电流方向成 $30^{\\circ}$ 角，导线受力为多大？","options":[{"id":"A","text":"$0.40\\ \\mathrm{N}$"},{"id":"B","text":"$0.20\\ \\mathrm{N}$"},{"id":"C","text":"$0.35\\ \\mathrm{N}$"},{"id":"D","text":"$0.10\\ \\mathrm{N}$"}],"answer":"B","explanation":"$F = BIL\\sin\\theta = 0.40\\times2.0\\times0.50\\times\\sin30^{\\circ} = 0.40\\times0.50 = 0.20\\ \\mathrm{N}$.","explanationZh":"$F=BIL\\sin\\theta=0.40\\times2.0\\times0.50\\times\\sin30^{\\circ}=0.40\\times0.50=0.20\\ \\mathrm{N}$","tags":["applied","electromagnetism","ampere-force"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-magnetic-field-force-006","unitId":"magnetic-field","topicId":"magnetic-field-force","difficulty":3,"stem":"A rectangular coil of area $0.50\\ \\mathrm{m^{2}}$ is placed in a uniform field of $0.80\\ \\mathrm{T}$. The normal to the coil makes an angle of $60^{\\circ}$ with the field. What is the magnetic flux through the coil?","stemZh":"面积 $0.50\\ \\mathrm{m^{2}}$ 的矩形线圈放在 $0.80\\ \\mathrm{T}$ 的匀强磁场中，线圈法线与磁场方向成 $60^{\\circ}$ 角，穿过线圈的磁通量为多少？","options":[{"id":"A","text":"$0.40\\ \\mathrm{Wb}$"},{"id":"B","text":"$0.20\\ \\mathrm{Wb}$"},{"id":"C","text":"$0.35\\ \\mathrm{Wb}$"},{"id":"D","text":"$0\\ \\mathrm{Wb}$"}],"answer":"B","explanation":"$\\Phi = BS\\cos\\theta$, where $\\theta$ is the angle between the normal and the $field: \\Phi = 0.80\\times0.50\\times\\cos60^{\\circ} = 0.20\\ \\mathrm{Wb}$.","explanationZh":"$\\Phi=BS\\cos\\theta，\\theta 为法线与磁场方向的夹角：\\Phi=0.80\\times0.50\\times\\cos60^{\\circ}=0.20\\ \\mathrm{Wb}。$","tags":["applied","electromagnetism","magnetic-flux"],"images":[{"src":"/physics/diagrams/magnetic-flux-coil.svg","alt":"A coil seen edge on, with magnetic field arrows at an angle theta to the coil's normal.","altZh":"线圈侧视图，磁场方向 B 与线圈法线 n 成 θ 角。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-magnetic-field-force-007","unitId":"magnetic-field","topicId":"magnetic-field-force","difficulty":3,"stem":"A flat coil of area $0.020\\ \\mathrm{m^{2}}$ is placed with its plane perpendicular to a uniform field of $0.50\\ \\mathrm{T}$. What is the flux through it?","stemZh":"面积 $0.020\\ \\mathrm{m^{2}}$ 的平面线圈，线圈平面与 $0.50\\ \\mathrm{T}$ 的匀强磁场垂直，穿过它的磁通量为多少？","options":[{"id":"A","text":"$0.025\\ \\mathrm{Wb}$"},{"id":"B","text":"$0\\ \\mathrm{Wb}$"},{"id":"C","text":"$0.010\\ \\mathrm{Wb}$"},{"id":"D","text":"$25\\ \\mathrm{Wb}$"}],"answer":"C","explanation":"Perpendicular to the field means the normal is parallel to it, so $\\Phi =$ BS $= 0.50\\times0.020 = 0.010\\ \\mathrm{Wb}$.","explanationZh":"线圈平面与磁场垂直，即法线与磁场平行，故 $\\Phi=BS=0.50\\times0.020=0.010\\ \\mathrm{Wb}。$","tags":["applied","electromagnetism","magnetic-flux"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-magnetic-field-force-008","unitId":"magnetic-field","topicId":"magnetic-field-force","difficulty":3,"stem":"A current-carrying wire is placed parallel to the direction of a uniform magnetic field. What is the force on it?","stemZh":"把通电直导线沿磁场方向（与磁场平行）放置，它受到的安培力为多少？","options":[{"id":"A","text":"BIL"},{"id":"B","text":"$\\dfrac{BIL}{2}$"},{"id":"C","text":"It depends on the length of the wire."},{"id":"D","text":"0"}],"answer":"D","explanation":"$F = BIL\\sin\\theta$, and $\\theta = 0$ when the wire is parallel to the field, so $F = 0$. The force is largest when the wire is perpendicular to the field.","explanationZh":"$F=BIL\\sin\\theta，导线与磁场平行时 \\theta=0，故 F=0。导线垂直磁场时受力最大。$","tags":["applied","electromagnetism","ampere-force"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-magnetic-field-force-009","unitId":"magnetic-field","topicId":"magnetic-field-force","difficulty":4,"stem":"A student is asked for the force on $a 0.50\\ \\mathrm{m}$ wire carrying $2.0\\ \\mathrm{A}$ in $a 0.40\\ \\mathrm{T} field$, with the field at $30^{\\circ}$ to the current, and answers $0.40\\ \\mathrm{N}$. What have they forgotten?","stemZh":"某同学计算“长 $0.50\\ \\mathrm{m}、通 2.0\\ \\mathrm{A}$ 的导线在 $0.40\\ \\mathrm{T}$ 磁场中，磁场与电流成 $30^{\\circ}”的安培力，得 0.40\\ \\mathrm{N}。他遗漏了什么？$","options":[{"id":"A","text":"Nothing $— 0.40\\ \\mathrm{N}$ is correct."},{"id":"B","text":"They used the wrong units."},{"id":"C","text":"They used $F =$ BIL and ignored the $\\sin\\theta$ factor."},{"id":"D","text":"They should have used $\\cos\\theta$."}],"answer":"C","explanation":"The full expression is $F = BIL\\sin\\theta$. Using $F =$ BIL gives $0.40\\ \\mathrm{N}$, but with $\\sin30^{\\circ} = 0.5$ the correct force is $0.20\\ \\mathrm{N}$.","explanationZh":"完整表达式为 $F=BIL\\sin\\theta。用 F=BIL 得 0.40\\ \\mathrm{N}，但 \\sin30^{\\circ}=0.5，正确结果为 0.20\\ \\mathrm{N}。$","tags":["trap","electromagnetism","ampere-force"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-magnetic-field-force-010","unitId":"magnetic-field","topicId":"magnetic-field-force","difficulty":4,"stem":"A flat coil of area $0.50\\ \\mathrm{m^{2}}$ is placed with its plane parallel to the field lines of a uniform $0.80\\ \\mathrm{T}$ field. What is the flux through the coil?","stemZh":"面积 $0.50\\ \\mathrm{m^{2}}$ 的平面线圈，线圈平面与 $0.80\\ \\mathrm{T}$ 匀强磁场的磁感线平行，穿过线圈的磁通量为多少？","options":[{"id":"A","text":"$0.40\\ \\mathrm{Wb}$"},{"id":"B","text":"$0.20\\ \\mathrm{Wb}$"},{"id":"C","text":"0"},{"id":"D","text":"$0.35\\ \\mathrm{Wb}$"}],"answer":"C","explanation":"If the plane is parallel to the field, the normal is perpendicular to it $(\\theta = 90^{\\circ})$, so $\\Phi = BS\\cos90^{\\circ} = 0$. No field lines pass through the coil.","explanationZh":"$线圈平面与磁场平行时，法线与磁场垂直（\\theta=90^{\\circ}），故 \\Phi=BS\\cos90^{\\circ}=0，没有磁感线穿过线圈。$","tags":["trap","electromagnetism","magnetic-flux"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-magnetic-field-force-011","unitId":"magnetic-field","topicId":"magnetic-field-force","difficulty":5,"stem":"A wire carrying current I in a uniform magnetic field experiences a force F directed to the left. The current is now reversed, with everything else unchanged. The force is now","stemZh":"通电导线在匀强磁场中所受安培力为 F，方向向左。若仅把电流方向反向，则此时导线受力为","options":[{"id":"A","text":"F, still to the left."},{"id":"B","text":"zero."},{"id":"C","text":"2F to the left."},{"id":"D","text":"F to the right."}],"answer":"D","explanation":"$F = BIL\\sin\\theta$ depends on the magnitude of the current, so reversing it does not change the size of the force. The direction reverses, because the force is perpendicular to both the current and the field.","explanationZh":"$F=BIL\\sin\\theta$ 只取决于电流大小，反向后力的大小不变。因安培力垂直于电流和磁场，方向随之反向。","tags":["trap","electromagnetism","ampere-force"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-magnetic-field-force-012","unitId":"magnetic-field","topicId":"magnetic-field-force","difficulty":5,"stem":"A coil of area S is turned in a uniform field B from a position where its plane is perpendicular to the field to one where its plane is parallel to the field. By how much does the flux through it change?","stemZh":"面积为 S 的线圈在磁感应强度 B 的匀强磁场中，由线圈平面与磁场垂直转到与磁场平行，磁通量变化了多少？","options":[{"id":"A","text":"BS"},{"id":"B","text":"0"},{"id":"C","text":"2BS"},{"id":"D","text":"$\\dfrac{BS}{2}$"}],"answer":"A","explanation":"Initially $\\Phi_1 = BS$; finally $\\Phi_2 = 0$. The change is $|\\Phi_2 - \\Phi_1| =$ BS.","explanationZh":"初态 $\\Phi_1=BS，末态 \\Phi_2=0，变化量 |\\Phi_2-\\Phi_1|=BS。$","tags":["trap","electromagnetism","magnetic-flux"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-faraday-lenz-001","unitId":"induction","topicId":"faraday-lenz","difficulty":1,"stem":"A coil of 50 turns has the magnetic flux through it changed uniformly from $0.20\\ \\mathrm{Wb}$ to $0.60\\ \\mathrm{Wb}$ in $0.10\\ \\mathrm{s}$. What is the magnitude of the induced emf?","stemZh":"50 匝线圈中的磁通量在 $0.10\\ \\mathrm{s}$ 内从 $0.20\\ \\mathrm{Wb}$ 均匀增至 $0.60\\ \\mathrm{Wb}，感应电动势的大小为多少？$","options":[{"id":"A","text":"$200\\ \\mathrm{V}$"},{"id":"B","text":"$4.0\\ \\mathrm{V}$"},{"id":"C","text":"$20\\ \\mathrm{V}$"},{"id":"D","text":"$2.0\\ \\mathrm{V}$"}],"answer":"A","explanation":"$E = n\\dfrac{\\Delta\\Phi}{\\Delta t} = 50\\times\\dfrac{0.60-0.20}{0.10} = 50\\times4.0 = 200\\ \\mathrm{V}$.","explanationZh":"$E=n\\dfrac{\\Delta\\Phi}{\\Delta t}=50\\times\\dfrac{0.60-0.20}{0.10}=50\\times4.0=200\\ \\mathrm{V}。$","tags":["basic","electromagnetism","faraday"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-faraday-lenz-002","unitId":"induction","topicId":"faraday-lenz","difficulty":1,"stem":"Lenz's law states that the direction of an induced current is such that it","stemZh":"楞次定律指出，感应电流的方向总是使其","options":[{"id":"A","text":"always flows in the same direction as the magnetic field."},{"id":"B","text":"increases the magnetic flux that produced it."},{"id":"C","text":"opposes the change in magnetic flux that produced it."},{"id":"D","text":"is decided only by the resistance of the circuit."}],"answer":"C","explanation":"Lenz's law is the minus sign in $E = -n\\dfrac{\\Delta\\Phi}{\\Delta t}$: the induced current produces a field that opposes the change in flux — not the flux itself.","explanationZh":"楞次定律对应 $E=-n\\dfrac{\\Delta\\Phi}{\\Delta t}$ 中的负号：感应电流的磁场阻碍磁通量的变化，而不是阻碍磁通量本身。","tags":["basic","electromagnetism","lenz"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-faraday-lenz-003","unitId":"induction","topicId":"faraday-lenz","difficulty":1,"stem":"In a single-turn coil the magnetic flux decreases uniformly at $2.0\\ \\mathrm{Wb}$ every second. What is the induced emf?","stemZh":"单匝线圈中的磁通量每秒均匀减少 $2.0\\ \\mathrm{Wb}，感应电动势为多少？$","options":[{"id":"A","text":"$2.0\\ \\mathrm{V}$, increasing with time."},{"id":"B","text":"$2.0\\ \\mathrm{V}$, constant."},{"id":"C","text":"$0\\ \\mathrm{V}$."},{"id":"D","text":"$2.0\\ \\mathrm{V}$, decreasing with time."}],"answer":"B","explanation":"$E = \\left|\\dfrac{\\Delta\\Phi}{\\Delta t}\\right| = 2.0\\ \\mathrm{V}$. Because the flux changes at a constant rate, the emf is constant — it neither grows nor decays.","explanationZh":"$E=\\left|\\dfrac{\\Delta\\Phi}{\\Delta t}\\right|=2.0\\ \\mathrm{V}。磁通量均匀变化，故感应电动势恒定，不随时间增大或减小。$","tags":["basic","electromagnetism","faraday"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-faraday-lenz-004","unitId":"induction","topicId":"faraday-lenz","difficulty":2,"stem":"A closed loop of wire is placed in a steady uniform magnetic field with its plane perpendicular to the field. Which of the following will produce an induced current?","stemZh":"闭合线框放在稳定的匀强磁场中，线框平面与磁场垂直。下列哪种情况会产生感应电流？","options":[{"id":"A","text":"Nothing will — the field is already passing through the loop."},{"id":"B","text":"Making the loop from thicker wire."},{"id":"C","text":"Moving the loop at constant speed along the field direction."},{"id":"D","text":"Rotating the loop about an axis lying in its own plane."}],"answer":"D","explanation":"An induced current needs a change of flux. Rotating the loop changes the effective area, so $\\Phi = BS\\cos\\theta$ changes. Translating the loop in a uniform field does not change the flux at all.","explanationZh":"$产生感应电流的条件是磁通量变化。转动线框使有效面积改变，\\Phi=BS\\cos\\theta$ 随之变化；在匀强磁场中平动线框，磁通量不变。","tags":["basic","electromagnetism","faraday"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-faraday-lenz-005","unitId":"induction","topicId":"faraday-lenz","difficulty":3,"stem":"A 100-turn coil experiences a change of flux of $0.050\\ \\mathrm{Wb}$ in $0.50\\ \\mathrm{s}$. What is the magnitude of the induced emf?","stemZh":"100 匝线圈中磁通量在 $0.50\\ \\mathrm{s}$ 内变化 $0.050\\ \\mathrm{Wb}，感应电动势的大小为多少？$","options":[{"id":"A","text":"$1.0\\ \\mathrm{V}$"},{"id":"B","text":"$250\\ \\mathrm{V}$"},{"id":"C","text":"$10\\ \\mathrm{V}$"},{"id":"D","text":"$0.50\\ \\mathrm{V}$"}],"answer":"C","explanation":"$E = n\\dfrac{\\Delta\\Phi}{\\Delta t} = 100\\times\\dfrac{0.050}{0.50} = 10\\ \\mathrm{V}$.","explanationZh":"$E=n\\dfrac{\\Delta\\Phi}{\\Delta t}=100\\times\\dfrac{0.050}{0.50}=10\\ \\mathrm{V}。$","tags":["applied","electromagnetism","faraday"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-faraday-lenz-006","unitId":"induction","topicId":"faraday-lenz","difficulty":3,"stem":"A single loop of area $0.50\\ \\mathrm{m^{2}}$ sits in a magnetic field whose flux density increases uniformly at $0.40\\ \\mathrm{T/s}$. What is the induced emf?","stemZh":"面积 $0.50\\ \\mathrm{m^{2}}$ 的单匝线圈处在磁感应强度以 $0.40\\ \\mathrm{T/s}$ 均匀增大的磁场中，感应电动势为多少？","options":[{"id":"A","text":"$0.20\\ \\mathrm{V}$"},{"id":"B","text":"$0.80\\ \\mathrm{V}$"},{"id":"C","text":"$1.25\\ \\mathrm{V}$"},{"id":"D","text":"$0\\ \\mathrm{V}$"}],"answer":"A","explanation":"$\\dfrac{\\Delta\\Phi}{\\Delta t} = S\\dfrac{\\Delta B}{\\Delta t} = 0.50\\times0.40 = 0.20\\ \\mathrm{V}$.","explanationZh":"$\\dfrac{\\Delta\\Phi}{\\Delta t}=S\\dfrac{\\Delta B}{\\Delta t}=0.50\\times0.40=0.20\\ \\mathrm{V}。$","tags":["applied","electromagnetism","faraday"],"images":[{"src":"/physics/diagrams/induction-coil-changing-b.svg","alt":"A circular coil in a uniform magnetic field directed into the page, with the field strength increasing.","altZh":"圆形线圈处在垂直纸面向里且正在增强的匀强磁场中。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-faraday-lenz-007","unitId":"induction","topicId":"faraday-lenz","difficulty":3,"stem":"A coil of resistance $5.0\\ \\mathrm{\\Omega}$ has an induced emf of $2.0\\ \\mathrm{V}$ in it. What is the induced current?","stemZh":"电阻 $5.0\\ \\mathrm{\\Omega}$ 的线圈中感应电动势为 $2.0\\ \\mathrm{V}，感应电流为多大？$","options":[{"id":"A","text":"$10\\ \\mathrm{A}$"},{"id":"B","text":"$2.5\\ \\mathrm{A}$"},{"id":"C","text":"$0.10\\ \\mathrm{A}$"},{"id":"D","text":"$0.40\\ \\mathrm{A}$"}],"answer":"D","explanation":"$I = \\dfrac{E}{R} = \\dfrac{2.0}{5.0} = 0.40\\ \\mathrm{A}$.","explanationZh":"$I=\\dfrac{E}{R}=\\dfrac{2.0}{5.0}=0.40\\ \\mathrm{A}。$","tags":["applied","electromagnetism","faraday"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-faraday-lenz-008","unitId":"induction","topicId":"faraday-lenz","difficulty":3,"stem":"The north pole of a bar magnet is pushed towards a coil. What happens at the face of the coil nearest the magnet?","stemZh":"把条形磁铁的 N 极推向线圈，线圈靠近磁铁的一面会出现什么情况？","options":[{"id":"A","text":"It becomes a south pole, attracting the magnet."},{"id":"B","text":"It becomes a north pole, repelling the magnet."},{"id":"C","text":"Nothing — a coil cannot behave like a magnet."},{"id":"D","text":"It becomes a north pole, but only after the magnet has passed through."}],"answer":"B","explanation":"The flux through the coil is increasing, so by Lenz's law the induced current creates a field that opposes the increase: the near face becomes a north pole and repels the approaching magnet.","explanationZh":"穿过线圈的磁通量增大，由楞次定律，感应电流的磁场要阻碍其增大，故靠近磁铁的一面成为 N 极，排斥靠近的磁铁。","tags":["applied","electromagnetism","lenz"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-faraday-lenz-009","unitId":"induction","topicId":"faraday-lenz","difficulty":4,"stem":"A coil of n turns has the flux through it changing at a constant rate. If the number of turns is doubled while the rate of change of flux through each turn stays the same, the induced emf","stemZh":"n 匝线圈中磁通量均匀变化。若匝数变为 2n，且每匝磁通量变化率不变，则感应电动势","options":[{"id":"A","text":"is unchanged."},{"id":"B","text":"is halved."},{"id":"C","text":"is quadrupled."},{"id":"D","text":"is doubled."}],"answer":"D","explanation":"$E = n\\dfrac{\\Delta\\Phi}{\\Delta t}$, so the emf is proportional to the number of turns. Doubling n doubles the emf.","explanationZh":"$E=n\\dfrac{\\Delta\\Phi}{\\Delta t}，感应电动势与匝数成正比，匝数加倍则电动势加倍。$","tags":["trap","electromagnetism","faraday"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-faraday-lenz-010","unitId":"induction","topicId":"faraday-lenz","difficulty":4,"stem":"A bar magnet is dropped so that it falls completely through a fixed horizontal coil. Which statement about the induced current is correct?","stemZh":"让条形磁铁从固定水平线圈中完全穿过，关于感应电流的说法正确的是","options":[{"id":"A","text":"It reverses direction as the magnet passes through the middle of the coil."},{"id":"B","text":"It flows in the same direction throughout."},{"id":"C","text":"It is zero throughout, because the magnet as a whole is neutral."},{"id":"D","text":"It reaches its maximum when the magnet is exactly at the centre of the coil."}],"answer":"A","explanation":"As the magnet approaches, the flux increases; as it leaves, the flux decreases. By Lenz's law the induced current must oppose each change, so it reverses direction. When the magnet is centred the flux is momentarily at a maximum, so $\\dfrac{\\Delta\\Phi}{\\Delta t} —$ and the emf — is zero.","explanationZh":"$磁铁靠近时磁通量增大，离开时减小，由楞次定律感应电流方向必须分别阻碍这两种变化，故方向反转。磁铁位于线圈中心瞬间磁通量最大，\\dfrac{\\Delta\\Phi}{\\Delta t}$ 为零，感应电动势为零。","tags":["trap","electromagnetism","lenz"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-faraday-lenz-011","unitId":"induction","topicId":"faraday-lenz","difficulty":5,"stem":"A rectangular conducting loop lies beside a long straight wire carrying a current that is steadily increasing. According to Lenz's law, the loop","stemZh":"矩形导电线框放在一根电流正在均匀增大的长直导线旁。根据楞次定律，线框将","options":[{"id":"A","text":"is attracted towards the wire."},{"id":"B","text":"rotates until its plane is parallel to the wire."},{"id":"C","text":"is pushed away from the wire."},{"id":"D","text":"experiences no force, because a magnetic field does no work."}],"answer":"C","explanation":"The growing current makes the flux through the loop increase. To oppose that increase the loop moves to where the field is weaker — that is, away from the wire. $(Because \\dfrac{\\Delta B}{\\Delta t}\\text{ is constant here, the induced emf and current are constant }too.)$","explanationZh":"电流增大使穿过线框的磁通量增大，为阻碍其增大，线框将向磁场较弱处运动，即远离导线。（因 $\\dfrac{\\Delta B}{\\Delta t}$ 恒定，感应电动势与感应电流也恒定。）","tags":["trap","electromagnetism","lenz"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-faraday-lenz-012","unitId":"induction","topicId":"faraday-lenz","difficulty":5,"stem":"A coil sits in a changing magnetic field, but its circuit is open. Which statement is correct?","stemZh":"线圈处在变化的磁场中，但电路断开。下列说法正确的是","options":[{"id":"A","text":"There is no induced emf and no induced current."},{"id":"B","text":"There is an induced emf but no induced current."},{"id":"C","text":"There is an induced emf and an induced current."},{"id":"D","text":"There is an induced current but no induced emf."}],"answer":"B","explanation":"Faraday's law gives an emf whenever the flux changes, whether or not the circuit is closed. A current, however, needs a complete circuit, so with the circuit open the emf exists but no current flows.","explanationZh":"只要磁通量变化就有感应电动势，与电路是否闭合无关；而感应电流需要闭合回路，故断路时有电动势无电流。","tags":["trap","electromagnetism","faraday"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-motional-emf-001","unitId":"induction","topicId":"motional-emf","difficulty":1,"stem":"A conducting rod of length $0.40\\ \\mathrm{m}$ moves at $5.0\\ \\mathrm{m/s}$ perpendicular to both its own length and a uniform field of $0.50\\ \\mathrm{T}$. What is the motional emf across its ends?","stemZh":"$长 0.40\\ \\mathrm{m}$ 的导体棒以 $5.0\\ \\mathrm{m/s}$ 的速度运动，速度方向既垂直于棒也垂直于 $0.50\\ \\mathrm{T}$ 的匀强磁场，棒两端的动生电动势为多少？","options":[{"id":"A","text":"$1.0\\ \\mathrm{V}$"},{"id":"B","text":"$0.10\\ \\mathrm{V}$"},{"id":"C","text":"$2.0\\ \\mathrm{V}$"},{"id":"D","text":"$10\\ \\mathrm{V}$"}],"answer":"A","explanation":"$E =$ BLv $= 0.50\\times0.40\\times5.0 = 1.0\\ \\mathrm{V}$.","explanationZh":"$E=BLv=0.50\\times0.40\\times5.0=1.0\\ \\mathrm{V}。$","tags":["basic","electromagnetism","motional-emf"],"images":[{"src":"/physics/diagrams/motional-emf-rod.svg","alt":"A conducting rod of length L sliding to the right along two rails in a magnetic field directed into the page, with a resistor R on the left.","altZh":"长为 L 的导体棒沿导轨向右滑动，磁场垂直纸面向里，左侧接有电阻 R。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-motional-emf-002","unitId":"induction","topicId":"motional-emf","difficulty":1,"stem":"The direction of the induced current in a straight conductor moving through a magnetic field is found using","stemZh":"判断导体切割磁感线时感应电流方向所用的定则是","options":[{"id":"A","text":"Fleming's left-hand rule."},{"id":"B","text":"Lenz's law only."},{"id":"C","text":"Coulomb's law."},{"id":"D","text":"Fleming's right-hand rule."}],"answer":"D","explanation":"For a generator — a conductor moving in a field — the direction follows from Fleming's right-hand rule. The left-hand rule is for the motor effect, the force on a current-carrying conductor.","explanationZh":"发电机情形（导体切割磁感线）用右手定则；左手定则用于电动机效应，即通电导体受力。","tags":["basic","electromagnetism","motional-emf"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-motional-emf-003","unitId":"induction","topicId":"motional-emf","difficulty":2,"stem":"A rod moving in a magnetic field generates an emf of $1.0\\ \\mathrm{V}$ in a circuit of resistance $2.0\\ \\Omega$. What is the induced current?","stemZh":"导体棒在磁场中运动产生 $1.0\\ \\mathrm{V}$ 的电动势，回路电阻 $2.0\\ \\Omega，感应电流为多大？$","options":[{"id":"A","text":"$2.0\\ \\mathrm{A}$"},{"id":"B","text":"$0.50\\ \\mathrm{A}$"},{"id":"C","text":"$0.20\\ \\mathrm{A}$"},{"id":"D","text":"$5.0\\ \\mathrm{A}$"}],"answer":"B","explanation":"$I = \\dfrac{E}{R} = \\dfrac{1.0}{2.0} = 0.50\\ \\mathrm{A}$.","explanationZh":"$I=\\dfrac{E}{R}=\\dfrac{1.0}{2.0}=0.50\\ \\mathrm{A}。$","tags":["basic","electromagnetism","motional-emf"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-motional-emf-004","unitId":"induction","topicId":"motional-emf","difficulty":2,"stem":"A rod of length $0.40\\ \\mathrm{m}$ carrying an induced current of $0.50\\ \\mathrm{A}$ sits perpendicular to a field of $0.50\\ \\mathrm{T}$. What magnetic force acts on it?","stemZh":"$长 0.40\\ \\mathrm{m}$ 的导体棒中感应电流 $0.50\\ \\mathrm{A}，棒垂直于 0.50\\ \\mathrm{T}$ 的磁场，棒受磁场力为多大、方向如何？","options":[{"id":"A","text":"$0.10\\ \\mathrm{N}$, in the direction of motion."},{"id":"B","text":"$0\\ \\mathrm{N}$."},{"id":"C","text":"$0.10\\ \\mathrm{N}$, opposing the motion."},{"id":"D","text":"$1.0\\ \\mathrm{N}$, opposing the motion."}],"answer":"C","explanation":"$F =$ BIL $= 0.50\\times0.50\\times0.40 = 0.10\\ \\mathrm{N}$. By Lenz's law this force opposes the motion that produced the current.","explanationZh":"$F=BIL=0.50\\times0.50\\times0.40=0.10\\ \\mathrm{N}。由楞次定律，该力阻碍引起感应电流的运动。$","tags":["basic","electromagnetism","motional-emf"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-motional-emf-005","unitId":"induction","topicId":"motional-emf","difficulty":3,"stem":"A rod of length $0.40\\ \\mathrm{m}$ moves at $5.0\\ \\mathrm{m/s}$ perpendicular to $a 0.50\\ \\mathrm{T}$ field in a circuit of resistance $2.0\\ \\Omega$. Ignoring friction, what force is needed to keep it moving at constant speed?","stemZh":"$长 0.40\\ \\mathrm{m}$ 的棒以 $5.0\\ \\mathrm{m/s}$ 垂直于 $0.50\\ \\mathrm{T}$ 磁场运动，回路电阻 $2.0\\ \\Omega。不计摩擦，保持匀速所需拉力为多大？$","options":[{"id":"A","text":"$0.10\\ \\mathrm{N}$"},{"id":"B","text":"$0.50\\ \\mathrm{N}$"},{"id":"C","text":"$1.0\\ \\mathrm{N}$"},{"id":"D","text":"$0.20\\ \\mathrm{N}$"}],"answer":"A","explanation":"$E =$ BLv $= 1.0\\ \\mathrm{V}$, so $I = \\dfrac{E}{R} = 0.50\\ \\mathrm{A}$ and the magnetic drag is $F =$ BIL $= 0.10\\ \\mathrm{N}$. Equivalently $F = \\dfrac{B^{2}L^{2}v}{R}$.","explanationZh":"$E=BLv=1.0\\ \\mathrm{V}，I=\\dfrac{E}{R}=0.50\\ \\mathrm{A}，磁阻力 F=BIL=0.10\\ \\mathrm{N}；也可直接用 F=\\dfrac{B^{2}L^{2}v}{R}。$","tags":["applied","electromagnetism","motional-emf"],"images":[{"src":"/physics/diagrams/motional-emf-rod.svg","alt":"A conducting rod of length L sliding to the right along two rails in a magnetic field directed into the page, with a resistor R on the left.","altZh":"长为 L 的导体棒沿导轨向右滑动，磁场垂直纸面向里，左侧接有电阻 R。"}]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-motional-emf-006","unitId":"induction","topicId":"motional-emf","difficulty":3,"stem":"In the situation above, a constant friction force of $0.050\\ \\mathrm{N}$ also opposes the motion. What pull force is now needed for constant speed?","stemZh":"上题中若还有 $0.050\\ \\mathrm{N}$ 的恒定摩擦阻力，保持匀速所需拉力为多大？","options":[{"id":"A","text":"$0.050\\ \\mathrm{N}$"},{"id":"B","text":"$0.10\\ \\mathrm{N}$"},{"id":"C","text":"$0.050\\ \\mathrm{N}$ less than before."},{"id":"D","text":"$0.15\\ \\mathrm{N}$"}],"answer":"D","explanation":"At constant speed the forces $balance: F_{pull} = F_{mag} + f = 0.10 + 0.050 = 0.15\\ \\mathrm{N}$.","explanationZh":"$匀速时受力平衡：F_{拉}=F_{磁}+f=0.10+0.050=0.15\\ \\mathrm{N}。$","tags":["applied","electromagnetism","motional-emf"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-motional-emf-007","unitId":"induction","topicId":"motional-emf","difficulty":3,"stem":"A rod is pulled at $5.0\\ \\mathrm{m/s}$ against a magnetic drag of $0.10\\ \\mathrm{N}$. What mechanical power must be supplied?","stemZh":"导体棒以 $5.0\\ \\mathrm{m/s}$ 匀速运动，需克服 $0.10\\ \\mathrm{N}$ 的磁阻力，需输入的机械功率为多少？","options":[{"id":"A","text":"$0.020\\ \\mathrm{W}$"},{"id":"B","text":"$2.0\\ \\mathrm{W}$"},{"id":"C","text":"$0.50\\ \\mathrm{W}$"},{"id":"D","text":"$50\\ \\mathrm{W}$"}],"answer":"C","explanation":"$P =$ Fv $= 0.10\\times5.0 = 0.50\\ \\mathrm{W}$, which matches the electrical power $\\dfrac{E^{2}}{R} = \\dfrac{1.0^{2}}{2.0} = 0.50\\ \\mathrm{W} —$ as it must, since energy is conserved.","explanationZh":"$P=Fv=0.10\\times5.0=0.50\\ \\mathrm{W}，与电功率 \\dfrac{E^{2}}{R}=\\dfrac{1.0^{2}}{2.0}=0.50\\ \\mathrm{W}$ 一致，符合能量守恒。","tags":["applied","electromagnetism","motional-emf"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-motional-emf-008","unitId":"induction","topicId":"motional-emf","difficulty":3,"stem":"If the speed of the rod is doubled and everything else is unchanged, the force needed to maintain that speed","stemZh":"若导体棒速度加倍、其余条件不变，维持该速度所需的拉力将","options":[{"id":"A","text":"is unchanged."},{"id":"B","text":"doubles."},{"id":"C","text":"quadruples."},{"id":"D","text":"is halved."}],"answer":"B","explanation":"$F = \\dfrac{B^{2}L^{2}v}{R}$ is proportional to v: doubling the speed doubles the emf and the current, and therefore doubles the magnetic drag.","explanationZh":"$F=\\dfrac{B^{2}L^{2}v}{R} 与 v$ 成正比：速度加倍使电动势和电流都加倍，磁阻力也随之加倍。","tags":["applied","electromagnetism","motional-emf"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-motional-emf-009","unitId":"induction","topicId":"motional-emf","difficulty":4,"stem":"The resistance of the circuit is increased while the rod is pulled at the same constant speed. What happens to the force needed?","stemZh":"在导体棒仍保持同样速度的情况下增大回路电阻，所需拉力如何变化？","options":[{"id":"A","text":"It decreases, because the current and the magnetic drag are smaller."},{"id":"B","text":"It increases, because more resistance needs more force."},{"id":"C","text":"It is unchanged, because the emf is unchanged."},{"id":"D","text":"It becomes zero."}],"answer":"A","explanation":"The emf $E =$ BLv is unchanged, but a larger R gives a smaller current, so the drag $F =$ BIL is smaller. The emf does not depend on R — that is the trap.","explanationZh":"电动势 $E=BLv$ 不变，但 R 增大使电流减小，磁阻力 $F=BIL$ 随之减小。电动势与 R 无关，这正是本题陷阱。","tags":["trap","electromagnetism","motional-emf"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-motional-emf-010","unitId":"induction","topicId":"motional-emf","difficulty":4,"stem":"A conducting rod moves along the direction of the magnetic field lines, i.e. parallel to B. What motional emf is produced across its ends?","stemZh":"导体棒沿磁感线方向（与磁场平行）运动，棒两端产生的动生电动势为多少？","options":[{"id":"A","text":"BLv"},{"id":"B","text":"$\\dfrac{BLv}{2}$"},{"id":"C","text":"It depends on the resistance of the circuit."},{"id":"D","text":"0"}],"answer":"D","explanation":"A motional emf appears only when the conductor cuts field lines. Moving parallel to B cuts none of them, so the emf is zero.","explanationZh":"只有导体切割磁感线时才产生动生电动势。沿磁场方向运动不切割磁感线，故电动势为零。","tags":["trap","electromagnetism","motional-emf"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-motional-emf-011","unitId":"induction","topicId":"motional-emf","difficulty":5,"stem":"A rod is pulled through a magnetic field at constant speed. Which statement about the energy is correct?","stemZh":"用外力拉动导体棒在磁场中匀速运动，关于能量的说法正确的是","options":[{"id":"A","text":"The mechanical work done is stored in the magnetic field."},{"id":"B","text":"The mechanical power supplied equals the electrical power dissipated in the circuit, plus any friction losses."},{"id":"C","text":"No work is needed, because a magnetic force cannot do work."},{"id":"D","text":"The electrical power is always greater than the mechanical power supplied."}],"answer":"B","explanation":"Energy is conserved: the work done against the magnetic drag reappears as electrical energy, which is then dissipated as heat in the resistance $(and$, if present, against friction).","explanationZh":"能量守恒：克服磁阻力所做的功转化为电能，再以焦耳热形式消耗在电阻上（若有摩擦则还有摩擦损耗）。","tags":["trap","electromagnetism","motional-emf"]},{"subject":"physics","moduleId":"electromagnetism","type":"single-choice","status":"published","id":"physics-motional-emf-012","unitId":"induction","topicId":"motional-emf","difficulty":5,"stem":"A rod is moving at constant speed through a magnetic field because it is being pulled. The pulling force is suddenly removed. What happens?","stemZh":"导体棒在拉力作用下于磁场中匀速运动，若突然撤去拉力，棒将","options":[{"id":"A","text":"keep moving at the same speed, because there is no friction."},{"id":"B","text":"speed up."},{"id":"C","text":"slow down and eventually stop."},{"id":"D","text":"reverse direction."}],"answer":"C","explanation":"The induced current produces a magnetic force that opposes the motion $(Lenz's law)$, so once the pull is removed the rod decelerates. The magnetic drag can never maintain the motion — it always resists it.","explanationZh":"感应电流产生的磁场力总是阻碍运动（楞次定律），撤去拉力后棒做减速运动直至停止。磁阻力不可能维持运动，它只起阻碍作用。","tags":["trap","electromagnetism","motional-emf"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-quantities-units-vectors-001","unitId":"physics-basics","topicId":"quantities-units-vectors","difficulty":1,"stem":"Which of the following is an SI base unit?","stemZh":"下列哪个是国际单位制（SI）的基本单位？","options":[{"id":"A","text":"$\\mathrm{m} (metre)$"},{"id":"B","text":"$\\mathrm{N} (newton)$"},{"id":"C","text":"$\\mathrm{J} (joule)$"},{"id":"D","text":"$\\mathrm{W} (watt)$"}],"optionsZh":[{"id":"A","text":"$\\mathrm{m}（米）$"},{"id":"B","text":"$\\mathrm{N}（牛顿）$"},{"id":"C","text":"$\\mathrm{J}（焦耳）$"},{"id":"D","text":"$\\mathrm{W}（瓦特）$"}],"answer":"A","explanation":"The metre is one of the seven SI base units $(m, kg, s, A, K, mol, cd)$. The newton, joule and watt are all derived units.","explanationZh":"米是七个 SI 基本单位之一（m、kg、s、A、K、mol、cd）。牛顿、焦耳、瓦特都是导出单位。","tags":["basic","units","si-base"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-quantities-units-vectors-002","unitId":"physics-basics","topicId":"quantities-units-vectors","difficulty":1,"stem":"Which of the following is a base quantity in the SI system?","stemZh":"下列哪个是 SI 基本量？","options":[{"id":"A","text":"Speed"},{"id":"B","text":"Length"},{"id":"C","text":"Force"},{"id":"D","text":"Energy"}],"optionsZh":[{"id":"A","text":"速度"},{"id":"B","text":"长度"},{"id":"C","text":"力"},{"id":"D","text":"能量"}],"answer":"B","explanation":"Length is an SI base quantity. Speed, force and energy are derived from base quantities.","explanationZh":"长度是 SI 基本量，速度、力、能量均由基本量导出。","tags":["basic","units","si-base"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-quantities-units-vectors-003","unitId":"physics-basics","topicId":"quantities-units-vectors","difficulty":1,"stem":"Convert $5\\ \\mathrm{km}$ into metres.","stemZh":"$将 5\\ \\mathrm{km}$ 换算为米。","options":[{"id":"A","text":"$50\\ \\mathrm{m}$"},{"id":"B","text":"$500\\ \\mathrm{m}$"},{"id":"C","text":"$5000\\ \\mathrm{m}$"},{"id":"D","text":"$50000\\ \\mathrm{m}$"}],"answer":"C","explanation":"The prefix $\\mathrm{k}$ means $10^{3}$, so $5\\ \\mathrm{km} = 5 \\times 10^{3}\\ \\mathrm{m} = 5000\\ \\mathrm{m}$.","explanationZh":"前缀 $\\mathrm{k}$ 表示 $10^{3}，故 5\\ \\mathrm{km}=5\\times10^{3}\\ \\mathrm{m}=5000\\ \\mathrm{m}。$","tags":["basic","units","prefix"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-quantities-units-vectors-004","unitId":"physics-basics","topicId":"quantities-units-vectors","difficulty":2,"stem":"The newton, the SI unit of force, is equivalent to which combination of base units?","stemZh":"力的 SI 单位牛顿等价于下列哪组基本单位组合？","options":[{"id":"A","text":"$\\mathrm{kg \\cdot m / s}$"},{"id":"B","text":"$\\mathrm{kg \\cdot m / s^{3}}$"},{"id":"C","text":"$\\mathrm{kg / (m \\cdot s^{2})}$"},{"id":"D","text":"$\\mathrm{kg \\cdot m / s^{2}}$"}],"answer":"D","explanation":"From $F = ma$, the unit is $\\mathrm{kg \\cdot m / s^{2}}$, i.e. one newton.","explanationZh":"$由 F=ma，单位为 \\mathrm{kg\\cdot m/s^{2}}，即 1$ 牛顿。","tags":["basic","units","derived"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-quantities-units-vectors-005","unitId":"physics-basics","topicId":"quantities-units-vectors","difficulty":2,"stem":"Which of the following is a vector quantity?","stemZh":"下列哪个是矢量？","options":[{"id":"A","text":"Displacement"},{"id":"B","text":"Distance"},{"id":"C","text":"Mass"},{"id":"D","text":"Time"}],"optionsZh":[{"id":"A","text":"位移"},{"id":"B","text":"路程"},{"id":"C","text":"质量"},{"id":"D","text":"时间"}],"answer":"A","explanation":"Displacement has both magnitude and direction, so it is a vector. Distance, mass and time are scalars.","explanationZh":"位移既有大小又有方向，是矢量；路程、质量、时间均为标量。","tags":["basic","vector","scalar-vs-vector"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-quantities-units-vectors-006","images":[{"src":"/physics/diagrams/vectors-resultant.svg","alt":"Two perpendicular vectors of 3 m and 4 m; their resultant is the hypotenuse of a 3-4-5 right triangle.","altZh":"大小分别为 3 m 与 4 m 的两个互相垂直的矢量；其合力为 3-4-5 直角三角形的斜边。"}],"unitId":"physics-basics","topicId":"quantities-units-vectors","difficulty":3,"stem":"Two displacement $vectors, \\vec{A} = 3\\hat{i}\\ \\mathrm{m}$ and $\\vec{B} = 4\\hat{j}\\ \\mathrm{m}$, are perpendicular. What is the magnitude of their resultant?","stemZh":"两个位移矢量 $\\vec{A}=3\\hat{i}\\ \\mathrm{m} 与 \\vec{B}=4\\hat{j}\\ \\mathrm{m}$ 互相垂直，合矢量的大小是多少？","options":[{"id":"A","text":"$1\\ \\mathrm{m}$"},{"id":"B","text":"$5\\ \\mathrm{m}$"},{"id":"C","text":"$7\\ \\mathrm{m}$"},{"id":"D","text":"$12\\ \\mathrm{m}$"}],"answer":"B","explanation":"Perpendicular vectors add by $Pythagoras: |\\vec{R}| = \\sqrt{3^{2} + 4^{2}} = 5\\ \\mathrm{m}$.","explanationZh":"$互相垂直的矢量按勾股定理合成：|\\vec{R}|=\\sqrt{3^{2}+4^{2}}=5\\ \\mathrm{m}。$","tags":["applied","vector","resultant","pythagoras"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-quantities-units-vectors-007","unitId":"physics-basics","topicId":"quantities-units-vectors","difficulty":3,"stem":"A vector of magnitude $10\\ \\mathrm{m}$ makes an angle of $30^{\\circ}$ with the x-axis. What is its x-component?","stemZh":"大小为 $10\\ \\mathrm{m}$ 的矢量与 x 轴夹角为 $30^{\\circ}，其 x$ 分量是多少？","options":[{"id":"A","text":"$5.0\\ \\mathrm{m}$"},{"id":"B","text":"$10\\ \\mathrm{m}$"},{"id":"C","text":"$8.7\\ \\mathrm{m}$"},{"id":"D","text":"$17.3\\ \\mathrm{m}$"}],"answer":"C","explanation":"The x-component is $v_{x} = v\\cos\\theta = 10 \\times \\cos 30^{\\circ} \\approx 8.7\\ \\mathrm{m}$.","explanationZh":"x 分量为 $v_{x}=v\\cos\\theta=10\\times\\cos30^{\\circ}\\approx8.7\\ \\mathrm{m}。$","tags":["applied","vector","components"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-quantities-units-vectors-008","unitId":"physics-basics","topicId":"quantities-units-vectors","difficulty":3,"stem":"The pascal $(\\mathrm{Pa})$ is the SI unit of pressure. Which base-unit combination equals one pascal?","stemZh":"$帕斯卡（\\mathrm{Pa}）是压强的$ SI 单位，下列哪个基本单位组合等于 1 帕？","options":[{"id":"A","text":"$\\mathrm{N \\cdot m}$"},{"id":"B","text":"$\\mathrm{N / m}$"},{"id":"C","text":"$\\mathrm{kg / m^{3}}$"},{"id":"D","text":"$\\mathrm{kg / (m \\cdot s^{2})}$"}],"answer":"D","explanation":"$1\\ \\mathrm{Pa} = 1\\ \\mathrm{N/m^{2}}$, and $1\\ \\mathrm{N} = 1\\ \\mathrm{kg \\cdot m / s^{2}}$, so $1\\ \\mathrm{Pa} = \\mathrm{kg / (m \\cdot s^{2})}$.","explanationZh":"$1\\ \\mathrm{Pa}=1\\ \\mathrm{N/m^{2}}，而 1\\ \\mathrm{N}=1\\ \\mathrm{kg\\cdot m/s^{2}}，故 1\\ \\mathrm{Pa}=\\mathrm{kg/(m\\cdot s^{2})}。$","tags":["applied","units","derived"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-quantities-units-vectors-009","unitId":"physics-basics","topicId":"quantities-units-vectors","difficulty":3,"stem":"Convert $2.5\\ \\mathrm{\\mu m}$ into metres.","stemZh":"$将 2.5\\ \\mathrm{\\mu m}$ 换算为米。","options":[{"id":"A","text":"$2.5 \\times 10^{-3}\\ \\mathrm{m}$"},{"id":"B","text":"$2.5 \\times 10^{-9}\\ \\mathrm{m}$"},{"id":"C","text":"$2.5 \\times 10^{-6}\\ \\mathrm{m}$"},{"id":"D","text":"$2.5 \\times 10^{6}\\ \\mathrm{m}$"}],"answer":"C","explanation":"The prefix $\\mu$ means $10^{-6}$, so $2.5\\ \\mathrm{\\mu m} = 2.5 \\times 10^{-6}\\ \\mathrm{m}$.","explanationZh":"前缀 $\\mu$ 表示 $10^{-6}，故 2.5\\ \\mathrm{\\mu m}=2.5\\times10^{-6}\\ \\mathrm{m}。$","tags":["applied","units","prefix"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-quantities-units-vectors-010","unitId":"physics-basics","topicId":"quantities-units-vectors","difficulty":3,"stem":"Two equal forces, each of magnitude $6\\ \\mathrm{N}$, act at an angle of $120^{\\circ}$ to each other. What is the magnitude of their resultant?","stemZh":"两个大小均为 $6\\ \\mathrm{N}$ 的力，夹角为 $120^{\\circ}，合力大小是多少？$","options":[{"id":"A","text":"$12\\ \\mathrm{N}$"},{"id":"B","text":"$6\\ \\mathrm{N}$"},{"id":"C","text":"$0\\ \\mathrm{N}$"},{"id":"D","text":"$3\\ \\mathrm{N}$"}],"answer":"B","explanation":"For two equal forces F at angle $\\theta: R = 2F\\cos\\dfrac{\\theta}{2} = 2 \\times 6 \\times \\cos 60^{\\circ} = 12 \\times 0.5 = 6\\ \\mathrm{N}$.","explanationZh":"两个等大、夹角 $\\theta 的力：R=2F\\cos\\dfrac{\\theta}{2}=2\\times6\\times\\cos60^{\\circ}=12\\times0.5=6\\ \\mathrm{N}。$","tags":["applied","vector","resultant","angle"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-quantities-units-vectors-011","unitId":"physics-basics","topicId":"quantities-units-vectors","difficulty":4,"stem":"A car travels $3\\ \\mathrm{km} east$, then $4\\ \\mathrm{km}$ north. What total distance has it travelled?","stemZh":"汽车向东行驶 $3\\ \\mathrm{km}，再向北行驶 4\\ \\mathrm{km}，它行驶的总路程是多少？$","options":[{"id":"A","text":"$7\\ \\mathrm{km}$"},{"id":"B","text":"$5\\ \\mathrm{km}$"},{"id":"C","text":"$1\\ \\mathrm{km}$"},{"id":"D","text":"$12\\ \\mathrm{km}$"}],"answer":"A","explanation":"Distance is the scalar path $length: 3 + 4 = 7\\ \\mathrm{km}$. Do not confuse it with the displacement of $5\\ \\mathrm{km} (the\\text{ straight-line }result)$.","explanationZh":"$路程是标量路径长：3+4=7\\ \\mathrm{km}。不要与位移 5\\ \\mathrm{km}（直线距离）混淆。$","tags":["trap","scalar-vs-vector","distance-vs-displacement"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-quantities-units-vectors-012","unitId":"physics-basics","topicId":"quantities-units-vectors","difficulty":4,"stem":"An object is acted on by two forces of magnitudes $3\\ \\mathrm{N}$ and $4\\ \\mathrm{N}$. Which of the following cannot be the magnitude of their resultant?","stemZh":"物体受两个大小分别为 $3\\ \\mathrm{N} 和 4\\ \\mathrm{N}$ 的力作用，下列哪个不可能是合力的大小？","options":[{"id":"A","text":"$1\\ \\mathrm{N}$"},{"id":"B","text":"$5\\ \\mathrm{N}$"},{"id":"C","text":"$7\\ \\mathrm{N}$"},{"id":"D","text":"$8\\ \\mathrm{N}$"}],"answer":"D","explanation":"The resultant magnitude lies between $|F_1 - F_2| = 1\\ \\mathrm{N}$ and $F_1 + F_2 = 7\\ \\mathrm{N}. A$ value of $8\\ \\mathrm{N}$ exceeds the sum and is impossible.","explanationZh":"合力大小在 $|F_1-F_2|=1\\ \\mathrm{N} 与 F_1+F_2=7\\ \\mathrm{N} 之间；8\\ \\mathrm{N}$ 超过两力之和，不可能。","tags":["trap","vector","resultant","range"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-quantities-units-vectors-013","unitId":"physics-basics","topicId":"quantities-units-vectors","difficulty":4,"stem":"A person walks $6\\ \\mathrm{m} north$, then $8\\ \\mathrm{m}$ south. What is the magnitude of the net displacement?","stemZh":"某人先向北走 $6\\ \\mathrm{m}，再向南走 8\\ \\mathrm{m}，合位移的大小是多少？$","options":[{"id":"A","text":"$2\\ \\mathrm{m}$"},{"id":"B","text":"$14\\ \\mathrm{m}$"},{"id":"C","text":"$10\\ \\mathrm{m}$"},{"id":"D","text":"$8\\ \\mathrm{m}$"}],"answer":"A","explanation":"Displacement is final minus initial $position: +6 - 8 = -2\\ \\mathrm{m}$; its magnitude is $2\\ \\mathrm{m}$. The path length (distance) would be $14\\ \\mathrm{m}$.","explanationZh":"$位移为末位置减初位置：+6-8=-2\\ \\mathrm{m}，大小为 2\\ \\mathrm{m}。总路程才是 14\\ \\mathrm{m}。$","tags":["trap","scalar-vs-vector","displacement"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-quantities-units-vectors-014","images":[{"src":"/physics/diagrams/vectors-resultant.svg","alt":"An aircraft velocity of 40 km/h east combines with a 30 km/h north wind; the resultant is the 30-40-50 hypotenuse.","altZh":"飞机速度 40 km/h 向东，叠加 30 km/h 向北的风；合力为 30-40-50 直角三角形的斜边。"}],"unitId":"physics-basics","topicId":"quantities-units-vectors","difficulty":5,"stem":"An aircraft flies east at $40\\ \\mathrm{km/h}$ while a wind blows north at $30\\ \\mathrm{km/h}$. What is the magnitude of the aircraft's resultant velocity relative to the ground?","stemZh":"飞机以 $40\\ \\mathrm{km/h}$ 向东飞行，同时风以 $30\\ \\mathrm{km/h}$ 向北吹。飞机相对地面的合速度大小是多少？","options":[{"id":"A","text":"$10\\ \\mathrm{km/h}$"},{"id":"B","text":"$50\\ \\mathrm{km/h}$"},{"id":"C","text":"$70\\ \\mathrm{km/h}$"},{"id":"D","text":"$35\\ \\mathrm{km/h}$"}],"answer":"B","explanation":"The two perpendicular velocities add by $Pythagoras: v = \\sqrt{40^{2} + 30^{2}} = \\sqrt{2500} = 50\\ \\mathrm{km/h}$. It is not the sum 70 nor the difference 10.","explanationZh":"$两个互相垂直的速度按勾股合成：v=\\sqrt{40^{2}+30^{2}}=\\sqrt{2500}=50\\ \\mathrm{km/h}，既不是和 70$ 也不是差 10。","tags":["trap","vector","resultant","relative-velocity"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-quantities-units-vectors-015","unitId":"physics-basics","topicId":"quantities-units-vectors","difficulty":5,"stem":"A device rated at $100\\ \\mathrm{W}$ is expressed in kilowatts. What is its power?","stemZh":"某设备功率为 $100\\ \\mathrm{W}，换算成千瓦是多少？$","options":[{"id":"A","text":"$10\\ \\mathrm{kW}$"},{"id":"B","text":"$1000\\ \\mathrm{kW}$"},{"id":"C","text":"$0.1\\ \\mathrm{kW}$"},{"id":"D","text":"$0.01\\ \\mathrm{kW}$"}],"answer":"C","explanation":"The prefix $\\mathrm{k}$ means $10^{3}$, so $100\\ \\mathrm{W} = \\dfrac{100}{1000}\\ \\mathrm{kW} = 0.1\\ \\mathrm{kW}$.","explanationZh":"前缀 $\\mathrm{k}$ 表示 $10^{3}，故 100\\ \\mathrm{W}=\\dfrac{100}{1000}\\ \\mathrm{kW}=0.1\\ \\mathrm{kW}。$","tags":["trap","units","prefix"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-particle-reference-displacement-001","unitId":"physics-basics","topicId":"particle-reference-displacement","difficulty":1,"stem":"In which situation is it most appropriate to model the object as a particle?","stemZh":"下列哪种情形最适合把物体看作质点？","options":[{"id":"A","text":"Earth orbiting the Sun"},{"id":"B","text":"A spinning top's rotation"},{"id":"C","text":"A rolling tyre's spin"},{"id":"D","text":"A gymnast's somersault"}],"optionsZh":[{"id":"A","text":"地球绕太阳公转"},{"id":"B","text":"陀螺的自转"},{"id":"C","text":"轮胎的滚动"},{"id":"D","text":"体操运动员的空翻"}],"answer":"A","explanation":"When studying orbital motion, Earth's own size is negligible compared with the orbit, so it can be treated as a particle. Rotation and spin require treating the body's size and orientation.","explanationZh":"研究公转时，地球自身大小相对轨道可忽略，可视为质点；自转、滚动、翻滚需考虑物体大小与姿态。","tags":["basic","particle-model"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-particle-reference-displacement-002","unitId":"physics-basics","topicId":"particle-reference-displacement","difficulty":1,"stem":"Distance travelled by an object is a:","stemZh":"物体运动的路程是：","options":[{"id":"A","text":"Vector quantity"},{"id":"B","text":"Scalar quantity"},{"id":"C","text":"Tensor quantity"},{"id":"D","text":"Dimensionless quantity"}],"optionsZh":[{"id":"A","text":"矢量"},{"id":"B","text":"标量"},{"id":"C","text":"张量"},{"id":"D","text":"无量纲量"}],"answer":"B","explanation":"Distance has magnitude only (it is the path $length)$, so it is a scalar. Displacement is the corresponding vector.","explanationZh":"路程只有大小（路径长度），是标量；位移才是对应的矢量。","tags":["basic","scalar-vs-vector","distance"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-particle-reference-displacement-003","unitId":"physics-basics","topicId":"particle-reference-displacement","difficulty":1,"stem":"A person walks $30\\ \\mathrm{m} east$, then $40\\ \\mathrm{m}$ west. What distance have they travelled?","stemZh":"某人向东走 $30\\ \\mathrm{m}，再向西走 40\\ \\mathrm{m}，走过的路程是多少？$","options":[{"id":"A","text":"$10\\ \\mathrm{m}$"},{"id":"B","text":"$40\\ \\mathrm{m}$"},{"id":"C","text":"$70\\ \\mathrm{m}$"},{"id":"D","text":"$50\\ \\mathrm{m}$"}],"answer":"C","explanation":"Distance is the total path $length: 30 + 40 = 70\\ \\mathrm{m}. (The\\text{ displacement would be }10\\ \\mathrm{m} west.)$","explanationZh":"$路程为总路径长：30+40=70\\ \\mathrm{m}。（位移为向西 10\\ \\mathrm{m}。）$","tags":["basic","distance"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-particle-reference-displacement-004","unitId":"physics-basics","topicId":"particle-reference-displacement","difficulty":2,"stem":"Whether an object is described as 'at rest' or 'in motion' depends on:","stemZh":"物体被描述为“静止”还是“运动”，取决于：","options":[{"id":"A","text":"Only the object itself"},{"id":"B","text":"The object's mass"},{"id":"C","text":"Time only"},{"id":"D","text":"The chosen reference frame"}],"optionsZh":[{"id":"A","text":"仅由物体本身决定"},{"id":"B","text":"物体的质量"},{"id":"C","text":"仅由时间决定"},{"id":"D","text":"所选取的参考系"}],"answer":"D","explanation":"Motion is relative: the same object can be at rest in one reference frame and moving in another. The description depends on the chosen reference frame.","explanationZh":"运动具有相对性：同一物体在一个参考系中静止，在另一参考系中运动。描述取决于所选参考系。","tags":["basic","reference-frame"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-particle-reference-displacement-005","unitId":"physics-basics","topicId":"particle-reference-displacement","difficulty":2,"stem":"A car travels $100\\ \\mathrm{m}$ in $20\\ \\mathrm{s}$ at constant speed. What is its average speed?","stemZh":"汽车以恒定速率在 $20\\ \\mathrm{s}$ 内行驶 $100\\ \\mathrm{m}，平均速度（平均速率）多大？$","options":[{"id":"A","text":"$5\\ \\mathrm{m/s}$"},{"id":"B","text":"$20\\ \\mathrm{m/s}$"},{"id":"C","text":"$0.2\\ \\mathrm{m/s}$"},{"id":"D","text":"$2000\\ \\mathrm{m/s}$"}],"answer":"A","explanation":"Average speed $= \\dfrac{total distance}{total time} = \\dfrac{100}{20} = 5\\ \\mathrm{m/s}$.","explanationZh":"平均速率 $=\\dfrac{\\text{总路程}}{\\text{总时间}}=\\dfrac{100}{20}=5\\ \\mathrm{m/s}。$","tags":["basic","average-speed"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-particle-reference-displacement-006","unitId":"physics-basics","topicId":"particle-reference-displacement","difficulty":3,"stem":"A person walks $30\\ \\mathrm{m} east$, then $40\\ \\mathrm{m} north$, taking $10\\ \\mathrm{s}$ altogether. What is the magnitude of their average velocity?","stemZh":"某人先向东走 $30\\ \\mathrm{m}，再向北走 40\\ \\mathrm{m}，共用 10\\ \\mathrm{s}，其平均速度的大小是多少？$","options":[{"id":"A","text":"$7\\ \\mathrm{m/s}$"},{"id":"B","text":"$5\\ \\mathrm{m/s}$"},{"id":"C","text":"$3\\ \\mathrm{m/s}$"},{"id":"D","text":"$0\\ \\mathrm{m/s}$"}],"answer":"B","explanation":"Average velocity uses displacement, not distance. The displacement is $\\sqrt{30^{2}+40^{2}} = 50\\ \\mathrm{m}$, so $|\\vec{v}_{avg}| = \\dfrac{50}{10} = 5\\ \\mathrm{m/s}. (Average\\text{ speed would be }7\\ \\mathrm{m/s}.)$","explanationZh":"平均速度用位移而非路程。位移 $\\sqrt{30^{2}+40^{2}}=50\\ \\mathrm{m}，故 |\\vec{v}_{\\text{avg}}|=\\dfrac{50}{10}=5\\ \\mathrm{m/s}。（平均速率为 7\\ \\mathrm{m/s}。）$","tags":["applied","average-velocity","displacement"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-particle-reference-displacement-007","unitId":"physics-basics","topicId":"particle-reference-displacement","difficulty":3,"stem":"A car drives once around a circular track of radius $R = 10\\ \\mathrm{m}$ and returns to its starting point. What is its displacement for the lap?","stemZh":"汽车绕半径 $R=10\\ \\mathrm{m}$ 的环形跑道行驶一圈回到起点，这一圈的位移是多少？","options":[{"id":"A","text":"$62.8\\ \\mathrm{m}$"},{"id":"B","text":"$20\\ \\mathrm{m}$"},{"id":"C","text":"$0\\ \\mathrm{m}$"},{"id":"D","text":"$31.4\\ \\mathrm{m}$"}],"answer":"C","explanation":"Displacement is the change in position. Since the car returns to the start, final and initial positions coincide, so displacement is $0\\ \\mathrm{m} (though\\text{ the distance travelled is the circumference }2\\pi R \\approx 62.8\\ \\mathrm{m})$.","explanationZh":"位移是位置的变化。汽车回到起点，末位置与初位置重合，位移为 $0\\ \\mathrm{m}（尽管路程等于周长 2\\pi R\\approx62.8\\ \\mathrm{m}）。$","tags":["applied","displacement","circular-motion"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-particle-reference-displacement-008","unitId":"physics-basics","topicId":"particle-reference-displacement","difficulty":3,"stem":"An object moves from position $x = +3\\ \\mathrm{m}$ to $x = -5\\ \\mathrm{m}$ along a line. What is its displacement?","stemZh":"物体沿直线从位置 $x=+3\\ \\mathrm{m}$ 运动到 $x=-5\\ \\mathrm{m}，其位移是多少？$","options":[{"id":"A","text":"$+8\\ \\mathrm{m}$"},{"id":"B","text":"$+2\\ \\mathrm{m}$"},{"id":"C","text":"$-2\\ \\mathrm{m}$"},{"id":"D","text":"$-8\\ \\mathrm{m}$"}],"answer":"D","explanation":"Displacement $\\Delta x = x_{final} - x_{initial} = -5 - (+3) = -8\\ \\mathrm{m}$. The minus sign shows the direction is toward negative x.","explanationZh":"位移 $\\Delta x=x_{\\text{末}}-x_{\\text{初}}=-5-(+3)=-8\\ \\mathrm{m}。负号表示沿 x$ 负方向。","tags":["applied","displacement","position"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-particle-reference-displacement-009","unitId":"physics-basics","topicId":"particle-reference-displacement","difficulty":3,"stem":"A car travels $60\\ \\mathrm{m}$ at $10\\ \\mathrm{m/s}$, then a further $60\\ \\mathrm{m}$ at $20\\ \\mathrm{m/s}$. What is its average speed for the whole trip?","stemZh":"汽车以 $10\\ \\mathrm{m/s}$ 行驶 $60\\ \\mathrm{m}，再以 20\\ \\mathrm{m/s}$ 行驶 $60\\ \\mathrm{m}，全程的平均速率是多少？$","options":[{"id":"A","text":"$15\\ \\mathrm{m/s}$"},{"id":"B","text":"$12\\ \\mathrm{m/s}$"},{"id":"C","text":"$13.3\\ \\mathrm{m/s}$"},{"id":"D","text":"$20\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"Use total distance over total time. $Times: t_1 = \\dfrac{60}{10} = 6\\ \\mathrm{s}, t_2 = \\dfrac{60}{20} = 3\\ \\mathrm{s}$. Average speed $= \\dfrac{120}{9} \\approx 13.3\\ \\mathrm{m/s}$, not the simple mean of the two speeds $(15\\ \\mathrm{m/s})$.","explanationZh":"$用总路程除以总时间。时间：t_1=\\dfrac{60}{10}=6\\ \\mathrm{s}，t_2=\\dfrac{60}{20}=3\\ \\mathrm{s}。平均速率 =\\dfrac{120}{9}\\approx13.3\\ \\mathrm{m/s}，不是两速率的算术平均（15\\ \\mathrm{m/s}）。$","tags":["applied","average-speed"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-particle-reference-displacement-010","unitId":"physics-basics","topicId":"particle-reference-displacement","difficulty":3,"stem":"A train moves east at $20\\ \\mathrm{m/s}. A$ passenger walks east at $2\\ \\mathrm{m/s}$ relative to the train. What is the passenger's speed relative to the ground?","stemZh":"列车以 $20\\ \\mathrm{m/s}$ 向东行驶，乘客相对列车以 $2\\ \\mathrm{m/s}$ 向东行走，乘客相对地面的速度多大？","options":[{"id":"A","text":"$18\\ \\mathrm{m/s}$"},{"id":"B","text":"$22\\ \\mathrm{m/s}$"},{"id":"C","text":"$2\\ \\mathrm{m/s}$"},{"id":"D","text":"$0\\ \\mathrm{m/s}$"}],"answer":"B","explanation":"Velocities in the same direction add in the ground $frame: v = 20 + 2 = 22\\ \\mathrm{m/s}$.","explanationZh":"$同方向速度在地面参考系中相加：v=20+2=22\\ \\mathrm{m/s}。$","tags":["applied","reference-frame","relative-velocity"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-particle-reference-displacement-011","unitId":"physics-basics","topicId":"particle-reference-displacement","difficulty":4,"stem":"A person walks $10\\ \\mathrm{m} east$, then $10\\ \\mathrm{m} west$, returning to the start. What is their displacement?","stemZh":"某人向东走 $10\\ \\mathrm{m}，再向西走 10\\ \\mathrm{m}$ 回到起点，其位移是多少？","options":[{"id":"A","text":"$0\\ \\mathrm{m}$"},{"id":"B","text":"$10\\ \\mathrm{m}$"},{"id":"C","text":"$20\\ \\mathrm{m}$"},{"id":"D","text":"$14.1\\ \\mathrm{m}$"}],"answer":"A","explanation":"The final position equals the initial position, so displacement is $0\\ \\mathrm{m}$. The distance travelled is $20\\ \\mathrm{m} — a$ common trap is to report the distance instead.","explanationZh":"末位置等于初位置，故位移为 $0\\ \\mathrm{m}。走过的总路程为 20\\ \\mathrm{m}——常见错误是答成路程。$","tags":["trap","displacement","distance-vs-displacement"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-particle-reference-displacement-012","unitId":"physics-basics","topicId":"particle-reference-displacement","difficulty":4,"stem":"A car drives $240\\ \\mathrm{km}$ east in $4\\ \\mathrm{h}$, then $240\\ \\mathrm{km}$ west in $4\\ \\mathrm{h}$, returning home. What is its average speed?","stemZh":"汽车以 $4\\ \\mathrm{h}$ 向东行驶 $240\\ \\mathrm{km}，再以 4\\ \\mathrm{h}$ 向西行驶 $240\\ \\mathrm{km}$ 回到家，其平均速率是多少？","options":[{"id":"A","text":"$0\\ \\mathrm{km/h}$"},{"id":"B","text":"$120\\ \\mathrm{km/h}$"},{"id":"C","text":"$30\\ \\mathrm{km/h}$"},{"id":"D","text":"$60\\ \\mathrm{km/h}$"}],"answer":"D","explanation":"Average speed $=$ total distance / total time $= 480\\ \\mathrm{km} / 8\\ \\mathrm{h} = 60\\ \\mathrm{km/h}$. The average velocity is $0 (back$ to start) — do not confuse the two.","explanationZh":"平均速率 $=$ 总路程 / 总时间 $=240\\times2/8=60\\ \\mathrm{km/h}。平均速度为零（回到起点）——二者勿混。$","tags":["trap","average-speed","average-velocity"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-particle-reference-displacement-013","unitId":"physics-basics","topicId":"particle-reference-displacement","difficulty":4,"stem":"A car drives $50\\ \\mathrm{m} forward$, then reverses $30\\ \\mathrm{m}$. What is the magnitude of its net displacement?","stemZh":"汽车向前行驶 $50\\ \\mathrm{m}，再倒退 30\\ \\mathrm{m}，其合位移的大小是多少？$","options":[{"id":"A","text":"$20\\ \\mathrm{m}$"},{"id":"B","text":"$80\\ \\mathrm{m}$"},{"id":"C","text":"$50\\ \\mathrm{m}$"},{"id":"D","text":"$30\\ \\mathrm{m}$"}],"answer":"A","explanation":"Net displacement $= +50 - 30 = 20\\ \\mathrm{m}$. The distance travelled is $80\\ \\mathrm{m}$; the trap is to add instead of subtract.","explanationZh":"合位移 $=+50-30=20\\ \\mathrm{m}。总路程为 80\\ \\mathrm{m}；易错点是把两者相加。$","tags":["trap","displacement","direction"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-particle-reference-displacement-014","unitId":"physics-basics","topicId":"particle-reference-displacement","difficulty":5,"stem":"An object moves from $x = 0$ to $x = 20\\ \\mathrm{m}$ in $5\\ \\mathrm{s}$, then back to $x = 0$ in a further $5\\ \\mathrm{s}$. What is its average velocity over the full $10\\ \\mathrm{s}$?","stemZh":"物体 $5\\ \\mathrm{s}$ 内从 $x=0$ 运动到 $x=20\\ \\mathrm{m}，再用 5\\ \\mathrm{s}$ 回到 $x=0。这 10\\ \\mathrm{s}$ 内的平均速度是多少？","options":[{"id":"A","text":"$4\\ \\mathrm{m/s}$"},{"id":"B","text":"$0\\ \\mathrm{m/s}$"},{"id":"C","text":"$2\\ \\mathrm{m/s}$"},{"id":"D","text":"$8\\ \\mathrm{m/s}$"}],"answer":"B","explanation":"Average velocity $=$ total displacement / total time. The object ends where it started, so total displacement $= 0$; hence average velocity $= 0$, even though the average speed is $2\\ \\mathrm{m/s}$.","explanationZh":"平均速度 $=$ 总位移 / 总时间。物体回到起点，总位移为 $0，故平均速度为 0，尽管平均速率为 2\\ \\mathrm{m/s}。$","tags":["trap","average-velocity","displacement"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-particle-reference-displacement-015","unitId":"physics-basics","topicId":"particle-reference-displacement","difficulty":5,"stem":"A ball is dropped inside a train moving at constant velocity. Relative to the ground, what happens to the ball's horizontal motion as it falls?","stemZh":"在匀速行驶的列车内释放一个小球，相对地面，小球下落过程中的水平运动如何？","options":[{"id":"A","text":"It stops horizontally"},{"id":"B","text":"It accelerates horizontally"},{"id":"C","text":"It continues at the train's horizontal speed"},{"id":"D","text":"It reverses direction"}],"optionsZh":[{"id":"A","text":"水平方向立即停止"},{"id":"B","text":"水平方向加速"},{"id":"C","text":"保持列车的水平速度"},{"id":"D","text":"水平方向反向"}],"answer":"C","explanation":"With no horizontal force in the ground frame, the ball keeps the train's horizontal velocity (Galilean relativity). It follows a parabola to the ground, not a vertical line.","explanationZh":"地面参考系中无水平力，小球保持列车的水平速度（伽利略相对性），相对地面作抛物线而非竖直下落。","tags":["trap","reference-frame","relative-motion"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-motion-graphs-001","unitId":"kinematics","topicId":"motion-graphs","difficulty":1,"stem":"On a displacement–time $(s–t) graph$, what physical quantity does the slope of the graph represent?","stemZh":"在位移—时间（s–t）图像中，图线的斜率表示什么物理量？","options":[{"id":"A","text":"Velocity"},{"id":"B","text":"Acceleration"},{"id":"C","text":"Distance"},{"id":"D","text":"Speed only"}],"optionsZh":[{"id":"A","text":"速度"},{"id":"B","text":"加速度"},{"id":"C","text":"路程"},{"id":"D","text":"仅为速率"}],"answer":"A","explanation":"The gradient of an s–t graph is $\\dfrac{ds}{dt}$, which equals velocity (including sign).","explanationZh":"s–t 图线的斜率是 $\\dfrac{ds}{dt}，等于速度（含方向）。$","tags":["basic","graph","slope","s-t"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-motion-graphs-002","unitId":"kinematics","topicId":"motion-graphs","difficulty":1,"stem":"On a velocity–time $(v–t) graph$, what does the slope of the graph represent?","stemZh":"在速度—时间（v–t）图像中，图线的斜率表示什么？","options":[{"id":"A","text":"Velocity"},{"id":"B","text":"Acceleration"},{"id":"C","text":"Displacement"},{"id":"D","text":"Distance"}],"optionsZh":[{"id":"A","text":"速度"},{"id":"B","text":"加速度"},{"id":"C","text":"位移"},{"id":"D","text":"路程"}],"answer":"B","explanation":"The gradient of a v–t graph is $\\dfrac{dv}{dt}$, which equals acceleration.","explanationZh":"v–t 图线的斜率是 $\\dfrac{dv}{dt}，等于加速度。$","tags":["basic","graph","slope","v-t"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-motion-graphs-003","unitId":"kinematics","topicId":"motion-graphs","difficulty":1,"stem":"On a velocity–time $(v–t) graph$, what does the area under the curve represent?","stemZh":"在速度—时间（v–t）图像中，曲线下的面积表示什么？","options":[{"id":"A","text":"Acceleration"},{"id":"B","text":"Velocity"},{"id":"C","text":"Displacement"},{"id":"D","text":"Force"}],"optionsZh":[{"id":"A","text":"加速度"},{"id":"B","text":"速度"},{"id":"C","text":"位移"},{"id":"D","text":"力"}],"answer":"C","explanation":"The area between the v–t curve and the time axis equals displacement $(\\int v\\,dt)$.","explanationZh":"v–t $曲线与时间轴围成的面积等于位移（\\int v\\,dt）。$","tags":["basic","graph","area","v-t"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-motion-graphs-004","unitId":"kinematics","topicId":"motion-graphs","difficulty":2,"stem":"A horizontal (flat) line on an s–t graph means the object is:","stemZh":"s–t 图像中的水平直线表示物体：","options":[{"id":"A","text":"Moving at constant velocity"},{"id":"B","text":"Accelerating"},{"id":"C","text":"Moving at constant speed"},{"id":"D","text":"At rest"}],"optionsZh":[{"id":"A","text":"匀速运动"},{"id":"B","text":"加速运动"},{"id":"C","text":"匀速（速率）运动"},{"id":"D","text":"静止"}],"answer":"D","explanation":"A horizontal s–t line has zero slope, so velocity is zero — the object is at rest (position is not changing).","explanationZh":"s–t 水平线斜率为零，速度为零，物体静止（位置不变）。","tags":["basic","graph","s-t","rest"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-motion-graphs-005","unitId":"kinematics","topicId":"motion-graphs","difficulty":2,"stem":"A straight line through the origin on a v–t graph with positive slope indicates:","stemZh":"v–t 图像中过原点、斜率为正的直线表示：","options":[{"id":"A","text":"Uniform (constant) acceleration"},{"id":"B","text":"Constant velocity"},{"id":"C","text":"At rest"},{"id":"D","text":"Zero acceleration"}],"optionsZh":[{"id":"A","text":"匀加速（加速度恒定）"},{"id":"B","text":"匀速运动"},{"id":"C","text":"静止"},{"id":"D","text":"加速度为零"}],"answer":"A","explanation":"A straight v–t line has constant slope, so acceleration is constant (uniform acceleration); starting from the origin means $v_0 = 0$.","explanationZh":"v–t 直线斜率恒定，故加速度恒定（匀加速）；过原点表示初速度 $v_0=0。$","tags":["basic","graph","v-t","constant-accel"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-motion-graphs-006","images":[{"src":"/physics/diagrams/vt-graph.svg","alt":"A velocity-time graph that rises as a straight line from $(0 s, 0 m/s)$ to $(4 s, 8 m/s)$.","altZh":"$v$–$t$ 图像：一条从 (0 s, 0 m/s) 到 (4 s, 8 m/s) 的直线。"}],"unitId":"kinematics","topicId":"motion-graphs","difficulty":3,"stem":"A v–t graph is a straight line from $(0\\ \\mathrm{s}, 0\\ \\mathrm{m/s})$ to $(4\\ \\mathrm{s}, 8\\ \\mathrm{m/s})$. What is the acceleration?","stemZh":"某 v–t 图像是一条从 $(0\\ \\mathrm{s},0\\ \\mathrm{m/s}) 到 (4\\ \\mathrm{s},8\\ \\mathrm{m/s})$ 的直线，加速度多大？","options":[{"id":"A","text":"$4\\ \\mathrm{m/s^{2}}$"},{"id":"B","text":"$2\\ \\mathrm{m/s^{2}}$"},{"id":"C","text":"$8\\ \\mathrm{m/s^{2}}$"},{"id":"D","text":"$0.5\\ \\mathrm{m/s^{2}}$"}],"answer":"B","explanation":"Slope $= \\dfrac{\\Delta v}{\\Delta t} = \\dfrac{8 - 0}{4 - 0} = 2\\ \\mathrm{m/s^{2}}$, which is the acceleration.","explanationZh":"斜率 $=\\dfrac{\\Delta v}{\\Delta t}=\\dfrac{8-0}{4-0}=2\\ \\mathrm{m/s^{2}}，即加速度。$","tags":["applied","graph","v-t","slope"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-motion-graphs-007","unitId":"kinematics","topicId":"motion-graphs","difficulty":3,"stem":"An s–t graph is a straight line from $(0\\ \\mathrm{s}, 0\\ \\mathrm{m})$ to $(5\\ \\mathrm{s}, 20\\ \\mathrm{m})$. What is the object's velocity?","stemZh":"某 s–t 图像是一条从 $(0\\ \\mathrm{s},0\\ \\mathrm{m}) 到 (5\\ \\mathrm{s},20\\ \\mathrm{m})$ 的直线，物体的速度多大？","options":[{"id":"A","text":"$5\\ \\mathrm{m/s}$"},{"id":"B","text":"$20\\ \\mathrm{m/s}$"},{"id":"C","text":"$4\\ \\mathrm{m/s}$"},{"id":"D","text":"$0.25\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"Slope $= \\dfrac{\\Delta s}{\\Delta t} = \\dfrac{20 - 0}{5 - 0} = 4\\ \\mathrm{m/s}$, which is the velocity.","explanationZh":"斜率 $=\\dfrac{\\Delta s}{\\Delta t}=\\dfrac{20-0}{5-0}=4\\ \\mathrm{m/s}，即速度。$","tags":["applied","graph","s-t","slope"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-motion-graphs-008","images":[{"src":"/physics/diagrams/vt-graph.svg","alt":"A velocity-time graph showing a constant 6 m/s from 0 s to $5 s (a$ horizontal rectangle).","altZh":"$v$–$t$ 图像：0 s 到 5 s 速度保持 6 m/s（水平矩形）。"}],"unitId":"kinematics","topicId":"motion-graphs","difficulty":3,"stem":"A v–t graph shows a constant velocity of $6\\ \\mathrm{m/s}$ from $t = 0$ to $t = 5\\ \\mathrm{s}$. What displacement occurs in this time?","stemZh":"某 v–t 图像显示速度在 $t=0 到 t=5\\ \\mathrm{s}$ 恒为 $6\\ \\mathrm{m/s}，这段时间的位移是多少？$","options":[{"id":"A","text":"$5\\ \\mathrm{m}$"},{"id":"B","text":"$6\\ \\mathrm{m}$"},{"id":"C","text":"$11\\ \\mathrm{m}$"},{"id":"D","text":"$30\\ \\mathrm{m}$"}],"answer":"D","explanation":"Displacement $=$ area under the graph $=$ rectangle $6 \\times 5 = 30\\ \\mathrm{m}$.","explanationZh":"位移 $=$ 图线下的面积 $=$ 矩形 $6\\times5=30\\ \\mathrm{m}。$","tags":["applied","graph","v-t","area"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-motion-graphs-009","images":[{"src":"/physics/diagrams/at-graph.svg","alt":"An acceleration-time graph showing a constant $3 m/s^{2}$ from 0 s to 4 s.","altZh":"$a$–$t$ 图像：0 s 到 4 s 加速度恒为 3 m/s^{2}。"}],"unitId":"kinematics","topicId":"motion-graphs","difficulty":3,"stem":"An a–t graph shows a constant acceleration of $3\\ \\mathrm{m/s^{2}}$ from $t = 0$ to $t = 4\\ \\mathrm{s}$, starting from rest. What is the final velocity?","stemZh":"某 a–t 图像显示加速度在 $t=0 到 t=4\\ \\mathrm{s}$ 恒为 $3\\ \\mathrm{m/s^{2}}，且初速度为零，末速度多大？$","options":[{"id":"A","text":"$3\\ \\mathrm{m/s}$"},{"id":"B","text":"$4\\ \\mathrm{m/s}$"},{"id":"C","text":"$12\\ \\mathrm{m/s}$"},{"id":"D","text":"$7\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"The area under an a–t graph equals the change in $velocity: \\Delta v = 3 \\times 4 = 12\\ \\mathrm{m/s}$. Starting from $rest, v = 12\\ \\mathrm{m/s}$.","explanationZh":"a–t $图线下的面积等于速度变化量：\\Delta v=3\\times4=12\\ \\mathrm{m/s}。初速为零，故 v=12\\ \\mathrm{m/s}。$","tags":["applied","graph","a-t","area"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-motion-graphs-010","images":[{"src":"/physics/diagrams/vt-graph.svg","alt":"A velocity-time graph shaped as a triangle: 0 to 10 m/s in 4 s, then back to 0 in 4 s.","altZh":"$v$–$t$ 图像呈三角形：4 s 内从 0 升到 10 m/s，再用 4 s 回到 0。"}],"unitId":"kinematics","topicId":"motion-graphs","difficulty":3,"stem":"A v–t graph rises linearly from 0 to $10\\ \\mathrm{m/s}$ in $4\\ \\mathrm{s}$, then falls linearly back to 0 in a further $4\\ \\mathrm{s}$. What is the total displacement?","stemZh":"某 v–t 图像在 $4\\ \\mathrm{s}$ 内由 0 线性升到 $10\\ \\mathrm{m/s}，再用 4\\ \\mathrm{s}$ 线性回降到 0，总位移是多少？","options":[{"id":"A","text":"$20\\ \\mathrm{m}$"},{"id":"B","text":"$40\\ \\mathrm{m}$"},{"id":"C","text":"$80\\ \\mathrm{m}$"},{"id":"D","text":"$10\\ \\mathrm{m}$"}],"answer":"B","explanation":"The graph is a triangle of base $8\\ \\mathrm{s}$ and height $10\\ \\mathrm{m/s}$. Area $= \\dfrac{1}{2} \\times 8 \\times 10 = 40\\ \\mathrm{m} =$ displacement.","explanationZh":"图线为底边 $8\\ \\mathrm{s}、高 10\\ \\mathrm{m/s}$ 的三角形。面积 $=\\dfrac{1}{2}\\times8\\times10=40\\ \\mathrm{m}，即位移。$","tags":["applied","graph","v-t","area","triangle"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-motion-graphs-011","unitId":"kinematics","topicId":"motion-graphs","difficulty":4,"stem":"A v–t graph is a straight line from $+6\\ \\mathrm{m/s}$ at $t = 0$ to $-6\\ \\mathrm{m/s}$ at $t = 6\\ \\mathrm{s}$, crossing zero at $t = 3\\ \\mathrm{s}$. What is the displacement over the $6\\ \\mathrm{s}$?","stemZh":"某 v–t 图像是一条从 $t=0 的 +6\\ \\mathrm{m/s}$ 线性降到 $t=6\\ \\mathrm{s} 的 -6\\ \\mathrm{m/s}、在 t=3\\ \\mathrm{s} 过零的直线。6\\ \\mathrm{s}$ 内的位移是多少？","options":[{"id":"A","text":"$0\\ \\mathrm{m}$"},{"id":"B","text":"$18\\ \\mathrm{m}$"},{"id":"C","text":"$9\\ \\mathrm{m}$"},{"id":"D","text":"$36\\ \\mathrm{m}$"}],"answer":"A","explanation":"Displacement $=$ signed area: positive triangle $\\dfrac{1}{2}\\times3\\times6 = 9$ minus negative triangle $\\dfrac{1}{2}\\times3\\times6 = 9$, giving 0. The distance travelled would be $18\\ \\mathrm{m} —$ the trap is to report distance.","explanationZh":"位移 $=$ 有符号面积：上方三角形 $\\dfrac{1}{2}\\times3\\times6=9$ 减去下方三角形 $9，得 0。走过的路程才是 18\\ \\mathrm{m}——易错点是答成路程。$","tags":["trap","graph","v-t","signed-area"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-motion-graphs-012","unitId":"kinematics","topicId":"motion-graphs","difficulty":4,"stem":"Position is given by $s = t^{2} (in metres)$. What is the average velocity between $t = 0$ and $t = 4\\ \\mathrm{s}$?","stemZh":"位置满足 $s=t^{2}（单位：米）。t=0 到 t=4\\ \\mathrm{s}$ 间的平均速度是多少？","options":[{"id":"A","text":"$8\\ \\mathrm{m/s}$"},{"id":"B","text":"$16\\ \\mathrm{m/s}$"},{"id":"C","text":"$2\\ \\mathrm{m/s}$"},{"id":"D","text":"$4\\ \\mathrm{m/s}$"}],"answer":"D","explanation":"Average velocity $=$ total displacement / total time $= (s(4)-s(0))/4 = (16-0)/4 = 4\\ \\mathrm{m/s}$. The instantaneous velocity at $t=4$ is $8\\ \\mathrm{m/s} —$ do not confuse average with instantaneous.","explanationZh":"平均速度 $=$ 总位移 / 总时间 $=(s(4)-s(0))/4=(16-0)/4=4\\ \\mathrm{m/s}。t=4$ 时的瞬时速度为 $8\\ \\mathrm{m/s}——勿将平均与瞬时混淆。$","tags":["trap","graph","average-vs-instantaneous"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-motion-graphs-013","unitId":"kinematics","topicId":"motion-graphs","difficulty":4,"stem":"On an s–t graph, a horizontal line segment (constant position) indicates that the object is:","stemZh":"在 s–t 图像中，水平线段（位置不变）表示物体：","options":[{"id":"A","text":"At rest"},{"id":"B","text":"Moving at constant velocity"},{"id":"C","text":"Moving with constant acceleration"},{"id":"D","text":"Moving backward"}],"optionsZh":[{"id":"A","text":"静止"},{"id":"B","text":"匀速运动"},{"id":"C","text":"匀加速运动"},{"id":"D","text":"向后运动"}],"answer":"A","explanation":"Zero slope on an s–t graph means zero velocity — the object is at rest. (A horizontal line on a v–t graph would instead mean constant velocity.)","explanationZh":"s–t 图斜率为零表示速度为零，物体静止。（若是 v–t 图的水平线，则表示匀速。）","tags":["trap","graph","s-t","interpretation"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-motion-graphs-014","unitId":"kinematics","topicId":"motion-graphs","difficulty":5,"stem":"A v–t graph rises linearly from 0 to $8\\ \\mathrm{m/s}$ in $4\\ \\mathrm{s}$, then falls linearly from $8\\ \\mathrm{m/s}$ to 0 in a further $2\\ \\mathrm{s}$. What is the total displacement?","stemZh":"某 v–t 图像在 $4\\ \\mathrm{s}$ 内由 0 线性升到 $8\\ \\mathrm{m/s}，再用 2\\ \\mathrm{s}$ 线性降到 0，总位移是多少？","options":[{"id":"A","text":"$8\\ \\mathrm{m}$"},{"id":"B","text":"$24\\ \\mathrm{m}$"},{"id":"C","text":"$32\\ \\mathrm{m}$"},{"id":"D","text":"$16\\ \\mathrm{m}$"}],"answer":"B","explanation":"Area $=$ first triangle $\\dfrac{1}{2}\\times4\\times8 = 16\\ \\mathrm{m}$ plus second triangle $\\dfrac{1}{2}\\times2\\times8 = 8\\ \\mathrm{m}$, total $24\\ \\mathrm{m}$. The trap is to treat the whole $6\\ \\mathrm{s}$ as one triangle of base 6, which would wrongly give 24 only by luck-free reasoning — check each region separately.","explanationZh":"面积 $=$ 第一个三角形 $\\dfrac{1}{2}\\times4\\times8=16\\ \\mathrm{m}$ 加第二个三角形 $\\dfrac{1}{2}\\times2\\times8=8\\ \\mathrm{m}，共 24\\ \\mathrm{m}。应分段计算面积。$","tags":["trap","graph","v-t","area","multi-region"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-motion-graphs-015","unitId":"kinematics","topicId":"motion-graphs","difficulty":5,"stem":"A v–t graph falls linearly from $10\\ \\mathrm{m/s}$ at $t = 0$ to 0 at $t = 5\\ \\mathrm{s}$. What is the acceleration (including sign)?","stemZh":"某 v–t 图像从 $t=0 的 10\\ \\mathrm{m/s}$ 线性降到 $t=5\\ \\mathrm{s} 的 0，加速度（含符号）是多少？$","options":[{"id":"A","text":"$+2\\ \\mathrm{m/s^{2}}$"},{"id":"B","text":"$0\\ \\mathrm{m/s^{2}}$"},{"id":"C","text":"$-2\\ \\mathrm{m/s^{2}}$"},{"id":"D","text":"$-10\\ \\mathrm{m/s^{2}}$"}],"answer":"C","explanation":"Slope $= \\dfrac{0 - 10}{5 - 0} = -2\\ \\mathrm{m/s^{2}}$. The negative sign shows the object is decelerating; the trap is to drop the sign and answer +2.","explanationZh":"斜率 $=\\dfrac{0-10}{5-0}=-2\\ \\mathrm{m/s^{2}}。负号表示减速；易错点是漏掉符号答成 +2。$","tags":["trap","graph","v-t","slope","sign"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-free-fall-001","unitId":"kinematics","topicId":"free-fall","difficulty":1,"stem":"An object is dropped from rest. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its speed after $3\\ \\mathrm{s}$?","stemZh":"物体由静止自由下落，取 $g=10\\ \\mathrm{m/s^{2}}，3\\ \\mathrm{s}$ 后的速度多大？","options":[{"id":"A","text":"$30\\ \\mathrm{m/s}$"},{"id":"B","text":"$10\\ \\mathrm{m/s}$"},{"id":"C","text":"$3\\ \\mathrm{m/s}$"},{"id":"D","text":"$15\\ \\mathrm{m/s}$"}],"answer":"A","explanation":"For a dropped $object, v =$ gt $= 10 \\times 3 = 30\\ \\mathrm{m/s}$.","explanationZh":"$自由下落物体，v=gt=10\\times3=30\\ \\mathrm{m/s}。$","tags":["basic","free-fall","velocity"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-free-fall-002","unitId":"kinematics","topicId":"free-fall","difficulty":1,"stem":"An object is dropped from rest. Taking $g = 10\\ \\mathrm{m/s^{2}}$, how far does it fall in $2\\ \\mathrm{s}$?","stemZh":"物体由静止自由下落，取 $g=10\\ \\mathrm{m/s^{2}}，2\\ \\mathrm{s}$ 内下落多高？","options":[{"id":"A","text":"$10\\ \\mathrm{m}$"},{"id":"B","text":"$20\\ \\mathrm{m}$"},{"id":"C","text":"$40\\ \\mathrm{m}$"},{"id":"D","text":"$5\\ \\mathrm{m}$"}],"answer":"B","explanation":"Distance $h = \\dfrac{1}{2}gt^{2} = \\dfrac{1}{2} \\times 10 \\times 2^{2} = 20\\ \\mathrm{m}$.","explanationZh":"下落高度 $h=\\dfrac{1}{2}gt^{2}=\\dfrac{1}{2}\\times10\\times2^{2}=20\\ \\mathrm{m}。$","tags":["basic","free-fall","height"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-free-fall-003","unitId":"kinematics","topicId":"free-fall","difficulty":1,"stem":"An object is dropped from a height of $45\\ \\mathrm{m}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its impact speed?","stemZh":"物体从 $45\\ \\mathrm{m}$ 高处自由下落，取 $g=10\\ \\mathrm{m/s^{2}}，落地速度多大？$","options":[{"id":"A","text":"$20\\ \\mathrm{m/s}$"},{"id":"B","text":"$45\\ \\mathrm{m/s}$"},{"id":"C","text":"$30\\ \\mathrm{m/s}$"},{"id":"D","text":"$90\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"Use $v^{2} = 2gh = 2 \\times 10 \\times 45 = 900$, so $v = 30\\ \\mathrm{m/s}$.","explanationZh":"$由 v^{2}=2gh=2\\times10\\times45=900，得 v=30\\ \\mathrm{m/s}。$","tags":["basic","free-fall","impact-speed"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-free-fall-004","unitId":"kinematics","topicId":"free-fall","difficulty":2,"stem":"An object is thrown vertically upward at $20\\ \\mathrm{m/s}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, how long does it take to reach the highest point?","stemZh":"物体以 $20\\ \\mathrm{m/s}$ 竖直上抛，取 $g=10\\ \\mathrm{m/s^{2}}，到达最高点需多长时间？$","options":[{"id":"A","text":"$1\\ \\mathrm{s}$"},{"id":"B","text":"$4\\ \\mathrm{s}$"},{"id":"C","text":"$0.5\\ \\mathrm{s}$"},{"id":"D","text":"$2\\ \\mathrm{s}$"}],"answer":"D","explanation":"At the top $v = 0$, so $t = \\dfrac{v_0}{g} = \\dfrac{20}{10} = 2\\ \\mathrm{s}$.","explanationZh":"最高点 $v=0，故 t=\\dfrac{v_0}{g}=\\dfrac{20}{10}=2\\ \\mathrm{s}。$","tags":["basic","free-fall","vertical-throw","time-to-top"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-free-fall-005","unitId":"kinematics","topicId":"free-fall","difficulty":2,"stem":"An object is thrown vertically upward at $20\\ \\mathrm{m/s}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the maximum height reached?","stemZh":"物体以 $20\\ \\mathrm{m/s}$ 竖直上抛，取 $g=10\\ \\mathrm{m/s^{2}}，能上升的最大高度是多少？$","options":[{"id":"A","text":"$20\\ \\mathrm{m}$"},{"id":"B","text":"$40\\ \\mathrm{m}$"},{"id":"C","text":"$10\\ \\mathrm{m}$"},{"id":"D","text":"$200\\ \\mathrm{m}$"}],"answer":"A","explanation":"Maximum height $H = \\dfrac{v_0^{2}}{2g} = \\dfrac{400}{20} = 20\\ \\mathrm{m}$.","explanationZh":"最大高度 $H=\\dfrac{v_0^{2}}{2g}=\\dfrac{400}{20}=20\\ \\mathrm{m}。$","tags":["basic","free-fall","vertical-throw","max-height"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-free-fall-006","unitId":"kinematics","topicId":"free-fall","difficulty":3,"stem":"An object is dropped from rest from a height of $80\\ \\mathrm{m}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, how long does it take to reach the ground?","stemZh":"物体从 $80\\ \\mathrm{m}$ 高处由静止自由下落，取 $g=10\\ \\mathrm{m/s^{2}}，到达地面需多长时间？$","options":[{"id":"A","text":"$8\\ \\mathrm{s}$"},{"id":"B","text":"$4\\ \\mathrm{s}$"},{"id":"C","text":"$2\\ \\mathrm{s}$"},{"id":"D","text":"$16\\ \\mathrm{s}$"}],"answer":"B","explanation":"From $h = \\dfrac{1}{2}gt^{2}: t = \\sqrt{\\dfrac{2h}{g}} = \\sqrt{\\dfrac{160}{10}} = \\sqrt{16} = 4\\ \\mathrm{s}$.","explanationZh":"$由 h=\\dfrac{1}{2}gt^{2} 得 t=\\sqrt{\\dfrac{2h}{g}}=\\sqrt{\\dfrac{160}{10}}=\\sqrt{16}=4\\ \\mathrm{s}。$","tags":["applied","free-fall","fall-time"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-free-fall-007","unitId":"kinematics","topicId":"free-fall","difficulty":3,"stem":"An object is thrown upward at $30\\ \\mathrm{m/s}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its height above the launch point after $1\\ \\mathrm{s}$?","stemZh":"物体以 $30\\ \\mathrm{m/s}$ 竖直上抛，取 $g=10\\ \\mathrm{m/s^{2}}，1\\ \\mathrm{s}$ 后相对抛出点的高度是多少？","options":[{"id":"A","text":"$30\\ \\mathrm{m}$"},{"id":"B","text":"$5\\ \\mathrm{m}$"},{"id":"C","text":"$25\\ \\mathrm{m}$"},{"id":"D","text":"$35\\ \\mathrm{m}$"}],"answer":"C","explanation":"Take upward as $positive: h = v_0 t - \\dfrac{1}{2}gt^{2} = 30 \\times 1 - \\dfrac{1}{2} \\times 10 \\times 1^{2} = 30 - 5 = 25\\ \\mathrm{m}$.","explanationZh":"$取向上为正：h=v_0t-\\dfrac{1}{2}gt^{2}=30\\times1-\\dfrac{1}{2}\\times10\\times1^{2}=30-5=25\\ \\mathrm{m}。$","tags":["applied","free-fall","vertical-throw","height"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-free-fall-008","unitId":"kinematics","topicId":"free-fall","difficulty":3,"stem":"An object is thrown vertically upward at $20\\ \\mathrm{m/s}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the total time of flight before it returns to the launch level?","stemZh":"物体以 $20\\ \\mathrm{m/s}$ 竖直上抛，取 $g=10\\ \\mathrm{m/s^{2}}，回到抛出点所需的总飞行时间是多少？$","options":[{"id":"A","text":"$1\\ \\mathrm{s}$"},{"id":"B","text":"$2\\ \\mathrm{s}$"},{"id":"C","text":"$20\\ \\mathrm{s}$"},{"id":"D","text":"$4\\ \\mathrm{s}$"}],"answer":"D","explanation":"Time up equals time down: total $T = \\dfrac{2v_0}{g} = \\dfrac{40}{10} = 4\\ \\mathrm{s}$.","explanationZh":"上升与下降时间相等：总时间 $T=\\dfrac{2v_0}{g}=\\dfrac{40}{10}=4\\ \\mathrm{s}。$","tags":["applied","free-fall","vertical-throw","time-of-flight"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-free-fall-009","unitId":"kinematics","topicId":"free-fall","difficulty":3,"stem":"An object is thrown downward with initial speed $10\\ \\mathrm{m/s}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its speed after $2\\ \\mathrm{s} (downward positive)$?","stemZh":"物体以 $10\\ \\mathrm{m/s}$ 的初速度向下抛出，取 $g=10\\ \\mathrm{m/s^{2}}，向下为正，2\\ \\mathrm{s}$ 后的速度多大？","options":[{"id":"A","text":"$10\\ \\mathrm{m/s}$"},{"id":"B","text":"$20\\ \\mathrm{m/s}$"},{"id":"C","text":"$30\\ \\mathrm{m/s}$"},{"id":"D","text":"$40\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"With downward $positive, v = v_0 +$ gt $= 10 + 10 \\times 2 = 30\\ \\mathrm{m/s}$.","explanationZh":"$取向下为正，v=v_0+gt=10+10\\times2=30\\ \\mathrm{m/s}。$","tags":["applied","free-fall","downward-throw","velocity"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-free-fall-010","unitId":"kinematics","topicId":"free-fall","difficulty":3,"stem":"An object is thrown upward at $20\\ \\mathrm{m/s}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its height above the launch point after $2\\ \\mathrm{s}$?","stemZh":"物体以 $20\\ \\mathrm{m/s}$ 竖直上抛，取 $g=10\\ \\mathrm{m/s^{2}}，2\\ \\mathrm{s}$ 后相对抛出点的高度是多少？","options":[{"id":"A","text":"$40\\ \\mathrm{m}$"},{"id":"B","text":"$20\\ \\mathrm{m}$"},{"id":"C","text":"$60\\ \\mathrm{m}$"},{"id":"D","text":"$0\\ \\mathrm{m}$"}],"answer":"B","explanation":"$h = v_0 t - \\dfrac{1}{2}gt^{2} = 20 \\times 2 - \\dfrac{1}{2} \\times 10 \\times 4 = 40 - 20 = 20\\ \\mathrm{m}. (At 2\\ \\mathrm{s}\\text{ it is exactly at the }top.)$","explanationZh":"$h=v_0t-\\dfrac{1}{2}gt^{2}=20\\times2-\\dfrac{1}{2}\\times10\\times4=40-20=20\\ \\mathrm{m}。（第 2\\ \\mathrm{s}$ 时恰好在最高点。）","tags":["applied","free-fall","vertical-throw","height"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-free-fall-011","unitId":"kinematics","topicId":"free-fall","difficulty":4,"stem":"A ball is thrown vertically upward at $30\\ \\mathrm{m/s}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its displacement from the launch point after $6\\ \\mathrm{s}$?","stemZh":"小球以 $30\\ \\mathrm{m/s}$ 竖直上抛，取 $g=10\\ \\mathrm{m/s^{2}}，6\\ \\mathrm{s}$ 后相对抛出点的位移是多少？","options":[{"id":"A","text":"$0\\ \\mathrm{m}$"},{"id":"B","text":"$45\\ \\mathrm{m}$"},{"id":"C","text":"$90\\ \\mathrm{m}$"},{"id":"D","text":"$180\\ \\mathrm{m}$"}],"answer":"A","explanation":"Time of flight $= \\dfrac{2v_0}{g} = \\dfrac{60}{10} = 6\\ \\mathrm{s}$, so at $t = 6\\ \\mathrm{s}$ it is back at the launch level: displacement is $0\\ \\mathrm{m}$. The trap is to compute $h = 30\\times6 - 5\\times36 = 0$ correctly but misread it as a height, or to report the max height $45\\ \\mathrm{m}$.","explanationZh":"飞行时间 $=\\dfrac{2v_0}{g}=\\dfrac{60}{10}=6\\ \\mathrm{s}，故 t=6\\ \\mathrm{s}$ 时回到抛出点，位移为 $0\\ \\mathrm{m}。易错点是误答最大高度 45\\ \\mathrm{m}$ 或把位移当作高度。","tags":["trap","free-fall","vertical-throw","displacement"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-free-fall-012","unitId":"kinematics","topicId":"free-fall","difficulty":4,"stem":"An object is dropped from rest. Taking $g = 10\\ \\mathrm{m/s^{2}}$, how far does it fall in $4\\ \\mathrm{s}$?","stemZh":"物体由静止自由下落，取 $g=10\\ \\mathrm{m/s^{2}}，4\\ \\mathrm{s}$ 内下落多高？","options":[{"id":"A","text":"$40\\ \\mathrm{m}$"},{"id":"B","text":"$10\\ \\mathrm{m}$"},{"id":"C","text":"$160\\ \\mathrm{m}$"},{"id":"D","text":"$80\\ \\mathrm{m}$"}],"answer":"D","explanation":"Distance $h = \\dfrac{1}{2}gt^{2} = \\dfrac{1}{2} \\times 10 \\times 16 = 80\\ \\mathrm{m}$. The trap is to use $v =$ gt $= 40$ and mistake it for the distance.","explanationZh":"下落高度 $h=\\dfrac{1}{2}gt^{2}=\\dfrac{1}{2}\\times10\\times16=80\\ \\mathrm{m}。易错点是误把速度 v=gt=40$ 当作位移。","tags":["trap","free-fall","height","v-vs-s"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-free-fall-013","unitId":"kinematics","topicId":"free-fall","difficulty":4,"stem":"A ball is thrown upward at $20\\ \\mathrm{m/s}$ and rises to its maximum height. Taking $g = 10\\ \\mathrm{m/s^{2}}$, with what speed does it pass the launch point again on the way down?","stemZh":"小球以 $20\\ \\mathrm{m/s}$ 上抛到达最高点后，取 $g=10\\ \\mathrm{m/s^{2}}，下落再次经过抛出点时的速度多大？$","options":[{"id":"A","text":"$20\\ \\mathrm{m/s}$"},{"id":"B","text":"$0\\ \\mathrm{m/s}$"},{"id":"C","text":"$10\\ \\mathrm{m/s}$"},{"id":"D","text":"$40\\ \\mathrm{m/s}$"}],"answer":"A","explanation":"By symmetry of free fall (no air $resistance)$, the speed returning to the launch level equals the launch $speed, 20\\ \\mathrm{m/s}$, but directed downward. The trap is to think it is $0 (that$ is only true at the top).","explanationZh":"无空气阻力时自由落体对称，回到抛出点的速率等于抛出速率 $20\\ \\mathrm{m/s}，但方向向下。易错点以为是 0（那只在最高点成立）。$","tags":["trap","free-fall","vertical-throw","symmetry"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-free-fall-014","unitId":"kinematics","topicId":"free-fall","difficulty":5,"stem":"A ball is thrown upward at $20\\ \\mathrm{m/s}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, at what time is it $15\\ \\mathrm{m}$ above the launch point on the way up?","stemZh":"小球以 $20\\ \\mathrm{m/s}$ 上抛，取 $g=10\\ \\mathrm{m/s^{2}}，在上升过程中何时距抛出点 15\\ \\mathrm{m}？$","options":[{"id":"A","text":"$3\\ \\mathrm{s}$"},{"id":"B","text":"$1\\ \\mathrm{s}$"},{"id":"C","text":"$2\\ \\mathrm{s}$"},{"id":"D","text":"$4\\ \\mathrm{s}$"}],"answer":"B","explanation":"Solve $15 = 20t - 5t^{2} \\Rightarrow 5t^{2} - 20t + 15 = 0 \\Rightarrow t^{2} - 4t + 3 = 0$, giving $t = 1\\ \\mathrm{s}$ or $t = 3\\ \\mathrm{s}$. On the way up the answer is $t = 1\\ \\mathrm{s}; 3\\ \\mathrm{s}$ is the same height on the way down.","explanationZh":"$解 15=20t-5t^{2}\\Rightarrow 5t^{2}-20t+15=0\\Rightarrow t^{2}-4t+3=0，得 t=1\\ \\mathrm{s} 或 t=3\\ \\mathrm{s}。上升过程中为 t=1\\ \\mathrm{s}；3\\ \\mathrm{s}$ 是下落经过同一高度时。","tags":["trap","free-fall","vertical-throw","quadratic"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-free-fall-015","unitId":"kinematics","topicId":"free-fall","difficulty":5,"stem":"A ball is thrown upward at $10\\ \\mathrm{m/s}$ from the top of $a 15\\ \\mathrm{m}$ cliff. Taking $g = 10\\ \\mathrm{m/s^{2}}$ and upward as positive, how long until it hits the ground below $(at y = -15\\ \\mathrm{m})$?","stemZh":"小球以 $10\\ \\mathrm{m/s} 从 15\\ \\mathrm{m}$ 高的悬崖顶上抛，取 $g=10\\ \\mathrm{m/s^{2}}、向上为正，多久后落到崖底（y=-15\\ \\mathrm{m}）？$","options":[{"id":"A","text":"$1\\ \\mathrm{s}$"},{"id":"B","text":"$5\\ \\mathrm{s}$"},{"id":"C","text":"$3\\ \\mathrm{s}$"},{"id":"D","text":"$2\\ \\mathrm{s}$"}],"answer":"C","explanation":"Use $y = v_0 t - \\dfrac{1}{2}gt^{2}$ with $y = -15: -15 = 10t - 5t^{2} \\Rightarrow 5t^{2} - 10t - 15 = 0 \\Rightarrow t^{2} - 2t - 3 = 0$, so $(t - 3)(t + 1) = 0$ and $t = 3\\ \\mathrm{s}$. The trap is to ignore the initial height and just use the throw time.","explanationZh":"$用 y=v_0t-\\dfrac{1}{2}gt^{2}，代入 y=-15：-15=10t-5t^{2}\\Rightarrow 5t^{2}-10t-15=0\\Rightarrow t^{2}-2t-3=0，得 (t-3)(t+1)=0，故 t=3\\ \\mathrm{s}。易错点是忽略初始高度只算上抛时间。$","tags":["trap","free-fall","vertical-throw","cliff"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-projectile-motion-001","unitId":"kinematics","topicId":"projectile-motion","difficulty":1,"stem":"For projectile motion with no air resistance, the horizontal component of velocity:","stemZh":"无空气阻力时，平抛/斜抛运动的水平分速度：","options":[{"id":"A","text":"Remains constant"},{"id":"B","text":"Increases with time"},{"id":"C","text":"Decreases with time"},{"id":"D","text":"Is always zero"}],"optionsZh":[{"id":"A","text":"保持不变"},{"id":"B","text":"随时间增大"},{"id":"C","text":"随时间减小"},{"id":"D","text":"始终为零"}],"answer":"A","explanation":"With no horizontal force, the horizontal velocity does not change; only the vertical motion is accelerated by gravity.","explanationZh":"水平方向不受力，水平速度不变；只有竖直方向受重力加速。","tags":["basic","projectile","horizontal-motion"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-projectile-motion-002","unitId":"kinematics","topicId":"projectile-motion","difficulty":1,"stem":"A projectile is launched at $20\\ \\mathrm{m/s}$ at $45^{\\circ}$ to the horizontal. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its range?","stemZh":"一抛体以 $20\\ \\mathrm{m/s}、与水平成 45^{\\circ}$ 抛出，取 $g=10\\ \\mathrm{m/s^{2}}，射程是多少？$","options":[{"id":"A","text":"$20\\ \\mathrm{m}$"},{"id":"B","text":"$40\\ \\mathrm{m}$"},{"id":"C","text":"$10\\ \\mathrm{m}$"},{"id":"D","text":"$80\\ \\mathrm{m}$"}],"answer":"B","explanation":"Range $R = \\dfrac{v^{2}\\sin 2\\theta}{g}$. At $45^{\\circ}, \\sin 90^{\\circ} = 1$, so $R = \\dfrac{20^{2}}{10} = 40\\ \\mathrm{m}$.","explanationZh":"射程 $R=\\dfrac{v^{2}\\sin2\\theta}{g}。\\theta=45^{\\circ} 时 \\sin90^{\\circ}=1，故 R=\\dfrac{20^{2}}{10}=40\\ \\mathrm{m}。$","tags":["basic","projectile","range"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-projectile-motion-003","unitId":"kinematics","topicId":"projectile-motion","difficulty":1,"stem":"A projectile is launched at $20\\ \\mathrm{m/s}$ at $30^{\\circ}$ to the horizontal. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its maximum height?","stemZh":"一抛体以 $20\\ \\mathrm{m/s}、与水平成 30^{\\circ}$ 抛出，取 $g=10\\ \\mathrm{m/s^{2}}，最大高度是多少？$","options":[{"id":"A","text":"$10\\ \\mathrm{m}$"},{"id":"B","text":"$20\\ \\mathrm{m}$"},{"id":"C","text":"$5\\ \\mathrm{m}$"},{"id":"D","text":"$15\\ \\mathrm{m}$"}],"answer":"C","explanation":"Vertical component $v_y = v\\sin\\theta = 20 \\times 0.5 = 10\\ \\mathrm{m/s}$. Max height $H = \\dfrac{v_y^{2}}{2g} = \\dfrac{100}{20} = 5\\ \\mathrm{m}$.","explanationZh":"竖直分速度 $v_y=v\\sin\\theta=20\\times0.5=10\\ \\mathrm{m/s}。最大高度 H=\\dfrac{v_y^{2}}{2g}=\\dfrac{100}{20}=5\\ \\mathrm{m}。$","tags":["basic","projectile","max-height"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-projectile-motion-004","unitId":"kinematics","topicId":"projectile-motion","difficulty":2,"stem":"A projectile is launched at $20\\ \\mathrm{m/s}$ at $30^{\\circ}$ to the horizontal. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its time of flight?","stemZh":"一抛体以 $20\\ \\mathrm{m/s}、与水平成 30^{\\circ}$ 抛出，取 $g=10\\ \\mathrm{m/s^{2}}，飞行时间是多少？$","options":[{"id":"A","text":"$1\\ \\mathrm{s}$"},{"id":"B","text":"$4\\ \\mathrm{s}$"},{"id":"C","text":"$0.5\\ \\mathrm{s}$"},{"id":"D","text":"$2\\ \\mathrm{s}$"}],"answer":"D","explanation":"Time of flight $T = \\dfrac{2v\\sin\\theta}{g} = \\dfrac{2 \\times 20 \\times 0.5}{10} = \\dfrac{20}{10} = 2\\ \\mathrm{s}$.","explanationZh":"飞行时间 $T=\\dfrac{2v\\sin\\theta}{g}=\\dfrac{2\\times20\\times0.5}{10}=\\dfrac{20}{10}=2\\ \\mathrm{s}。$","tags":["basic","projectile","time-of-flight"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-projectile-motion-005","unitId":"kinematics","topicId":"projectile-motion","difficulty":2,"stem":"For a given launch speed, the maximum horizontal range is achieved at a launch angle of:","stemZh":"给定初速度大小时，最大水平射程对应的抛射角是：","options":[{"id":"A","text":"$45^{\\circ}$"},{"id":"B","text":"$30^{\\circ}$"},{"id":"C","text":"$60^{\\circ}$"},{"id":"D","text":"$90^{\\circ}$"}],"optionsZh":[{"id":"A","text":"$45^{\\circ}$"},{"id":"B","text":"$30^{\\circ}$"},{"id":"C","text":"$60^{\\circ}$"},{"id":"D","text":"$90^{\\circ}$"}],"answer":"A","explanation":"Range $R = \\dfrac{v^{2}\\sin 2\\theta}{g}$ is largest when $\\sin 2\\theta = 1$, i.e. $\\theta = 45^{\\circ}$.","explanationZh":"射程 $R=\\dfrac{v^{2}\\sin2\\theta}{g} 在 \\sin2\\theta=1$ 时最大，即 $\\theta=45^{\\circ}。$","tags":["basic","projectile","range","optimal-angle"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-projectile-motion-006","unitId":"kinematics","topicId":"projectile-motion","difficulty":3,"stem":"A projectile is launched at $20\\ \\mathrm{m/s}$ at $30^{\\circ}$ to the horizontal. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its range?","stemZh":"一抛体以 $20\\ \\mathrm{m/s}、与水平成 30^{\\circ}$ 抛出，取 $g=10\\ \\mathrm{m/s^{2}}，射程是多少？$","options":[{"id":"A","text":"$40\\ \\mathrm{m}$"},{"id":"B","text":"$34.6\\ \\mathrm{m}$"},{"id":"C","text":"$20\\ \\mathrm{m}$"},{"id":"D","text":"$17.3\\ \\mathrm{m}$"}],"answer":"B","explanation":"$R = \\dfrac{v^{2}\\sin 2\\theta}{g} = \\dfrac{400 \\times \\sin 60^{\\circ}}{10} = 40 \\times 0.866 \\approx 34.6\\ \\mathrm{m}$.","explanationZh":"$R=\\dfrac{v^{2}\\sin2\\theta}{g}=\\dfrac{400\\times\\sin60^{\\circ}}{10}=40\\times0.866\\approx34.6\\ \\mathrm{m}。$","tags":["applied","projectile","range"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-projectile-motion-007","unitId":"kinematics","topicId":"projectile-motion","difficulty":3,"stem":"A projectile is launched at $20\\ \\mathrm{m/s}$ at $45^{\\circ}$ to the horizontal. What is its horizontal velocity component?","stemZh":"一抛体以 $20\\ \\mathrm{m/s}、与水平成 45^{\\circ}$ 抛出，其水平速度分量是多少？","options":[{"id":"A","text":"$20\\ \\mathrm{m/s}$"},{"id":"B","text":"$10\\ \\mathrm{m/s}$"},{"id":"C","text":"$14.1\\ \\mathrm{m/s}$"},{"id":"D","text":"$28.3\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"$v_x = v\\cos\\theta = 20 \\times \\cos 45^{\\circ} = 20 \\times 0.707 \\approx 14.1\\ \\mathrm{m/s} (constant\\text{ throughout the }flight)$.","explanationZh":"$v_x=v\\cos\\theta=20\\times\\cos45^{\\circ}=20\\times0.707\\approx14.1\\ \\mathrm{m/s}（全程不变）。$","tags":["applied","projectile","components"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-projectile-motion-008","unitId":"kinematics","topicId":"projectile-motion","difficulty":3,"stem":"A projectile is launched at $20\\ \\mathrm{m/s}$ at $37^{\\circ}$ to the horizontal $(take \\sin 37^{\\circ} \\approx 0.6)$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its time of flight?","stemZh":"一抛体以 $20\\ \\mathrm{m/s}、与水平成 37^{\\circ}$ 抛出（取 $\\sin37^{\\circ}\\approx0.6），取 g=10\\ \\mathrm{m/s^{2}}，飞行时间是多少？$","options":[{"id":"A","text":"$1.2\\ \\mathrm{s}$"},{"id":"B","text":"$4\\ \\mathrm{s}$"},{"id":"C","text":"$2\\ \\mathrm{s}$"},{"id":"D","text":"$2.4\\ \\mathrm{s}$"}],"answer":"D","explanation":"$T = \\dfrac{2v\\sin\\theta}{g} = \\dfrac{2 \\times 20 \\times 0.6}{10} = \\dfrac{24}{10} = 2.4\\ \\mathrm{s}$.","explanationZh":"$T=\\dfrac{2v\\sin\\theta}{g}=\\dfrac{2\\times20\\times0.6}{10}=\\dfrac{24}{10}=2.4\\ \\mathrm{s}。$","tags":["applied","projectile","time-of-flight"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-projectile-motion-009","unitId":"kinematics","topicId":"projectile-motion","difficulty":3,"stem":"A projectile is launched at $20\\ \\mathrm{m/s}$ at $60^{\\circ}$ to the horizontal. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its maximum height?","stemZh":"一抛体以 $20\\ \\mathrm{m/s}、与水平成 60^{\\circ}$ 抛出，取 $g=10\\ \\mathrm{m/s^{2}}，最大高度是多少？$","options":[{"id":"A","text":"$30\\ \\mathrm{m}$"},{"id":"B","text":"$7.5\\ \\mathrm{m}$"},{"id":"C","text":"$15\\ \\mathrm{m}$"},{"id":"D","text":"$60\\ \\mathrm{m}$"}],"answer":"C","explanation":"$v_y = v\\sin\\theta = 20 \\times \\sin 60^{\\circ} = 20 \\times 0.866 \\approx 17.32\\ \\mathrm{m/s}. H = \\dfrac{v_y^{2}}{2g} = \\dfrac{300}{20} = 15\\ \\mathrm{m}$.","explanationZh":"$v_y=v\\sin\\theta=20\\times\\sin60^{\\circ}=20\\times0.866\\approx17.32\\ \\mathrm{m/s}。H=\\dfrac{v_y^{2}}{2g}=\\dfrac{300}{20}=15\\ \\mathrm{m}。$","tags":["applied","projectile","max-height"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-projectile-motion-010","unitId":"kinematics","topicId":"projectile-motion","difficulty":3,"stem":"A projectile is launched at $30\\ \\mathrm{m/s}$ at $45^{\\circ}$ to the horizontal. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its range?","stemZh":"一抛体以 $30\\ \\mathrm{m/s}、与水平成 45^{\\circ}$ 抛出，取 $g=10\\ \\mathrm{m/s^{2}}，射程是多少？$","options":[{"id":"A","text":"$45\\ \\mathrm{m}$"},{"id":"B","text":"$90\\ \\mathrm{m}$"},{"id":"C","text":"$30\\ \\mathrm{m}$"},{"id":"D","text":"$180\\ \\mathrm{m}$"}],"answer":"B","explanation":"At $45^{\\circ}, R = \\dfrac{v^{2}}{g} = \\dfrac{30^{2}}{10} = \\dfrac{900}{10} = 90\\ \\mathrm{m}$.","explanationZh":"$\\theta=45^{\\circ} 时，R=\\dfrac{v^{2}}{g}=\\dfrac{30^{2}}{10}=\\dfrac{900}{10}=90\\ \\mathrm{m}。$","tags":["applied","projectile","range"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-projectile-motion-011","unitId":"kinematics","topicId":"projectile-motion","difficulty":4,"stem":"A projectile is launched at $20\\ \\mathrm{m/s}$ at $45^{\\circ}$. At the highest point of its path, what is the magnitude of its velocity?","stemZh":"一抛体以 $20\\ \\mathrm{m/s}、与水平成 45^{\\circ}$ 抛出，在其轨迹最高点时，速度大小是多少？","options":[{"id":"A","text":"$14.1\\ \\mathrm{m/s}$"},{"id":"B","text":"$0\\ \\mathrm{m/s}$"},{"id":"C","text":"$20\\ \\mathrm{m/s}$"},{"id":"D","text":"$10\\ \\mathrm{m/s}$"}],"answer":"A","explanation":"At the apex the vertical velocity is zero, but the horizontal component is $unchanged: v = v\\cos 45^{\\circ} = 20 \\times 0.707 \\approx 14.1\\ \\mathrm{m/s}$. The trap is to think the whole velocity is zero at the top.","explanationZh":"$最高点时竖直分速度为零，但水平分量不变：v=v\\cos45^{\\circ}=20\\times0.707\\approx14.1\\ \\mathrm{m/s}。易错点是以为顶点处速度整体为零。$","tags":["trap","projectile","apex","horizontal-motion"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-projectile-motion-012","unitId":"kinematics","topicId":"projectile-motion","difficulty":4,"stem":"A projectile is launched at $20\\ \\mathrm{m/s}$ at $30^{\\circ}$. Taking $g = 10\\ \\mathrm{m/s^{2}}, a$ second projectile is launched at the same speed but at $60^{\\circ}$. How do their ranges compare?","stemZh":"一抛体以 $20\\ \\mathrm{m/s}、与水平成 30^{\\circ}$ 抛出，取 $g=10\\ \\mathrm{m/s^{2}}；另一抛体以相同速率、60^{\\circ}$ 抛出。两者射程相比如何？","options":[{"id":"A","text":"The $30^{\\circ}$ shot goes twice as far"},{"id":"B","text":"The $60^{\\circ}$ shot goes twice as far"},{"id":"C","text":"The $60^{\\circ}$ shot goes four times as far"},{"id":"D","text":"They have the same range"}],"optionsZh":[{"id":"A","text":"$30^{\\circ}$ 射程远一倍"},{"id":"B","text":"$60^{\\circ}$ 射程远一倍"},{"id":"C","text":"$60^{\\circ}$ 射程远三倍"},{"id":"D","text":"两者射程相同"}],"answer":"D","explanation":"Complementary angles $(30^{\\circ}\\text{ and }60^{\\circ})$ give the same range because $\\sin 2(30^{\\circ}) = \\sin 60^{\\circ} = \\sin 2(60^{\\circ}) = \\sin 120^{\\circ}$. Each range is $\\approx 34.6\\ \\mathrm{m}$.","explanationZh":"$互余角（30^{\\circ} 与 60^{\\circ}）射程相同，因为 \\sin2(30^{\\circ})=\\sin60^{\\circ}=\\sin2(60^{\\circ})=\\sin120^{\\circ}。各自射程均约 34.6\\ \\mathrm{m}。$","tags":["trap","projectile","range","complementary-angles"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-projectile-motion-013","unitId":"kinematics","topicId":"projectile-motion","difficulty":4,"stem":"A bullet is fired horizontally from a height h, and at the same instant a second bullet is dropped from the same height. Ignoring air resistance, which hits the ground first?","stemZh":"一颗子弹从高度 h 水平射出，同时另一颗子弹从同一高度自由下落。忽略空气阻力，哪颗先落地？","options":[{"id":"A","text":"They hit at the same time"},{"id":"B","text":"The fired bullet hits first"},{"id":"C","text":"The dropped bullet hits first"},{"id":"D","text":"It depends on the firing speed"}],"optionsZh":[{"id":"A","text":"同时落地"},{"id":"B","text":"射出的子弹先落地"},{"id":"C","text":"下落的子弹先落地"},{"id":"D","text":"取决于发射速度"}],"answer":"A","explanation":"Both have zero initial vertical velocity, so their vertical motion is $identical: h = \\dfrac{1}{2}gt^{2}$. The horizontal launch speed does not affect the fall time.","explanationZh":"$两者初竖直速度均为零，竖直运动相同：h=\\dfrac{1}{2}gt^{2}。水平发射速度不影响下落时间。$","tags":["trap","projectile","horizontal-launch","independent-motion"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-projectile-motion-014","unitId":"kinematics","topicId":"projectile-motion","difficulty":5,"stem":"A projectile is launched at $20\\ \\mathrm{m/s}$ at $30^{\\circ}$ to the horizontal. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the magnitude of its velocity at the highest point?","stemZh":"一抛体以 $20\\ \\mathrm{m/s}、与水平成 30^{\\circ}$ 抛出，取 $g=10\\ \\mathrm{m/s^{2}}，其在最高点速度的大小是多少？$","options":[{"id":"A","text":"$20\\ \\mathrm{m/s}$"},{"id":"B","text":"$17.3\\ \\mathrm{m/s}$"},{"id":"C","text":"$0\\ \\mathrm{m/s}$"},{"id":"D","text":"$10\\ \\mathrm{m/s}$"}],"answer":"B","explanation":"At the apex the vertical part is zero, so speed $=$ horizontal component $= v\\cos 30^{\\circ} = 20 \\times 0.866 \\approx 17.3\\ \\mathrm{m/s}$.","explanationZh":"顶点处竖直分速度为零，故速率 $=$ 水平分量 $=v\\cos30^{\\circ}=20\\times0.866\\approx17.3\\ \\mathrm{m/s}。$","tags":["trap","projectile","apex","velocity-magnitude"]},{"subject":"physics","moduleId":"mechanics","type":"single-choice","status":"published","id":"physics-projectile-motion-015","unitId":"kinematics","topicId":"projectile-motion","difficulty":5,"stem":"A projectile is fired at $20\\ \\mathrm{m/s}$ at $37^{\\circ} (take \\sin 74^{\\circ} \\approx 0.96)$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its range?","stemZh":"一抛体以 $20\\ \\mathrm{m/s}、与水平成 37^{\\circ}$ 射出（取 $\\sin74^{\\circ}\\approx0.96），取 g=10\\ \\mathrm{m/s^{2}}，射程是多少？$","options":[{"id":"A","text":"$76.8\\ \\mathrm{m}$"},{"id":"B","text":"$19.2\\ \\mathrm{m}$"},{"id":"C","text":"$38.4\\ \\mathrm{m}$"},{"id":"D","text":"$12.0\\ \\mathrm{m}$"}],"answer":"C","explanation":"$R = \\dfrac{v^{2}\\sin 2\\theta}{g} = \\dfrac{400 \\times \\sin 74^{\\circ}}{10} \\approx \\dfrac{400 \\times 0.96}{10} = 38.4\\ \\mathrm{m}$. The trap is to forget the factor of 2 in $2\\theta$ and use $\\sin 37^{\\circ}$ instead.","explanationZh":"$R=\\dfrac{v^{2}\\sin2\\theta}{g}=\\dfrac{400\\times\\sin74^{\\circ}}{10}\\approx\\dfrac{400\\times0.96}{10}=38.4\\ \\mathrm{m}。易错点是漏掉 2\\theta$ 中的 $2，误用 \\sin37^{\\circ}。$","tags":["trap","projectile","range","double-angle"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-force-analysis-equilibrium-001","moduleId":"mechanics","unitId":"forces-newton","topicId":"force-analysis-equilibrium","difficulty":1,"stem":"$A 4\\ \\mathrm{kg}$ block rests on a horizontal table. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the normal force from the table?","stemZh":"质量 $4\\ \\mathrm{kg}$ 的物块静止在水平桌面上，取 $g=10\\ \\mathrm{m/s^{2}}，桌面对它的支持力多大？$","options":[{"id":"A","text":"$40\\ \\mathrm{N}$"},{"id":"B","text":"$4\\ \\mathrm{N}$"},{"id":"C","text":"$0\\ \\mathrm{N}$"},{"id":"D","text":"$10\\ \\mathrm{N}$"}],"answer":"A","explanation":"Vertically the block is in equilibrium, so $N =$ mg $= 4 \\times 10 = 40\\ \\mathrm{N}$.","explanationZh":"竖直方向平衡，故 $N=mg=4\\times10=40\\ \\mathrm{N}。$","tags":["basic","forces","normal-force","equilibrium"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-force-analysis-equilibrium-002","moduleId":"mechanics","unitId":"forces-newton","topicId":"force-analysis-equilibrium","difficulty":1,"stem":"A block is pulled by $a 5\\ \\mathrm{N}$ force to the right and $a 3\\ \\mathrm{N}$ force to the left, both horizontal. What is the net force?","stemZh":"物块同时受到向右 $5\\ \\mathrm{N}$ 与向左 $3\\ \\mathrm{N}$ 的水平力，合力多大、方向如何？","options":[{"id":"A","text":"$8\\ \\mathrm{N}$ to the right"},{"id":"B","text":"$2\\ \\mathrm{N}$ to the right"},{"id":"C","text":"$2\\ \\mathrm{N}$ to the left"},{"id":"D","text":"0"}],"optionsZh":[{"id":"A","text":"$8\\ \\mathrm{N}，向右$"},{"id":"B","text":"$2\\ \\mathrm{N}，向右$"},{"id":"C","text":"$2\\ \\mathrm{N}，向左$"},{"id":"D","text":"0"}],"answer":"B","explanation":"Forces oppose, so the net force is $5 - 3 = 2\\ \\mathrm{N}$ to the right.","explanationZh":"两力反向，合力为 $5-3=2\\ \\mathrm{N}，方向向右。$","tags":["basic","forces","net-force"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-force-analysis-equilibrium-003","moduleId":"mechanics","unitId":"forces-newton","topicId":"force-analysis-equilibrium","difficulty":1,"stem":"A lamp of weight $20\\ \\mathrm{N}$ hangs at rest from a single vertical cord. What is the tension in the cord?","stemZh":"重量 $20\\ \\mathrm{N}$ 的灯用一根竖直细绳悬挂静止，绳的张力多大？","options":[{"id":"A","text":"$10\\ \\mathrm{N}$"},{"id":"B","text":"$40\\ \\mathrm{N}$"},{"id":"C","text":"$20\\ \\mathrm{N}$"},{"id":"D","text":"$0\\ \\mathrm{N}$"}],"answer":"C","explanation":"Vertically the lamp is in equilibrium, so the upward tension balances the $weight: T = 20\\ \\mathrm{N}$.","explanationZh":"$竖直方向平衡，向上的张力与重力平衡：T=20\\ \\mathrm{N}。$","tags":["basic","forces","tension","equilibrium"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-force-analysis-equilibrium-004","images":[{"src":"/physics/diagrams/incline-smooth-30.svg","alt":"Block resting on a smooth 30-degree incline; weight mg and the normal reaction N (no friction).","altZh":"光滑 30° 斜面上的静止物块；受力为重力 mg 与支持力 N（无摩擦）。"}],"moduleId":"mechanics","unitId":"forces-newton","topicId":"force-analysis-equilibrium","difficulty":2,"stem":"$A 5\\ \\mathrm{kg}$ block rests on a smooth incline of angle $\\theta = 30^{\\circ}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the normal reaction?","stemZh":"质量 $5\\ \\mathrm{kg}$ 的物块静止在倾角 $\\theta=30^{\\circ}$ 的光滑斜面上，取 $g=10\\ \\mathrm{m/s^{2}}，支持力多大？$","options":[{"id":"A","text":"$50\\ \\mathrm{N}$"},{"id":"B","text":"$25\\ \\mathrm{N}$"},{"id":"C","text":"$25\\sqrt{3}\\ \\mathrm{N}$"},{"id":"D","text":"$43.3\\ \\mathrm{N}$"}],"answer":"D","explanation":"Perpendicular to the plane, forces $balance: N = mg\\cos\\theta = 5 \\times 10 \\times \\cos 30^{\\circ} \\approx 43.3\\ \\mathrm{N}$. Note $N \\neq$ mg.","explanationZh":"$垂直斜面方向平衡：N=mg\\cos\\theta=5\\times10\\times\\cos30^{\\circ}\\approx43.3\\ \\mathrm{N}。注意 N\\neq$ mg。","tags":["basic","forces","incline","normal-force"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-force-analysis-equilibrium-005","images":[{"src":"/physics/diagrams/force-triangle.svg","alt":"Three forces of 6 N, 8 N and 10 N acting at a point can form a closed triangle of forces.","altZh":"大小为 6 N、8 N、10 N 的三个共点力可构成封闭的力的三角形。"}],"moduleId":"mechanics","unitId":"forces-newton","topicId":"force-analysis-equilibrium","difficulty":2,"stem":"An object is acted on by three forces of magnitude $6\\ \\mathrm{N}, 8\\ \\mathrm{N}$ and $10\\ \\mathrm{N}$ at a point. Can they keep it in equilibrium?","stemZh":"物体受三个大小分别为 $6\\ \\mathrm{N}、8\\ \\mathrm{N}、10\\ \\mathrm{N}$ 的共点力作用，它们能否使物体平衡？","options":[{"id":"A","text":"No — three forces can never balance"},{"id":"B","text":"Yes — they can form a closed triangle $(6-8-10$ is $right-angled)$"},{"id":"C","text":"Only if all three point the same way"},{"id":"D","text":"Only if a friction force is also present"}],"optionsZh":[{"id":"A","text":"不能——三个力不可能平衡"},{"id":"B","text":"能——可构成封闭三角形（6-8-10 为直角三角形）"},{"id":"C","text":"只有三力同向才行"},{"id":"D","text":"须有摩擦力才行"}],"answer":"B","explanation":"If the vectors can be arranged head-to-tail into a closed triangle, the resultant is zero. A 6-8-10 set is right-angled, so it can close and balance.","explanationZh":"若三力矢量能首尾相接成封闭三角形，则合力为零。6-8-10 为直角三角形，可以封闭而平衡。","tags":["basic","forces","three-force","force-triangle"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-force-analysis-equilibrium-006","moduleId":"mechanics","unitId":"forces-newton","topicId":"force-analysis-equilibrium","difficulty":3,"stem":"An object is in equilibrium under two perpendicular forces of $6\\ \\mathrm{N}$ and $8\\ \\mathrm{N}$. What single third force would balance them?","stemZh":"物体在两个互相垂直的 $6\\ \\mathrm{N}、8\\ \\mathrm{N}$ 力作用下平衡，需要多大的单个第三力与之平衡？","options":[{"id":"A","text":"$14\\ \\mathrm{N}$"},{"id":"B","text":"$2\\ \\mathrm{N}$"},{"id":"C","text":"$10\\ \\mathrm{N}$"},{"id":"D","text":"$48\\ \\mathrm{N}$"}],"answer":"C","explanation":"The two forces have a resultant $\\sqrt{6^{2} + 8^{2}} = 10\\ \\mathrm{N}$. To balance, the third force must be equal and $opposite: 10\\ \\mathrm{N}$.","explanationZh":"两力合力为 $\\sqrt{6^{2}+8^{2}}=10\\ \\mathrm{N}，第三力须等大反向，为 10\\ \\mathrm{N}。$","tags":["applied","forces","resultant","equilibrium"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-force-analysis-equilibrium-007","moduleId":"mechanics","unitId":"forces-newton","topicId":"force-analysis-equilibrium","difficulty":3,"stem":"A force of $100\\ \\mathrm{N}$ acts at $30^{\\circ}$ above the horizontal. What is its horizontal component?","stemZh":"大小为 $100\\ \\mathrm{N}$ 的力与水平方向成 $30^{\\circ}$ 角，它的水平分力多大？","options":[{"id":"A","text":"$50\\ \\mathrm{N}$"},{"id":"B","text":"$100\\ \\mathrm{N}$"},{"id":"C","text":"$70.7\\ \\mathrm{N}$"},{"id":"D","text":"$86.6\\ \\mathrm{N}$"}],"answer":"D","explanation":"The horizontal component is $F\\cos\\theta = 100 \\times \\cos 30^{\\circ} = 100 \\times 0.866 \\approx 86.6\\ \\mathrm{N}$.","explanationZh":"水平分力为 $F\\cos\\theta=100\\times\\cos30^{\\circ}=100\\times0.866\\approx86.6\\ \\mathrm{N}。$","tags":["applied","forces","resolve","components"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-force-analysis-equilibrium-008","moduleId":"mechanics","unitId":"forces-newton","topicId":"force-analysis-equilibrium","difficulty":3,"stem":"$A 3\\ \\mathrm{kg}$ block is released from rest on a smooth incline of angle $30^{\\circ}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its acceleration?","stemZh":"质量 $3\\ \\mathrm{kg}$ 的物块由静止在倾角 $30^{\\circ}$ 的光滑斜面上释放，取 $g=10\\ \\mathrm{m/s^{2}}，加速度多大？$","options":[{"id":"A","text":"$5\\ \\mathrm{m/s^{2}}$"},{"id":"B","text":"$8.66\\ \\mathrm{m/s^{2}}$"},{"id":"C","text":"$10\\ \\mathrm{m/s^{2}}$"},{"id":"D","text":"$2.5\\ \\mathrm{m/s^{2}}$"}],"answer":"A","explanation":"Along the slope only $mg\\sin\\theta acts$, so $a = g\\sin 30^{\\circ} = 10 \\times 0.5 = 5\\ \\mathrm{m/s^{2}} (independent\\text{ of }mass)$.","explanationZh":"沿斜面只有 $mg\\sin\\theta，故 a=g\\sin30^{\\circ}=10\\times0.5=5\\ \\mathrm{m/s^{2}}（与质量无关）。$","tags":["applied","forces","incline","acceleration"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-force-analysis-equilibrium-009","moduleId":"mechanics","unitId":"forces-newton","topicId":"force-analysis-equilibrium","difficulty":3,"stem":"$A 10\\ \\mathrm{kg}$ mass hangs at rest from a vertical rope. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the rope tension?","stemZh":"质量 $10\\ \\mathrm{kg}$ 的物体用竖直细绳悬挂静止，取 $g=10\\ \\mathrm{m/s^{2}}，绳的张力多大？$","options":[{"id":"A","text":"$10\\ \\mathrm{N}$"},{"id":"B","text":"$100\\ \\mathrm{N}$"},{"id":"C","text":"$50\\ \\mathrm{N}$"},{"id":"D","text":"$0\\ \\mathrm{N}$"}],"answer":"B","explanation":"At rest the tension balances the $weight: T =$ mg $= 10 \\times 10 = 100\\ \\mathrm{N}$.","explanationZh":"$静止时张力与重力平衡：T=mg=10\\times10=100\\ \\mathrm{N}。$","tags":["applied","forces","tension","equilibrium"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-force-analysis-equilibrium-010","moduleId":"mechanics","unitId":"forces-newton","topicId":"force-analysis-equilibrium","difficulty":3,"stem":"$A 10\\ \\mathrm{kg}$ block sits on a horizontal floor with $\\mu = 0.3$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what minimum horizontal force starts it moving?","stemZh":"质量 $10\\ \\mathrm{kg} 的物块在水平面上，\\mu=0.3，取 g=10\\ \\mathrm{m/s^{2}}，使其开始运动的最小水平力多大？$","options":[{"id":"A","text":"$100\\ \\mathrm{N}$"},{"id":"B","text":"$10\\ \\mathrm{N}$"},{"id":"C","text":"$30\\ \\mathrm{N}$"},{"id":"D","text":"$3\\ \\mathrm{N}$"}],"answer":"C","explanation":"The maximum static friction is about $\\mu$ mg $= 0.3 \\times 10 \\times 10 = 30\\ \\mathrm{N}$. The applied force must reach this to start motion.","explanationZh":"最大静摩擦力约 $\\mu mg=0.3\\times10\\times10=30\\ \\mathrm{N}，外力需达到此值才能启动。$","tags":["applied","forces","friction","static"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-force-analysis-equilibrium-011","moduleId":"mechanics","unitId":"forces-newton","topicId":"force-analysis-equilibrium","difficulty":4,"stem":"$A 2\\ \\mathrm{kg}$ block is held at rest on a smooth $30^{\\circ}$ incline by a force parallel to the slope. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what force is required?","stemZh":"质量 $2\\ \\mathrm{kg}$ 的物块被平行于斜面的力拉住，静止在倾角 $30^{\\circ}$ 的光滑斜面上，取 $g=10\\ \\mathrm{m/s^{2}}，所需力多大？$","options":[{"id":"A","text":"$20\\ \\mathrm{N}$"},{"id":"B","text":"$17.3\\ \\mathrm{N}$"},{"id":"C","text":"$0\\ \\mathrm{N}$"},{"id":"D","text":"$10\\ \\mathrm{N}$"}],"answer":"D","explanation":"Only the down-slope component $mg\\sin\\theta$ must be $balanced: F = mg\\sin 30^{\\circ} = 2 \\times 10 \\times 0.5 = 10\\ \\mathrm{N}$. Using mg or $mg\\cos\\theta$ is wrong here.","explanationZh":"只需平衡沿斜面向下的分力 $mg\\sin\\theta：F=mg\\sin30^{\\circ}=2\\times10\\times0.5=10\\ \\mathrm{N}。用$ mg $或 mg\\cos\\theta$ 都错。","tags":["trap","forces","incline","resolve"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-force-analysis-equilibrium-012","images":[{"src":"/physics/diagrams/incline-rough.svg","alt":"Block at rest on a rough incline; static friction acts up the slope to oppose the downslope component of weight.","altZh":"粗糙斜面上的静止物块；静摩擦力沿斜面向上以平衡重力的下滑分力。"}],"moduleId":"mechanics","unitId":"forces-newton","topicId":"force-analysis-equilibrium","difficulty":4,"stem":"A block is at rest on a rough incline. What is the direction of the static friction on the block?","stemZh":"物块静止在粗糙斜面上，物块所受静摩擦力的方向如何？","options":[{"id":"A","text":"Up the slope"},{"id":"B","text":"Down the slope"},{"id":"C","text":"Perpendicular to the slope"},{"id":"D","text":"Zero"}],"optionsZh":[{"id":"A","text":"沿斜面向上"},{"id":"B","text":"沿斜面向下"},{"id":"C","text":"垂直斜面"},{"id":"D","text":"为零"}],"answer":"A","explanation":"Weight has a down-slope component, so to keep the block at rest static friction must oppose this tendency: it acts up the slope.","explanationZh":"重力有沿斜面向下的分力，为保持静止，静摩擦力须阻碍该趋势，故沿斜面向上。","tags":["trap","forces","friction","incline","direction"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-force-analysis-equilibrium-013","images":[{"src":"/physics/diagrams/force-triangle.svg","alt":"A hanging weight balanced by a horizontal string and a slanted string; the three forces close to a 3-4-5 triangle.","altZh":"悬挂重物由水平绳与斜绳平衡；三力构成 3-4-5 三角形。"}],"moduleId":"mechanics","unitId":"forces-newton","topicId":"force-analysis-equilibrium","difficulty":4,"stem":"A weight of $30\\ \\mathrm{N}$ is held by a horizontal string $(tension T_{h})$ and a second string at an angle whose $\\sin\\theta = \\dfrac{3}{5}, \\cos\\theta = \\dfrac{4}{5}$. What is $T_{h}$?","stemZh":"重量 $30\\ \\mathrm{N}$ 由水平绳（张力 $T_h）与另一根绳平衡，该绳方向满足 \\sin\\theta=\\dfrac{3}{5}、\\cos\\theta=\\dfrac{4}{5}，求 T_h。$","options":[{"id":"A","text":"$50\\ \\mathrm{N}$"},{"id":"B","text":"$30\\ \\mathrm{N}$"},{"id":"C","text":"$40\\ \\mathrm{N}$"},{"id":"D","text":"$75\\ \\mathrm{N}$"}],"answer":"C","explanation":"Vertically $T_{s}\\sin\\theta = 30$ gives $T_{s} = 50\\ \\mathrm{N}$. Horizontally $T_{h} = T_{s}\\cos\\theta = 50 \\times \\dfrac{4}{5} = 40\\ \\mathrm{N}$.","explanationZh":"竖直方向 $T_s\\sin\\theta=30 得 T_s=50\\ \\mathrm{N}；水平方向 T_h=T_s\\cos\\theta=50\\times\\dfrac{4}{5}=40\\ \\mathrm{N}。$","tags":["trap","forces","three-force","force-triangle"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-force-analysis-equilibrium-014","images":[{"src":"/physics/diagrams/stacked-blocks.svg","alt":"A 2 kg block resting on a 6 kg block on a smooth floor, pulled by a horizontal force on the lower block; friction drives the upper block.","altZh":"光滑地面上 2 kg 物块叠放在 6 kg 物块上，对下层施加水平力；摩擦力带动上层。"}],"moduleId":"mechanics","unitId":"forces-newton","topicId":"force-analysis-equilibrium","difficulty":5,"stem":"$A 2\\ \\mathrm{kg}$ block rests on top of $a 6\\ \\mathrm{kg}$ block on a smooth floor. A horizontal force $F = 16\\ \\mathrm{N}$ pulls the lower block and the two move together. What friction acts on the upper block?","stemZh":"质量 $2\\ \\mathrm{kg}$ 的物块叠放在 $6\\ \\mathrm{kg}$ 物块上，地面光滑。用 $F=16\\ \\mathrm{N}$ 的水平力拉下面物块，两者一起运动。上面物块受到的摩擦力多大？","options":[{"id":"A","text":"$16\\ \\mathrm{N}$"},{"id":"B","text":"$4\\ \\mathrm{N}$"},{"id":"C","text":"$8\\ \\mathrm{N}$"},{"id":"D","text":"$2\\ \\mathrm{N}$"}],"answer":"B","explanation":"Treat the pair as one $body: a = \\dfrac{F}{m+M} = \\dfrac{16}{8} = 2\\ \\mathrm{m/s^{2}}$. The only horizontal force on the upper block is friction, so $f = m a = 2 \\times 2 = 4\\ \\mathrm{N}$.","explanationZh":"$整体法：a=\\dfrac{F}{m+M}=\\dfrac{16}{8}=2\\ \\mathrm{m/s^{2}}。上面物块水平方向只受摩擦力，f=ma=2\\times2=4\\ \\mathrm{N}。$","tags":["trap","forces","friction","connected","whole-body"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-force-analysis-equilibrium-015","images":[{"src":"/physics/diagrams/incline-rough.svg","alt":"Block on a rough 30-degree incline with mu $= 0.5$; first decide whether it slides, then find acceleration.","altZh":"μ=0.5 的粗糙 30° 斜面上的物块；先判断是否下滑，再求加速度。"}],"moduleId":"mechanics","unitId":"forces-newton","topicId":"force-analysis-equilibrium","difficulty":5,"stem":"$A 4\\ \\mathrm{kg}$ block is released on an incline of $\\theta = 30^{\\circ}$ with $\\mu = 0.5$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what happens?","stemZh":"质量 $4\\ \\mathrm{kg}$ 的物块在倾角 $\\theta=30^{\\circ}、\\mu=0.5$ 的斜面上由静止释放，取 $g=10\\ \\mathrm{m/s^{2}}，结果如何？$","options":[{"id":"A","text":"It stays at rest"},{"id":"B","text":"It accelerates down at $5.0\\ \\mathrm{m/s^{2}}$"},{"id":"C","text":"It accelerates down at $3.3\\ \\mathrm{m/s^{2}}$"},{"id":"D","text":"It accelerates down at about $0.67\\ \\mathrm{m/s^{2}}$"}],"optionsZh":[{"id":"A","text":"保持静止"},{"id":"B","text":"$以 5.0\\ \\mathrm{m/s^{2}}$ 沿斜面向下加速"},{"id":"C","text":"$以 3.3\\ \\mathrm{m/s^{2}}$ 沿斜面向下加速"},{"id":"D","text":"以约 $0.67\\ \\mathrm{m/s^{2}}$ 沿斜面向下加速"}],"answer":"D","explanation":"Down-slope component $mg\\sin\\theta = 20\\ \\mathrm{N}$ exceeds the static limit $\\mu mg\\cos\\theta = 0.5 \\times 34.6 \\approx 17.3\\ \\mathrm{N}$, so it slides. Then $a = g(\\sin\\theta - \\mu\\cos\\theta) = 10(0.5 - 0.433) \\approx 0.67\\ \\mathrm{m/s^{2}}$, much less than the smooth value.","explanationZh":"下滑分力 $mg\\sin\\theta=20\\ \\mathrm{N}$ 大于最大静摩擦力 $\\mu mg\\cos\\theta=0.5\\times34.6\\approx17.3\\ \\mathrm{N}，故下滑。a=g(\\sin\\theta-\\mu\\cos\\theta)=10(0.5-0.433)\\approx0.67\\ \\mathrm{m/s^{2}}，远小于光滑斜面的值。$","tags":["trap","forces","incline","friction","decide-then-solve"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-impulse-momentum-001","moduleId":"mechanics","unitId":"momentum","topicId":"impulse-momentum","difficulty":1,"stem":"What is the momentum of $a 2\\ \\mathrm{kg}$ object moving at $3\\ \\mathrm{m/s}$?","stemZh":"质量 $2\\ \\mathrm{kg}$ 的物体以 $3\\ \\mathrm{m/s}$ 运动，其动量多大？","options":[{"id":"A","text":"$6\\ \\mathrm{kg\\cdot m/s}$"},{"id":"B","text":"$5\\ \\mathrm{kg\\cdot m/s}$"},{"id":"C","text":"$1.5\\ \\mathrm{kg\\cdot m/s}$"},{"id":"D","text":"$2\\ \\mathrm{kg\\cdot m/s}$"}],"answer":"A","explanation":"Momentum is $p =$ mv $= 2 \\times 3 = 6\\ \\mathrm{kg\\cdot m/s}$.","explanationZh":"动量 $p=mv=2\\times3=6\\ \\mathrm{kg\\cdot m/s}。$","tags":["basic","momentum","p-mv"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-impulse-momentum-002","moduleId":"mechanics","unitId":"momentum","topicId":"impulse-momentum","difficulty":1,"stem":"A constant force of $10\\ \\mathrm{N}$ acts for $2\\ \\mathrm{s}$. What impulse is delivered?","stemZh":"恒力 $10\\ \\mathrm{N}$ 作用 $2\\ \\mathrm{s}，传递的冲量多大？$","options":[{"id":"A","text":"$5\\ \\mathrm{N\\cdot s}$"},{"id":"B","text":"$20\\ \\mathrm{N\\cdot s}$"},{"id":"C","text":"$12\\ \\mathrm{N\\cdot s}$"},{"id":"D","text":"$200\\ \\mathrm{N\\cdot s}$"}],"answer":"B","explanation":"Impulse is $J = F\\Delta t = 10 \\times 2 = 20\\ \\mathrm{N\\cdot s}$.","explanationZh":"冲量 $J=F\\Delta t=10\\times2=20\\ \\mathrm{N\\cdot s}。$","tags":["basic","impulse","j-ft"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-impulse-momentum-003","moduleId":"mechanics","unitId":"momentum","topicId":"impulse-momentum","difficulty":1,"stem":"$A 4\\ \\mathrm{kg}$ object speeds up from $2\\ \\mathrm{m/s}$ to $5\\ \\mathrm{m/s}$. What is the change in its momentum?","stemZh":"质量 $4\\ \\mathrm{kg}$ 的物体速度由 $2\\ \\mathrm{m/s}$ 增至 $5\\ \\mathrm{m/s}，动量变化多大？$","options":[{"id":"A","text":"$8\\ \\mathrm{kg\\cdot m/s}$"},{"id":"B","text":"$20\\ \\mathrm{kg\\cdot m/s}$"},{"id":"C","text":"$12\\ \\mathrm{kg\\cdot m/s}$"},{"id":"D","text":"$3\\ \\mathrm{kg\\cdot m/s}$"}],"answer":"C","explanation":"$\\Delta p = m\\Delta v = 4 \\times (5 - 2) = 12\\ \\mathrm{kg\\cdot m/s}$.","explanationZh":"$\\Delta p=m\\Delta v=4\\times(5-2)=12\\ \\mathrm{kg\\cdot m/s}。$","tags":["basic","momentum","delta-p"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-impulse-momentum-004","moduleId":"mechanics","unitId":"momentum","topicId":"impulse-momentum","difficulty":2,"stem":"$A 0.5\\ \\mathrm{kg}$ ball moving at $20\\ \\mathrm{m/s}$ is brought to rest in $0.1\\ \\mathrm{s}$. What average force acts on it?","stemZh":"$0.5\\ \\mathrm{kg}$ 的球以 $20\\ \\mathrm{m/s}$ 运动，在 $0.1\\ \\mathrm{s}$ 内停下，所受平均力多大？","options":[{"id":"A","text":"$10\\ \\mathrm{N}$"},{"id":"B","text":"$1\\ \\mathrm{N}$"},{"id":"C","text":"$20\\ \\mathrm{N}$"},{"id":"D","text":"$100\\ \\mathrm{N}$"}],"answer":"D","explanation":"$\\Delta p = 0.5 \\times 20 = 10\\ \\mathrm{kg\\cdot m/s}$, so $F = \\dfrac{\\Delta p}{\\Delta t} = \\dfrac{10}{0.1} = 100\\ \\mathrm{N}$.","explanationZh":"$\\Delta p=0.5\\times20=10\\ \\mathrm{kg\\cdot m/s}，故 F=\\dfrac{\\Delta p}{\\Delta t}=\\dfrac{10}{0.1}=100\\ \\mathrm{N}。$","tags":["basic","impulse","average-force"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-impulse-momentum-005","moduleId":"mechanics","unitId":"momentum","topicId":"impulse-momentum","difficulty":2,"stem":"Which quantity is equivalent to an impulse of $1\\ \\mathrm{N\\cdot s}$?","stemZh":"冲量 $1\\ \\mathrm{N\\cdot s}$ 等价于下列哪一量？","options":[{"id":"A","text":"$1\\ \\mathrm{J}$"},{"id":"B","text":"$1\\ \\mathrm{kg\\cdot m/s}$"},{"id":"C","text":"$1\\ \\mathrm{W}$"},{"id":"D","text":"$1\\ \\mathrm{Pa}$"}],"answer":"B","explanation":"Since $1\\ \\mathrm{N} = 1\\ \\mathrm{kg\\cdot m/s^{2}}$, an impulse $1\\ \\mathrm{N\\cdot s} = 1\\ \\mathrm{kg\\cdot m/s}$, the unit of momentum.","explanationZh":"$因 1\\ \\mathrm{N}=1\\ \\mathrm{kg\\cdot m/s^{2}}，故 1\\ \\mathrm{N\\cdot s}=1\\ \\mathrm{kg\\cdot m/s}，即动量的单位。$","tags":["basic","impulse","units"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-impulse-momentum-006","moduleId":"mechanics","unitId":"momentum","topicId":"impulse-momentum","difficulty":3,"stem":"$A 0.2\\ \\mathrm{kg}$ baseball moving at $30\\ \\mathrm{m/s}$ is hit and leaves at $20\\ \\mathrm{m/s}$ in the opposite direction. What is the magnitude of the change in momentum?","stemZh":"$0.2\\ \\mathrm{kg}$ 的棒球以 $30\\ \\mathrm{m/s}$ 运动，被击打后以 $20\\ \\mathrm{m/s}$ 反向飞出，动量变化量的大小是多少？","options":[{"id":"A","text":"$1\\ \\mathrm{kg\\cdot m/s}$"},{"id":"B","text":"$6\\ \\mathrm{kg\\cdot m/s}$"},{"id":"C","text":"$10\\ \\mathrm{kg\\cdot m/s}$"},{"id":"D","text":"$2\\ \\mathrm{kg\\cdot m/s}$"}],"answer":"C","explanation":"Take the initial direction as $positive: \\Delta p = 0.2(-20 - 30) = -10\\ \\mathrm{kg\\cdot m/s}$, magnitude $10\\ \\mathrm{kg\\cdot m/s}$.","explanationZh":"$取初速度方向为正：\\Delta p=0.2(-20-30)=-10\\ \\mathrm{kg\\cdot m/s}，大小为 10\\ \\mathrm{kg\\cdot m/s}。$","tags":["applied","momentum","change","direction"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-impulse-momentum-007","moduleId":"mechanics","unitId":"momentum","topicId":"impulse-momentum","difficulty":3,"stem":"A force of $50\\ \\mathrm{N}$ acts for $0.04\\ \\mathrm{s}$ on $a 2\\ \\mathrm{kg}$ object initially at rest. What is its final speed?","stemZh":"$50\\ \\mathrm{N}$ 的力作用 $0.04\\ \\mathrm{s}$ 于静止的 $2\\ \\mathrm{kg}$ 物体，末速度多大？","options":[{"id":"A","text":"$0.04\\ \\mathrm{m/s}$"},{"id":"B","text":"$2\\ \\mathrm{m/s}$"},{"id":"C","text":"$100\\ \\mathrm{m/s}$"},{"id":"D","text":"$1\\ \\mathrm{m/s}$"}],"answer":"D","explanation":"$J = F\\Delta t = 50 \\times 0.04 = 2\\ \\mathrm{N\\cdot s} = \\Delta p = m\\Delta v = 2v$, so $v = 1\\ \\mathrm{m/s}$.","explanationZh":"$J=F\\Delta t=50\\times0.04=2\\ \\mathrm{N\\cdot s}=\\Delta p=m\\Delta v=2v，故 v=1\\ \\mathrm{m/s}。$","tags":["applied","impulse","impulse-momentum"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-impulse-momentum-008","moduleId":"mechanics","unitId":"momentum","topicId":"impulse-momentum","difficulty":3,"stem":"$A 5\\ \\mathrm{kg}$ object's speed increases from $2\\ \\mathrm{m/s}$ to $4\\ \\mathrm{m/s}$ under a constant force acting for $1\\ \\mathrm{s}$. What is the force?","stemZh":"$5\\ \\mathrm{kg}$ 物体的速度在 $1\\ \\mathrm{s}$ 内由 $2\\ \\mathrm{m/s}$ 增至 $4\\ \\mathrm{m/s}，所受恒力多大？$","options":[{"id":"A","text":"$10\\ \\mathrm{N}$"},{"id":"B","text":"$5\\ \\mathrm{N}$"},{"id":"C","text":"$20\\ \\mathrm{N}$"},{"id":"D","text":"$2\\ \\mathrm{N}$"}],"answer":"A","explanation":"$\\Delta p = 5 \\times (4 - 2) = 10\\ \\mathrm{kg\\cdot m/s}$, so $F = \\dfrac{\\Delta p}{\\Delta t} = \\dfrac{10}{1} = 10\\ \\mathrm{N}$.","explanationZh":"$\\Delta p=5\\times(4-2)=10\\ \\mathrm{kg\\cdot m/s}，故 F=\\dfrac{\\Delta p}{\\Delta t}=\\dfrac{10}{1}=10\\ \\mathrm{N}。$","tags":["applied","impulse","force"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-impulse-momentum-009","moduleId":"mechanics","unitId":"momentum","topicId":"impulse-momentum","difficulty":3,"stem":"An impulse of $24\\ \\mathrm{N\\cdot s}$ is delivered to $a 3\\ \\mathrm{kg}$ object initially moving at $4\\ \\mathrm{m/s}$ in the direction of the impulse. What is its final speed?","stemZh":"大小为 $24\\ \\mathrm{N\\cdot s}$ 的冲量作用于初速 $4\\ \\mathrm{m/s} 的 3\\ \\mathrm{kg}$ 物体（冲量方向与运动方向相同），末速度多大？","options":[{"id":"A","text":"$8\\ \\mathrm{m/s}$"},{"id":"B","text":"$12\\ \\mathrm{m/s}$"},{"id":"C","text":"$20\\ \\mathrm{m/s}$"},{"id":"D","text":"$4\\ \\mathrm{m/s}$"}],"answer":"B","explanation":"$\\Delta p = 24 = m(v - 4)$, so $v - 4 = 8$ and $v = 12\\ \\mathrm{m/s}$.","explanationZh":"$\\Delta p=24=m(v-4)，得 v-4=8，故 v=12\\ \\mathrm{m/s}。$","tags":["applied","impulse","impulse-momentum"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-impulse-momentum-010","moduleId":"mechanics","unitId":"momentum","topicId":"impulse-momentum","difficulty":3,"stem":"$A 2\\ \\mathrm{kg}$ cart has its velocity changed from $-3\\ \\mathrm{m/s}$ to $+3\\ \\mathrm{m/s}$. What impulse was required?","stemZh":"$2\\ \\mathrm{kg}$ 的小车速度由 $-3\\ \\mathrm{m/s}$ 变为 $+3\\ \\mathrm{m/s}，所需冲量多大？$","options":[{"id":"A","text":"$0\\ \\mathrm{N\\cdot s}$"},{"id":"B","text":"$6\\ \\mathrm{N\\cdot s}$"},{"id":"C","text":"$12\\ \\mathrm{N\\cdot s}$"},{"id":"D","text":"$18\\ \\mathrm{N\\cdot s}$"}],"answer":"C","explanation":"$\\Delta p = m\\Delta v = 2 \\times (3 - (-3)) = 2 \\times 6 = 12\\ \\mathrm{N\\cdot s}$.","explanationZh":"$\\Delta p=m\\Delta v=2\\times(3-(-3))=2\\times6=12\\ \\mathrm{N\\cdot s}。$","tags":["applied","impulse","delta-v"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-impulse-momentum-011","moduleId":"mechanics","unitId":"momentum","topicId":"impulse-momentum","difficulty":4,"stem":"$A 0.5\\ \\mathrm{kg}$ ball moving at $4\\ \\mathrm{m/s}$ is stopped by a wall. Taking the initial direction as positive, what impulse does the wall deliver to the ball?","stemZh":"$0.5\\ \\mathrm{kg}$ 的球以 $4\\ \\mathrm{m/s}$ 运动被墙挡停。取初速度方向为正，墙对球的冲量是多少？","options":[{"id":"A","text":"$+2\\ \\mathrm{N\\cdot s}$"},{"id":"B","text":"$+8\\ \\mathrm{N\\cdot s}$"},{"id":"C","text":"$-8\\ \\mathrm{N\\cdot s}$"},{"id":"D","text":"$-2\\ \\mathrm{N\\cdot s}$"}],"answer":"D","explanation":"$\\Delta p = 0 -$ mv $= -0.5 \\times 4 = -2\\ \\mathrm{N\\cdot s}$. The impulse has the opposite sign to the motion — a common sign-error trap.","explanationZh":"$\\Delta p=0-mv=-0.5\\times4=-2\\ \\mathrm{N\\cdot s}。冲量方向与运动方向相反，这是常见的符号陷阱。$","tags":["trap","impulse","sign"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-impulse-momentum-012","moduleId":"mechanics","unitId":"momentum","topicId":"impulse-momentum","difficulty":4,"stem":"$A 0.2\\ \\mathrm{kg}$ ball strikes a wall at $10\\ \\mathrm{m/s}$ and rebounds at $8\\ \\mathrm{m/s}$ in the opposite direction. What is the magnitude of the impulse on the ball?","stemZh":"$0.2\\ \\mathrm{kg}$ 的球以 $10\\ \\mathrm{m/s}$ 撞墙后以 $8\\ \\mathrm{m/s}$ 反向弹回，球所受冲量的大小是多少？","options":[{"id":"A","text":"$3.6\\ \\mathrm{N\\cdot s}$"},{"id":"B","text":"$0.4\\ \\mathrm{N\\cdot s}$"},{"id":"C","text":"$1.8\\ \\mathrm{N\\cdot s}$"},{"id":"D","text":"$36\\ \\mathrm{N\\cdot s}$"}],"answer":"A","explanation":"Take the incoming direction as $positive: \\Delta p = 0.2(-8 - 10) = -3.6\\ \\mathrm{N\\cdot s}$, so the magnitude is $3.6\\ \\mathrm{N\\cdot s}$. Ignoring the reversal gives a wrong $0.2\\times10=2$ or $0.2\\times8=1.6$.","explanationZh":"$取入射方向为正：\\Delta p=0.2(-8-10)=-3.6\\ \\mathrm{N\\cdot s}，大小为 3.6\\ \\mathrm{N\\cdot s}。忽略反向会错算成 0.2\\times10=2 或 0.2\\times8=1.6。$","tags":["trap","impulse","rebound"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-impulse-momentum-013","moduleId":"mechanics","unitId":"momentum","topicId":"impulse-momentum","difficulty":4,"stem":"$A 1000\\ \\mathrm{kg}$ car travelling at $15\\ \\mathrm{m/s}$ is brought to rest by a wall in $0.5\\ \\mathrm{s}$. What is the average force on the car?","stemZh":"质量 $1000\\ \\mathrm{kg}$ 的汽车以 $15\\ \\mathrm{m/s}$ 行驶，在 $0.5\\ \\mathrm{s}$ 内被墙撞停，汽车所受平均力多大？","options":[{"id":"A","text":"$3\\times10^{3}\\ \\mathrm{N}$"},{"id":"B","text":"$7.5\\times10^{3}\\ \\mathrm{N}$"},{"id":"C","text":"$3\\times10^{4}\\ \\mathrm{N}$"},{"id":"D","text":"$1.5\\times10^{4}\\ \\mathrm{N}$"}],"answer":"C","explanation":"$\\Delta p = 1000 \\times 15 = 1.5\\times10^{4}\\ \\mathrm{kg\\cdot m/s}$, so $F = \\dfrac{\\Delta p}{\\Delta t} = \\dfrac{1.5\\times10^{4}}{0.5} = 3\\times10^{4}\\ \\mathrm{N}$. Forgetting to divide by $\\Delta t$ gives the distractor $1.5\\times10^{4}\\ \\mathrm{N}$.","explanationZh":"$\\Delta p=1000\\times15=1.5\\times10^{4}\\ \\mathrm{kg\\cdot m/s}，故 F=\\dfrac{\\Delta p}{\\Delta t}=\\dfrac{1.5\\times10^{4}}{0.5}=3\\times10^{4}\\ \\mathrm{N}。忘记除以 \\Delta t$ 会得干扰项 $1.5\\times10^{4}\\ \\mathrm{N}。$","tags":["trap","impulse","average-force"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-impulse-momentum-014","moduleId":"mechanics","unitId":"momentum","topicId":"impulse-momentum","difficulty":5,"stem":"To stop a moving object, increasing the stopping time (while keeping the change in momentum fixed) will reduce which quantity?","stemZh":"要使运动的物体停下，在动量变化量不变的前提下延长作用时间，会减小下列哪个量？","options":[{"id":"A","text":"The total impulse"},{"id":"B","text":"The average force"},{"id":"C","text":"The change in momentum"},{"id":"D","text":"The final momentum"}],"optionsZh":[{"id":"A","text":"总冲量"},{"id":"B","text":"平均作用力"},{"id":"C","text":"动量变化量"},{"id":"D","text":"末动量"}],"answer":"B","explanation":"Since $J = \\Delta p = F_{avg}\\Delta t$ and $\\Delta p$ is fixed, a longer $\\Delta t$ lowers the average force. The impulse and momentum change stay the same.","explanationZh":"$因 J=\\Delta p=F_{\\text{avg}}\\Delta t 且 \\Delta p$ 固定，时间 $\\Delta t$ 越长平均力越小；冲量与动量变化量不变。","tags":["trap","impulse","conceptual"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-impulse-momentum-015","moduleId":"mechanics","unitId":"momentum","topicId":"impulse-momentum","difficulty":5,"stem":"$A 2\\ \\mathrm{kg}$ object moving at $3\\ \\mathrm{m/s}$ is acted on by a constant $6\\ \\mathrm{N}$ force opposite to its motion for $1\\ \\mathrm{s}$. What is its final velocity?","stemZh":"$2\\ \\mathrm{kg}$ 的物体以 $3\\ \\mathrm{m/s}$ 运动，受到与其运动反向的恒力 $6\\ \\mathrm{N}$ 作用 $1\\ \\mathrm{s}，末速度多大？$","options":[{"id":"A","text":"$6\\ \\mathrm{m/s}$"},{"id":"B","text":"$3\\ \\mathrm{m/s}$"},{"id":"C","text":"$-3\\ \\mathrm{m/s}$"},{"id":"D","text":"$0\\ \\mathrm{m/s}$"}],"answer":"D","explanation":"The impulse is $-6 \\times 1 = -6\\ \\mathrm{N\\cdot s}$, so $\\Delta v = \\dfrac{-6}{2} = -3\\ \\mathrm{m/s}$ and $v = 3 - 3 = 0$: it just comes to rest. A sign or $v=v_0+at$ slip can give a wrong nonzero value.","explanationZh":"冲量为 $-6\\times1=-6\\ \\mathrm{N\\cdot s}，故 \\Delta v=\\dfrac{-6}{2}=-3\\ \\mathrm{m/s}，v=3-3=0：恰好停下。符号或公式用错会得非零的错误值。$","tags":["trap","impulse","sign","stop"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-momentum-conservation-001","moduleId":"mechanics","unitId":"momentum","topicId":"momentum-conservation","difficulty":1,"stem":"Two $1\\ \\mathrm{kg}$ carts move on a smooth track; one travels at $2\\ \\mathrm{m/s}$, the other is at rest. They stick together. What is their common final speed?","stemZh":"两辆 $1\\ \\mathrm{kg}$ 小车在光滑轨道上运动，一辆以 $2\\ \\mathrm{m/s}$ 行驶，另一辆静止，两者粘在一起，共同末速度多大？","options":[{"id":"A","text":"$1\\ \\mathrm{m/s}$"},{"id":"B","text":"$2\\ \\mathrm{m/s}$"},{"id":"C","text":"$0.5\\ \\mathrm{m/s}$"},{"id":"D","text":"$4\\ \\mathrm{m/s}$"}],"answer":"A","explanation":"Conserve $momentum: (1\\times2 + 1\\times0) = (1+1)v$, so $v = \\dfrac{2}{2} = 1\\ \\mathrm{m/s}$.","explanationZh":"$动量守恒：(1\\times2+1\\times0)=(1+1)v，故 v=\\dfrac{2}{2}=1\\ \\mathrm{m/s}。$","tags":["basic","momentum","inelastic","stick"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-momentum-conservation-002","moduleId":"mechanics","unitId":"momentum","topicId":"momentum-conservation","difficulty":1,"stem":"$A 50\\ \\mathrm{kg}$ person standing on frictionless ice throws $a 5\\ \\mathrm{kg}$ ball forward at $4\\ \\mathrm{m/s}$. What is the person's recoil speed?","stemZh":"质量 $50\\ \\mathrm{kg}$ 的人站在无摩擦冰面上，向前抛出 $5\\ \\mathrm{kg}$ 的球，球速 $4\\ \\mathrm{m/s}，人的反冲速度多大？$","options":[{"id":"A","text":"$4\\ \\mathrm{m/s}$"},{"id":"B","text":"$0.4\\ \\mathrm{m/s}$"},{"id":"C","text":"$0.5\\ \\mathrm{m/s}$"},{"id":"D","text":"$40\\ \\mathrm{m/s}$"}],"answer":"B","explanation":"Initial momentum is zero, so $50v + 5\\times4 = 0$, giving $v = -\\dfrac{20}{50} = -0.4\\ \\mathrm{m/s} (opposite\\text{ the }throw)$.","explanationZh":"初动量为零，故 $50v+5\\times4=0，得 v=-\\dfrac{20}{50}=-0.4\\ \\mathrm{m/s}（与抛出方向相反）。$","tags":["basic","momentum","recoil"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-momentum-conservation-003","moduleId":"mechanics","unitId":"momentum","topicId":"momentum-conservation","difficulty":1,"stem":"Two equal $2\\ \\mathrm{kg}$ masses move toward each other, one at $+3\\ \\mathrm{m/s}$ and the other at $-3\\ \\mathrm{m/s}$. What is the total momentum of the system?","stemZh":"两个质量均为 $2\\ \\mathrm{kg}$ 的物体相向运动，速度分别为 $+3\\ \\mathrm{m/s} 和 -3\\ \\mathrm{m/s}，系统总动量是多少？$","options":[{"id":"A","text":"$12\\ \\mathrm{kg\\cdot m/s}$"},{"id":"B","text":"$6\\ \\mathrm{kg\\cdot m/s}$"},{"id":"C","text":"0"},{"id":"D","text":"$-6\\ \\mathrm{kg\\cdot m/s}$"}],"answer":"C","explanation":"Total momentum $= 2\\times3 + 2\\times(-3) = 6 - 6 = 0$.","explanationZh":"总动量 $=2\\times3+2\\times(-3)=6-6=0。$","tags":["basic","momentum","total"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-momentum-conservation-004","moduleId":"mechanics","unitId":"momentum","topicId":"momentum-conservation","difficulty":2,"stem":"$A 3\\ \\mathrm{kg}$ cart moving at $4\\ \\mathrm{m/s}$ collides and sticks with $a 1\\ \\mathrm{kg}$ cart at rest. What is their common final speed?","stemZh":"$3\\ \\mathrm{kg}$ 的小车以 $4\\ \\mathrm{m/s}$ 运动，与静止的 $1\\ \\mathrm{kg}$ 小车碰撞并粘在一起，共同末速度多大？","options":[{"id":"A","text":"$4\\ \\mathrm{m/s}$"},{"id":"B","text":"$1\\ \\mathrm{m/s}$"},{"id":"C","text":"$2\\ \\mathrm{m/s}$"},{"id":"D","text":"$3\\ \\mathrm{m/s}$"}],"answer":"D","explanation":"Conserve $momentum: 3\\times4 = (3+1)v$, so $v = \\dfrac{12}{4} = 3\\ \\mathrm{m/s}$.","explanationZh":"$动量守恒：3\\times4=(3+1)v，故 v=\\dfrac{12}{4}=3\\ \\mathrm{m/s}。$","tags":["basic","momentum","inelastic"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-momentum-conservation-005","moduleId":"mechanics","unitId":"momentum","topicId":"momentum-conservation","difficulty":2,"stem":"$A 6\\ \\mathrm{kg}$ object at rest explodes into two $pieces: a 4\\ \\mathrm{kg}$ piece moves away at $3\\ \\mathrm{m/s}$. What is the speed of the $2\\ \\mathrm{kg}$ piece?","stemZh":"静止的 $6\\ \\mathrm{kg} 物体炸裂成两块：4\\ \\mathrm{kg}$ 的一块以 $3\\ \\mathrm{m/s}$ 飞出，另一块 $2\\ \\mathrm{kg}$ 的速度大小是多少？","options":[{"id":"A","text":"$3\\ \\mathrm{m/s}$"},{"id":"B","text":"$6\\ \\mathrm{m/s}$"},{"id":"C","text":"$2\\ \\mathrm{m/s}$"},{"id":"D","text":"$12\\ \\mathrm{m/s}$"}],"answer":"B","explanation":"Initial momentum is zero, so $4\\times3 + 2v = 0$, giving $v = -6\\ \\mathrm{m/s}$; the speed is $6\\ \\mathrm{m/s}$ in the opposite direction.","explanationZh":"初动量为零，故 $4\\times3+2v=0，得 v=-6\\ \\mathrm{m/s}；速度大小为 6\\ \\mathrm{m/s}，方向相反。$","tags":["basic","momentum","explosion"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-momentum-conservation-006","moduleId":"mechanics","unitId":"momentum","topicId":"momentum-conservation","difficulty":3,"stem":"$A 2\\ \\mathrm{kg}$ cart moving at $4\\ \\mathrm{m/s}$ has a perfectly elastic head-on collision with an identical $2\\ \\mathrm{kg}$ cart at rest. What is the speed of the second cart after the collision?","stemZh":"$2\\ \\mathrm{kg}$ 的小车以 $4\\ \\mathrm{m/s}$ 与静止的相同 $2\\ \\mathrm{kg}$ 小车发生一维完全弹性正碰，碰后第二辆车速度多大？","options":[{"id":"A","text":"$2\\ \\mathrm{m/s}$"},{"id":"B","text":"$8\\ \\mathrm{m/s}$"},{"id":"C","text":"$4\\ \\mathrm{m/s}$"},{"id":"D","text":"$0\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"For equal masses in a 1D elastic collision, the velocities are exchanged, so the struck cart moves off at $4\\ \\mathrm{m/s}$.","explanationZh":"等质量一维弹性碰撞中速度互换，故被撞小车以 $4\\ \\mathrm{m/s}$ 离开。","tags":["applied","momentum","elastic","equal-mass"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-momentum-conservation-007","moduleId":"mechanics","unitId":"momentum","topicId":"momentum-conservation","difficulty":3,"stem":"$A 1\\ \\mathrm{kg}$ cart moving at $6\\ \\mathrm{m/s}$ collides and sticks with $a 2\\ \\mathrm{kg}$ cart at rest. What is their common final speed?","stemZh":"$1\\ \\mathrm{kg}$ 的小车以 $6\\ \\mathrm{m/s}$ 与静止的 $2\\ \\mathrm{kg}$ 小车碰撞并粘在一起，共同末速度多大？","options":[{"id":"A","text":"$6\\ \\mathrm{m/s}$"},{"id":"B","text":"$3\\ \\mathrm{m/s}$"},{"id":"C","text":"$1\\ \\mathrm{m/s}$"},{"id":"D","text":"$2\\ \\mathrm{m/s}$"}],"answer":"D","explanation":"Conserve $momentum: 1\\times6 = (1+2)v$, so $v = \\dfrac{6}{3} = 2\\ \\mathrm{m/s}$.","explanationZh":"$动量守恒：1\\times6=(1+2)v，故 v=\\dfrac{6}{3}=2\\ \\mathrm{m/s}。$","tags":["applied","momentum","inelastic"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-momentum-conservation-008","moduleId":"mechanics","unitId":"momentum","topicId":"momentum-conservation","difficulty":3,"stem":"$A 2\\ \\mathrm{kg}$ object moving at $3\\ \\mathrm{m/s}$ has a perfectly elastic head-on collision with $a 4\\ \\mathrm{kg}$ object at rest. What is the speed of the $4\\ \\mathrm{kg}$ object after the collision?","stemZh":"$2\\ \\mathrm{kg}$ 的物体以 $3\\ \\mathrm{m/s}$ 与静止的 $4\\ \\mathrm{kg}$ 物体发生一维完全弹性正碰，碰后 $4\\ \\mathrm{kg}$ 物体速度多大？","options":[{"id":"A","text":"$2\\ \\mathrm{m/s}$"},{"id":"B","text":"$3\\ \\mathrm{m/s}$"},{"id":"C","text":"$1\\ \\mathrm{m/s}$"},{"id":"D","text":"$4\\ \\mathrm{m/s}$"}],"answer":"A","explanation":"For a 1D elastic collision with the target at $rest, v_2' = \\dfrac{2m_1}{m_1+m_2}v_1 = \\dfrac{2\\times2}{2+4}\\times3 = 2\\ \\mathrm{m/s}$.","explanationZh":"$一维弹性碰撞且靶静止时，v_2'=\\dfrac{2m_1}{m_1+m_2}v_1=\\dfrac{2\\times2}{2+4}\\times3=2\\ \\mathrm{m/s}。$","tags":["applied","momentum","elastic"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-momentum-conservation-009","moduleId":"mechanics","unitId":"momentum","topicId":"momentum-conservation","difficulty":3,"stem":"$A 100\\ \\mathrm{kg}$ cannon fires $a 2\\ \\mathrm{kg}$ shell at $200\\ \\mathrm{m/s}$. What is the recoil speed of the cannon?","stemZh":"质量 $100\\ \\mathrm{kg}$ 的炮发射 $2\\ \\mathrm{kg}$ 的炮弹，炮弹出膛速度 $200\\ \\mathrm{m/s}，炮的反冲速度多大？$","options":[{"id":"A","text":"$200\\ \\mathrm{m/s}$"},{"id":"B","text":"$4\\ \\mathrm{m/s}$"},{"id":"C","text":"$2\\ \\mathrm{m/s}$"},{"id":"D","text":"$100\\ \\mathrm{m/s}$"}],"answer":"B","explanation":"Momentum is $conserved: 100v + 2\\times200 = 0$, so $v = -\\dfrac{400}{100} = -4\\ \\mathrm{m/s}$; the recoil speed is $4\\ \\mathrm{m/s}$.","explanationZh":"$动量守恒：100v+2\\times200=0，得 v=-\\dfrac{400}{100}=-4\\ \\mathrm{m/s}；反冲速度大小为 4\\ \\mathrm{m/s}。$","tags":["applied","momentum","recoil","cannon"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-momentum-conservation-010","moduleId":"mechanics","unitId":"momentum","topicId":"momentum-conservation","difficulty":3,"stem":"$A 2\\ \\mathrm{kg}$ cart moving right at $3\\ \\mathrm{m/s}$ collides and sticks with $a 4\\ \\mathrm{kg}$ cart moving left at $1\\ \\mathrm{m/s}$. What is their common final speed (right positive)?","stemZh":"$2\\ \\mathrm{kg}$ 小车以 $3\\ \\mathrm{m/s}$ 向右运动，与以 $1\\ \\mathrm{m/s}$ 向左运动的 $4\\ \\mathrm{kg}$ 小车碰撞并粘在一起（取向右为正），共同末速度多大？","options":[{"id":"A","text":"$1\\ \\mathrm{m/s}$"},{"id":"B","text":"$0.5\\ \\mathrm{m/s}$"},{"id":"C","text":"$0.33\\ \\mathrm{m/s}$"},{"id":"D","text":"$1.33\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"Total momentum $= 2\\times3 + 4\\times(-1) = 6 - 4 = 2\\ \\mathrm{kg\\cdot m/s}$. Total mass $6\\ \\mathrm{kg}$, so $v = \\dfrac{2}{6} \\approx 0.33\\ \\mathrm{m/s} (rightward)$.","explanationZh":"总动量 $=2\\times3+4\\times(-1)=6-4=2\\ \\mathrm{kg\\cdot m/s}，总质量 6\\ \\mathrm{kg}，故 v=\\dfrac{2}{6}\\approx0.33\\ \\mathrm{m/s}（向右）。$","tags":["applied","momentum","inelastic","two-direction"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-momentum-conservation-011","moduleId":"mechanics","unitId":"momentum","topicId":"momentum-conservation","difficulty":4,"stem":"A ball is dropped and collides with the ground. During the short impact with the ground, is the ball's momentum conserved?","stemZh":"球被释放下落并与地面碰撞。在与地面接触的短暂冲击过程中，球的动量是否守恒？","options":[{"id":"A","text":"Yes — momentum is always conserved"},{"id":"B","text":"Yes — kinetic energy is conserved"},{"id":"C","text":"Only the speed is conserved"},{"id":"D","text":"No — the ground exerts a large external force"}],"optionsZh":[{"id":"A","text":"是——动量总是守恒"},{"id":"B","text":"是——动能守恒"},{"id":"C","text":"只有速率守恒"},{"id":"D","text":"否——地面对球施加了很大的外力"}],"answer":"D","explanation":"Momentum conservation applies to an isolated system. During impact the ground exerts a large external force, so the ball's own momentum is not conserved (it reverses).","explanationZh":"动量守恒只适用于孤立系统。碰撞时地面对球施加很大的外力，故球自身的动量不守恒（发生反向）。","tags":["trap","momentum","external-force","conceptual"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-momentum-conservation-012","moduleId":"mechanics","unitId":"momentum","topicId":"momentum-conservation","difficulty":4,"stem":"$A 2\\ \\mathrm{kg}$ cart at $3\\ \\mathrm{m/s}$ collides inelastically with an identical $2\\ \\mathrm{kg}$ cart at rest and they stick. How much kinetic energy is lost?","stemZh":"$2\\ \\mathrm{kg}$ 的小车以 $3\\ \\mathrm{m/s}$ 与静止的相同 $2\\ \\mathrm{kg}$ 小车发生完全非弹性碰撞并粘在一起，损失了多少动能？","options":[{"id":"A","text":"$4.5\\ \\mathrm{J}$"},{"id":"B","text":"$9\\ \\mathrm{J}$"},{"id":"C","text":"$0\\ \\mathrm{J}$"},{"id":"D","text":"$13.5\\ \\mathrm{J}$"}],"answer":"A","explanation":"Initial KE $= \\dfrac{1}{2}\\times2\\times3^{2} = 9\\ \\mathrm{J}$. Final speed $= \\dfrac{2\\times3}{4} = 1.5\\ \\mathrm{m/s}$, final KE $= \\dfrac{1}{2}\\times4\\times1.5^{2} = 4.5\\ \\mathrm{J}$. Lost $= 9 - 4.5 = 4.5\\ \\mathrm{J}$.","explanationZh":"初动能 $=\\dfrac{1}{2}\\times2\\times3^{2}=9\\ \\mathrm{J}。末速 =\\dfrac{2\\times3}{4}=1.5\\ \\mathrm{m/s}，末动能 =\\dfrac{1}{2}\\times4\\times1.5^{2}=4.5\\ \\mathrm{J}。损失 =9-4.5=4.5\\ \\mathrm{J}。$","tags":["trap","momentum","inelastic","ke-lost"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-momentum-conservation-013","moduleId":"mechanics","unitId":"momentum","topicId":"momentum-conservation","difficulty":4,"stem":"$A 1\\ \\mathrm{kg}$ cart moving at $4\\ \\mathrm{m/s}$ has a perfectly elastic head-on collision with $a 3\\ \\mathrm{kg}$ cart at rest. What is the velocity of the $1\\ \\mathrm{kg}$ cart after the collision?","stemZh":"$1\\ \\mathrm{kg}$ 的小车以 $4\\ \\mathrm{m/s}$ 与静止的 $3\\ \\mathrm{kg}$ 小车发生一维完全弹性正碰，碰后 $1\\ \\mathrm{kg}$ 小车的速度是多少？","options":[{"id":"A","text":"$2\\ \\mathrm{m/s}$"},{"id":"B","text":"$1\\ \\mathrm{m/s}$"},{"id":"C","text":"$-2\\ \\mathrm{m/s}$"},{"id":"D","text":"$4\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"For 1D elastic $collision, v_1' = \\dfrac{m_1-m_2}{m_1+m_2}v_1 = \\dfrac{1-3}{1+3}\\times4 = -2\\ \\mathrm{m/s} —$ it rebounds. The light cart does not keep moving forward.","explanationZh":"$一维弹性碰撞：v_1'=\\dfrac{m_1-m_2}{m_1+m_2}v_1=\\dfrac{1-3}{1+3}\\times4=-2\\ \\mathrm{m/s}——发生反弹，轻车并不继续前进。$","tags":["trap","momentum","elastic","rebound"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-momentum-conservation-014","moduleId":"mechanics","unitId":"momentum","topicId":"momentum-conservation","difficulty":5,"stem":"A bullet of mass $0.01\\ \\mathrm{kg}$ travelling at $400\\ \\mathrm{m/s}$ embeds in $a 0.99\\ \\mathrm{kg}$ block initially at rest on a frictionless surface. What is the speed of the combined mass?","stemZh":"质量 $0.01\\ \\mathrm{kg}$ 的子弹以 $400\\ \\mathrm{m/s}$ 射入静止在光滑面上的 $0.99\\ \\mathrm{kg}$ 木块并嵌入，共同速度多大？","options":[{"id":"A","text":"$400\\ \\mathrm{m/s}$"},{"id":"B","text":"$4\\ \\mathrm{m/s}$"},{"id":"C","text":"$40\\ \\mathrm{m/s}$"},{"id":"D","text":"$0.4\\ \\mathrm{m/s}$"}],"answer":"B","explanation":"Conserve $momentum: 0.01\\times400 = (0.01+0.99)V$, so $V = \\dfrac{4}{1} = 4\\ \\mathrm{m/s}$. Forgetting the block gives the wrong $400\\ \\mathrm{m/s}$.","explanationZh":"$动量守恒：0.01\\times400=(0.01+0.99)V，故 V=\\dfrac{4}{1}=4\\ \\mathrm{m/s}。忘记木块质量会错得 400\\ \\mathrm{m/s}。$","tags":["trap","momentum","ballistic","embed"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-momentum-conservation-015","moduleId":"mechanics","unitId":"momentum","topicId":"momentum-conservation","difficulty":5,"stem":"$A 10\\ \\mathrm{kg}$ object at rest explodes into two pieces. One $4\\ \\mathrm{kg}$ piece moves right at $5\\ \\mathrm{m/s}$. What is the speed of the other $6\\ \\mathrm{kg}$ piece?","stemZh":"静止的 $10\\ \\mathrm{kg}$ 物体炸裂成两块，其中 $4\\ \\mathrm{kg}$ 的一块以 $5\\ \\mathrm{m/s}$ 向右飞出，另一块 $6\\ \\mathrm{kg}$ 的速度大小是多少？","options":[{"id":"A","text":"$5\\ \\mathrm{m/s}$"},{"id":"B","text":"$2\\ \\mathrm{m/s}$"},{"id":"C","text":"$8.33\\ \\mathrm{m/s}$"},{"id":"D","text":"$3.33\\ \\mathrm{m/s}$"}],"answer":"D","explanation":"Initial momentum is zero, so $4\\times5 + 6v = 0$, giving $v = -\\dfrac{20}{6} \\approx -3.33\\ \\mathrm{m/s}$; the speed is $3.33\\ \\mathrm{m/s}$ opposite.","explanationZh":"初动量为零，故 $4\\times5+6v=0，得 v=-\\dfrac{20}{6}\\approx-3.33\\ \\mathrm{m/s}；速度大小 3.33\\ \\mathrm{m/s}，方向相反。$","tags":["trap","momentum","explosion","two-piece"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-work-power-001","moduleId":"mechanics","unitId":"work-energy","topicId":"work-power","difficulty":1,"stem":"A constant $10\\ \\mathrm{N}$ force moves an object $5\\ \\mathrm{m}$ in the direction of the force. How much work is done?","stemZh":"恒力 $10\\ \\mathrm{N}$ 使物体沿力的方向移动 $5\\ \\mathrm{m}，做的功是多少？$","options":[{"id":"A","text":"$50\\ \\mathrm{J}$"},{"id":"B","text":"$2\\ \\mathrm{J}$"},{"id":"C","text":"$15\\ \\mathrm{J}$"},{"id":"D","text":"$500\\ \\mathrm{J}$"}],"answer":"A","explanation":"Work is $W = Fs\\cos 0^{\\circ} = 10 \\times 5 = 50\\ \\mathrm{J}$.","explanationZh":"$功 W=Fs\\cos0^{\\circ}=10\\times5=50\\ \\mathrm{J}。$","tags":["basic","work","w-fs"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-work-power-002","moduleId":"mechanics","unitId":"work-energy","topicId":"work-power","difficulty":1,"stem":"$A 20\\ \\mathrm{N}$ force acts at $60^{\\circ}$ to the direction of $a 4\\ \\mathrm{m}$ displacement. How much work is done?","stemZh":"$20\\ \\mathrm{N}$ 的力与 $4\\ \\mathrm{m}$ 位移方向成 $60^{\\circ}$ 角，做的功是多少？","options":[{"id":"A","text":"$80\\ \\mathrm{J}$"},{"id":"B","text":"$40\\ \\mathrm{J}$"},{"id":"C","text":"$20\\ \\mathrm{J}$"},{"id":"D","text":"$8\\ \\mathrm{J}$"}],"answer":"B","explanation":"$W = Fs\\cos\\theta = 20 \\times 4 \\times \\cos 60^{\\circ} = 80 \\times 0.5 = 40\\ \\mathrm{J}$.","explanationZh":"$W=Fs\\cos\\theta=20\\times4\\times\\cos60^{\\circ}=80\\times0.5=40\\ \\mathrm{J}。$","tags":["basic","work","angle"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-work-power-003","moduleId":"mechanics","unitId":"work-energy","topicId":"work-power","difficulty":1,"stem":"An engine does $200\\ \\mathrm{J}$ of work in $10\\ \\mathrm{s}$. What is its average power?","stemZh":"发动机在 $10\\ \\mathrm{s}$ 内做功 $200\\ \\mathrm{J}，平均功率多大？$","options":[{"id":"A","text":"$2000\\ \\mathrm{W}$"},{"id":"B","text":"$10\\ \\mathrm{W}$"},{"id":"C","text":"$20\\ \\mathrm{W}$"},{"id":"D","text":"$2\\ \\mathrm{W}$"}],"answer":"C","explanation":"Power is $P = \\dfrac{W}{t} = \\dfrac{200}{10} = 20\\ \\mathrm{W}$.","explanationZh":"功率 $P=\\dfrac{W}{t}=\\dfrac{200}{10}=20\\ \\mathrm{W}。$","tags":["basic","power","p-wt"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-work-power-004","moduleId":"mechanics","unitId":"work-energy","topicId":"work-power","difficulty":2,"stem":"A block slides horizontally. How much work does the upward normal force do on it?","stemZh":"物块水平滑动，向上的支持力对它做多少功？","options":[{"id":"A","text":"Positive work"},{"id":"B","text":"Negative work"},{"id":"C","text":"Maximum work"},{"id":"D","text":"Zero work"}],"optionsZh":[{"id":"A","text":"正功"},{"id":"B","text":"负功"},{"id":"C","text":"最大功"},{"id":"D","text":"零功"}],"answer":"D","explanation":"Work is $Fs\\cos\\theta$; the normal force is perpendicular to the horizontal displacement, so $\\cos 90^{\\circ} = 0$ and the work is zero.","explanationZh":"功为 $Fs\\cos\\theta；支持力与水平位移垂直，\\cos90^{\\circ}=0，故做功为零。$","tags":["basic","work","normal-zero"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-work-power-005","moduleId":"mechanics","unitId":"work-energy","topicId":"work-power","difficulty":2,"stem":"A force of $40\\ \\mathrm{N}$ moves an object at a steady $5\\ \\mathrm{m/s}$. What power is being delivered?","stemZh":"$40\\ \\mathrm{N}$ 的力使物体以恒定 $5\\ \\mathrm{m/s}$ 运动，输出功率多大？","options":[{"id":"A","text":"$8\\ \\mathrm{W}$"},{"id":"B","text":"$200\\ \\mathrm{W}$"},{"id":"C","text":"$45\\ \\mathrm{W}$"},{"id":"D","text":"$1000\\ \\mathrm{W}$"}],"answer":"B","explanation":"Power is $P =$ Fv $= 40 \\times 5 = 200\\ \\mathrm{W} (force\\text{ parallel to }velocity)$.","explanationZh":"功率 $P=Fv=40\\times5=200\\ \\mathrm{W}（力与速度同向）。$","tags":["basic","power","p-fv"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-work-power-006","moduleId":"mechanics","unitId":"work-energy","topicId":"work-power","difficulty":3,"stem":"Lifting $a 5\\ \\mathrm{kg}$ mass by $2\\ \\mathrm{m} vertically$, taking $g = 10\\ \\mathrm{m/s^{2}}$. How much work is done against gravity?","stemZh":"$将 5\\ \\mathrm{kg}$ 的物体竖直提升 $2\\ \\mathrm{m}，取 g=10\\ \\mathrm{m/s^{2}}，克服重力做多少功？$","options":[{"id":"A","text":"$10\\ \\mathrm{J}$"},{"id":"B","text":"$50\\ \\mathrm{J}$"},{"id":"C","text":"$100\\ \\mathrm{J}$"},{"id":"D","text":"$250\\ \\mathrm{J}$"}],"answer":"C","explanation":"Work against gravity $=$ mgh $= 5 \\times 10 \\times 2 = 100\\ \\mathrm{J}$.","explanationZh":"克服重力做功 $=mgh=5\\times10\\times2=100\\ \\mathrm{J}。$","tags":["applied","work","gravity"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-work-power-007","images":[{"src":"/physics/diagrams/incline-smooth-30.svg","alt":"Block pulled 3 m up a smooth 30-degree incline; work against gravity is mgh where $h = 3 \\sin 30$ degrees.","altZh":"沿光滑 30° 斜面拉动 3 m；克服重力做功为 mgh，其中 h=3 \\sin30°。"}],"moduleId":"mechanics","unitId":"work-energy","topicId":"work-power","difficulty":3,"stem":"$A 2\\ \\mathrm{kg}$ block is pulled $3\\ \\mathrm{m}$ up a smooth $30^{\\circ}$ incline. Taking $g = 10\\ \\mathrm{m/s^{2}}$, how much work is done against gravity?","stemZh":"质量 $2\\ \\mathrm{kg}$ 的物块沿光滑 $30^{\\circ}$ 斜面被拉动 $3\\ \\mathrm{m}，取 g=10\\ \\mathrm{m/s^{2}}，克服重力做多少功？$","options":[{"id":"A","text":"$60\\ \\mathrm{J}$"},{"id":"B","text":"$15\\ \\mathrm{J}$"},{"id":"C","text":"$52\\ \\mathrm{J}$"},{"id":"D","text":"$30\\ \\mathrm{J}$"}],"answer":"D","explanation":"Vertical rise is $h = 3\\sin 30^{\\circ} = 1.5\\ \\mathrm{m}$, so work against gravity $=$ mgh $= 2 \\times 10 \\times 1.5 = 30\\ \\mathrm{J} (independent\\text{ of }path)$.","explanationZh":"竖直升高 $h=3\\sin30^{\\circ}=1.5\\ \\mathrm{m}，克服重力做功 =mgh=2\\times10\\times1.5=30\\ \\mathrm{J}（与路径无关）。$","tags":["applied","work","incline","gravity"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-work-power-008","moduleId":"mechanics","unitId":"work-energy","topicId":"work-power","difficulty":3,"stem":"An engine does $600\\ \\mathrm{J}$ of work in $20\\ \\mathrm{s}$. What is its average power output?","stemZh":"发动机在 $20\\ \\mathrm{s}$ 内做功 $600\\ \\mathrm{J}，平均输出功率多大？$","options":[{"id":"A","text":"$30\\ \\mathrm{W}$"},{"id":"B","text":"$12\\ \\mathrm{kW}$"},{"id":"C","text":"$120\\ \\mathrm{W}$"},{"id":"D","text":"$3\\ \\mathrm{W}$"}],"answer":"A","explanation":"$P = \\dfrac{W}{t} = \\dfrac{600}{20} = 30\\ \\mathrm{W}$.","explanationZh":"$P=\\dfrac{W}{t}=\\dfrac{600}{20}=30\\ \\mathrm{W}。$","tags":["applied","power","average"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-work-power-009","moduleId":"mechanics","unitId":"work-energy","topicId":"work-power","difficulty":3,"stem":"A car engine delivers $1000\\ \\mathrm{W}$ while the car moves at $20\\ \\mathrm{m/s}$. What forward force does the engine provide?","stemZh":"汽车发动机输出功率 $1000\\ \\mathrm{W}，汽车以 20\\ \\mathrm{m/s}$ 行驶，发动机提供的向前推力多大？","options":[{"id":"A","text":"$20000\\ \\mathrm{N}$"},{"id":"B","text":"$50\\ \\mathrm{N}$"},{"id":"C","text":"$20\\ \\mathrm{N}$"},{"id":"D","text":"$500\\ \\mathrm{N}$"}],"answer":"B","explanation":"From $P =$ Fv (force parallel to $motion), F = \\dfrac{P}{v} = \\dfrac{1000}{20} = 50\\ \\mathrm{N}$.","explanationZh":"$由 P=Fv（力与运动同向），F=\\dfrac{P}{v}=\\dfrac{1000}{20}=50\\ \\mathrm{N}。$","tags":["applied","power","force"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-work-power-010","moduleId":"mechanics","unitId":"work-energy","topicId":"work-power","difficulty":3,"stem":"$A 10\\ \\mathrm{N}$ friction force opposes an object's motion over $4\\ \\mathrm{m}$. What work does friction do?","stemZh":"$10\\ \\mathrm{N}$ 的滑动摩擦力阻碍物体运动，物体移动 $4\\ \\mathrm{m}，摩擦力做多少功？$","options":[{"id":"A","text":"$40\\ \\mathrm{J}$"},{"id":"B","text":"$400\\ \\mathrm{J}$"},{"id":"C","text":"$-40\\ \\mathrm{J}$"},{"id":"D","text":"$0\\ \\mathrm{J}$"}],"answer":"C","explanation":"Friction opposes displacement, so the work is $negative: W =$ -Fs $= -10 \\times 4 = -40\\ \\mathrm{J}$.","explanationZh":"$摩擦力与位移反向，故做负功：W=-Fs=-10\\times4=-40\\ \\mathrm{J}。$","tags":["applied","work","friction","negative"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-work-power-011","moduleId":"mechanics","unitId":"work-energy","topicId":"work-power","difficulty":4,"stem":"A mass moves in a circle at constant speed. What work is done by the centripetal force over one full revolution?","stemZh":"物体以恒定速率做圆周运动，向心力在一整圈内做了多少功？","options":[{"id":"A","text":"Positive work"},{"id":"B","text":"Negative work"},{"id":"C","text":"Equal to $mv^{2}$"},{"id":"D","text":"0"}],"optionsZh":[{"id":"A","text":"正功"},{"id":"B","text":"负功"},{"id":"C","text":"等于 $mv^{2}$"},{"id":"D","text":"0"}],"answer":"D","explanation":"The centripetal force is always perpendicular to the instantaneous displacement, so $\\cos 90^{\\circ} = 0$ and the total work over a full circle is zero.","explanationZh":"向心力始终与瞬时位移垂直，故 $\\cos90^{\\circ}=0，整圈总功为零。$","tags":["trap","work","centripetal","perpendicular"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-work-power-012","moduleId":"mechanics","unitId":"work-energy","topicId":"work-power","difficulty":4,"stem":"$A 3\\ \\mathrm{kg}$ mass is lifted vertically by $4\\ \\mathrm{m}$ by any path. Taking $g = 10\\ \\mathrm{m/s^{2}}$, how much work is done against gravity?","stemZh":"$3\\ \\mathrm{kg}$ 的物体沿任意路径竖直提升 $4\\ \\mathrm{m}，取 g=10\\ \\mathrm{m/s^{2}}，克服重力做多少功？$","options":[{"id":"A","text":"$120\\ \\mathrm{J}$"},{"id":"B","text":"$30\\ \\mathrm{J}$"},{"id":"C","text":"It depends on the path length"},{"id":"D","text":"$12\\ \\mathrm{J}$"}],"answer":"A","explanation":"Work against gravity depends only on the vertical $height: W =$ mgh $= 3 \\times 10 \\times 4 = 120\\ \\mathrm{J}$, not on the path taken.","explanationZh":"$克服重力做功只取决于竖直高度：W=mgh=3\\times10\\times4=120\\ \\mathrm{J}，与路径无关。$","tags":["trap","work","gravity","path-independent"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-work-power-013","moduleId":"mechanics","unitId":"work-energy","topicId":"work-power","difficulty":4,"stem":"$A 4\\ \\mathrm{kg}$ mass falls freely through $5\\ \\mathrm{m}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what work is done by gravity?","stemZh":"质量 $4\\ \\mathrm{kg}$ 的物体自由下落 $5\\ \\mathrm{m}，取 g=10\\ \\mathrm{m/s^{2}}，重力做多少功？$","options":[{"id":"A","text":"$-200\\ \\mathrm{J}$"},{"id":"B","text":"$0\\ \\mathrm{J}$"},{"id":"C","text":"$+200\\ \\mathrm{J}$"},{"id":"D","text":"$+20\\ \\mathrm{J}$"}],"answer":"C","explanation":"Gravity acts downward, the same direction as the displacement, so it does positive $work: W =$ mgh $= 4 \\times 10 \\times 5 = +200\\ \\mathrm{J}. (Work\\text{ against gravity would be }-200\\ \\mathrm{J}.)$","explanationZh":"$重力向下，与位移同向，做正功：W=mgh=4\\times10\\times5=+200\\ \\mathrm{J}（克服重力做功才是 -200\\ \\mathrm{J}）。$","tags":["trap","work","gravity","sign"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-work-power-014","moduleId":"mechanics","unitId":"work-energy","topicId":"work-power","difficulty":5,"stem":"$A 50\\ \\mathrm{kg}$ person climbs a height of $3\\ \\mathrm{m}$ in $5\\ \\mathrm{s}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is the average power output?","stemZh":"质量 $50\\ \\mathrm{kg}$ 的人用 $5\\ \\mathrm{s}$ 爬升 $3\\ \\mathrm{m}，取 g=10\\ \\mathrm{m/s^{2}}，平均输出功率多大？$","options":[{"id":"A","text":"$150\\ \\mathrm{W}$"},{"id":"B","text":"$300\\ \\mathrm{W}$"},{"id":"C","text":"$1500\\ \\mathrm{W}$"},{"id":"D","text":"$30\\ \\mathrm{W}$"}],"answer":"B","explanation":"Work against gravity $=$ mgh $= 50 \\times 10 \\times 3 = 1500\\ \\mathrm{J}$, so average power $= \\dfrac{1500}{5} = 300\\ \\mathrm{W}$. Forgetting h or g gives a wrong value.","explanationZh":"克服重力做功 $=mgh=50\\times10\\times3=1500\\ \\mathrm{J}，平均功率 =\\dfrac{1500}{5}=300\\ \\mathrm{W}。漏掉 h 或 g$ 会算错。","tags":["trap","power","climb","average"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-work-power-015","moduleId":"mechanics","unitId":"work-energy","topicId":"work-power","difficulty":5,"stem":"$A 30\\ \\mathrm{N}$ force pushes a box $2\\ \\mathrm{m}$ but is directed $30^{\\circ}$ above the horizontal. How much work is done by the force?","stemZh":"$30\\ \\mathrm{N}$ 的力推箱子前进 $2\\ \\mathrm{m}，但力与水平方向成 30^{\\circ}$ 向上，该力做多少功？","options":[{"id":"A","text":"$60\\ \\mathrm{J}$"},{"id":"B","text":"$30\\ \\mathrm{J}$"},{"id":"C","text":"$15\\ \\mathrm{J}$"},{"id":"D","text":"$52\\ \\mathrm{J}$"}],"answer":"D","explanation":"$W = Fs\\cos\\theta = 30 \\times 2 \\times \\cos 30^{\\circ} = 60 \\times 0.866 \\approx 52\\ \\mathrm{J}$. Using Fs alone gives the wrong $60\\ \\mathrm{J}$.","explanationZh":"$W=Fs\\cos\\theta=30\\times2\\times\\cos30^{\\circ}=60\\times0.866\\approx52\\ \\mathrm{J}。直接用$ Fs 会错得 $60\\ \\mathrm{J}。$","tags":["trap","work","angle","\\cos"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-energy-theorem-001","moduleId":"mechanics","unitId":"work-energy","topicId":"energy-theorem","difficulty":1,"stem":"What is the kinetic energy of $a 2\\ \\mathrm{kg}$ object moving at $3\\ \\mathrm{m/s}$?","stemZh":"质量 $2\\ \\mathrm{kg}$ 的物体以 $3\\ \\mathrm{m/s}$ 运动，动能多大？","options":[{"id":"A","text":"$9\\ \\mathrm{J}$"},{"id":"B","text":"$6\\ \\mathrm{J}$"},{"id":"C","text":"$3\\ \\mathrm{J}$"},{"id":"D","text":"$18\\ \\mathrm{J}$"}],"answer":"A","explanation":"Kinetic energy is $E_k = \\dfrac{1}{2}mv^{2} = \\dfrac{1}{2} \\times 2 \\times 3^{2} = 9\\ \\mathrm{J}$.","explanationZh":"动能 $E_k=\\dfrac{1}{2}mv^{2}=\\dfrac{1}{2}\\times2\\times3^{2}=9\\ \\mathrm{J}。$","tags":["basic","energy","kinetic"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-energy-theorem-002","moduleId":"mechanics","unitId":"work-energy","topicId":"energy-theorem","difficulty":1,"stem":"What is the kinetic energy of $a 4\\ \\mathrm{kg}$ object moving at $2\\ \\mathrm{m/s}$?","stemZh":"质量 $4\\ \\mathrm{kg}$ 的物体以 $2\\ \\mathrm{m/s}$ 运动，动能多大？","options":[{"id":"A","text":"$4\\ \\mathrm{J}$"},{"id":"B","text":"$8\\ \\mathrm{J}$"},{"id":"C","text":"$16\\ \\mathrm{J}$"},{"id":"D","text":"$2\\ \\mathrm{J}$"}],"answer":"B","explanation":"$E_k = \\dfrac{1}{2}mv^{2} = \\dfrac{1}{2} \\times 4 \\times 2^{2} = 8\\ \\mathrm{J}$.","explanationZh":"$E_k=\\dfrac{1}{2}mv^{2}=\\dfrac{1}{2}\\times4\\times2^{2}=8\\ \\mathrm{J}。$","tags":["basic","energy","kinetic"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-energy-theorem-003","moduleId":"mechanics","unitId":"work-energy","topicId":"energy-theorem","difficulty":1,"stem":"What is the gravitational potential energy, relative to the floor, of $a 5\\ \\mathrm{kg}$ mass raised $3\\ \\mathrm{m}$? Take $g = 10\\ \\mathrm{m/s^{2}}$.","stemZh":"质量 $5\\ \\mathrm{kg}$ 的物体升高 $3\\ \\mathrm{m}，取 g=10\\ \\mathrm{m/s^{2}}，相对地面的重力势能多大？$","options":[{"id":"A","text":"$15\\ \\mathrm{J}$"},{"id":"B","text":"$50\\ \\mathrm{J}$"},{"id":"C","text":"$150\\ \\mathrm{J}$"},{"id":"D","text":"$500\\ \\mathrm{J}$"}],"answer":"C","explanation":"Potential energy is $E_p =$ mgh $= 5 \\times 10 \\times 3 = 150\\ \\mathrm{J}$.","explanationZh":"重力势能 $E_p=mgh=5\\times10\\times3=150\\ \\mathrm{J}。$","tags":["basic","energy","potential"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-energy-theorem-004","moduleId":"mechanics","unitId":"work-energy","topicId":"energy-theorem","difficulty":2,"stem":"What is the kinetic energy of $a 3\\ \\mathrm{kg}$ object moving at $2\\ \\mathrm{m/s}$?","stemZh":"质量 $3\\ \\mathrm{kg}$ 的物体以 $2\\ \\mathrm{m/s}$ 运动，动能多大？","options":[{"id":"A","text":"$12\\ \\mathrm{J}$"},{"id":"B","text":"$3\\ \\mathrm{J}$"},{"id":"C","text":"$9\\ \\mathrm{J}$"},{"id":"D","text":"$6\\ \\mathrm{J}$"}],"answer":"D","explanation":"$E_k = \\dfrac{1}{2}mv^{2} = \\dfrac{1}{2} \\times 3 \\times 2^{2} = 6\\ \\mathrm{J}$.","explanationZh":"$E_k=\\dfrac{1}{2}mv^{2}=\\dfrac{1}{2}\\times3\\times2^{2}=6\\ \\mathrm{J}。$","tags":["basic","energy","kinetic"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-energy-theorem-005","moduleId":"mechanics","unitId":"work-energy","topicId":"energy-theorem","difficulty":2,"stem":"$A 2\\ \\mathrm{kg}$ object is accelerated from rest to $5\\ \\mathrm{m/s}$. By the work-energy theorem, what net work was done on it?","stemZh":"$2\\ \\mathrm{kg}$ 的物体由静止加速到 $5\\ \\mathrm{m/s}，由动能定理，其合外力做功是多少？$","options":[{"id":"A","text":"$10\\ \\mathrm{J}$"},{"id":"B","text":"$25\\ \\mathrm{J}$"},{"id":"C","text":"$50\\ \\mathrm{J}$"},{"id":"D","text":"$5\\ \\mathrm{J}$"}],"answer":"B","explanation":"Net work equals the change in kinetic $energy: W_{net} = \\Delta E_k = \\dfrac{1}{2} \\times 2 \\times 5^{2} = 25\\ \\mathrm{J}$.","explanationZh":"$合外力做功等于动能变化：W_{\\text{net}}=\\Delta E_k=\\dfrac{1}{2}\\times2\\times5^{2}=25\\ \\mathrm{J}。$","tags":["basic","energy","work-energy-theorem"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-energy-theorem-006","moduleId":"mechanics","unitId":"work-energy","topicId":"energy-theorem","difficulty":3,"stem":"$A 3\\ \\mathrm{kg}$ object is accelerated from rest to $4\\ \\mathrm{m/s}$. What net work was done on it?","stemZh":"$3\\ \\mathrm{kg}$ 的物体由静止加速到 $4\\ \\mathrm{m/s}，合外力做了多少功？$","options":[{"id":"A","text":"$12\\ \\mathrm{J}$"},{"id":"B","text":"$48\\ \\mathrm{J}$"},{"id":"C","text":"$24\\ \\mathrm{J}$"},{"id":"D","text":"$8\\ \\mathrm{J}$"}],"answer":"C","explanation":"$W_{net} = \\Delta E_k = \\dfrac{1}{2} \\times 3 \\times 4^{2} = 24\\ \\mathrm{J}$.","explanationZh":"$W_{\\text{net}}=\\Delta E_k=\\dfrac{1}{2}\\times3\\times4^{2}=24\\ \\mathrm{J}。$","tags":["applied","energy","work-energy-theorem"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-energy-theorem-007","moduleId":"mechanics","unitId":"work-energy","topicId":"energy-theorem","difficulty":3,"stem":"$A 2\\ \\mathrm{kg}$ object slows from $6\\ \\mathrm{m/s}$ to $2\\ \\mathrm{m/s}$. What is the net work done on it?","stemZh":"$2\\ \\mathrm{kg}$ 的物体速度由 $6\\ \\mathrm{m/s}$ 减到 $2\\ \\mathrm{m/s}，合外力做功是多少？$","options":[{"id":"A","text":"$32\\ \\mathrm{J}$"},{"id":"B","text":"$16\\ \\mathrm{J}$"},{"id":"C","text":"$0\\ \\mathrm{J}$"},{"id":"D","text":"$-32\\ \\mathrm{J}$"}],"answer":"D","explanation":"$W_{net} = \\Delta E_k = \\dfrac{1}{2} \\times 2 \\times (2^{2} - 6^{2}) = 4 - 36 = -32\\ \\mathrm{J}$. Slowing means negative net work.","explanationZh":"$W_{\\text{net}}=\\Delta E_k=\\dfrac{1}{2}\\times2\\times(2^{2}-6^{2})=4-36=-32\\ \\mathrm{J}。减速意味着合外力做负功。$","tags":["applied","energy","work-energy-theorem","negative"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-energy-theorem-008","moduleId":"mechanics","unitId":"work-energy","topicId":"energy-theorem","difficulty":3,"stem":"$A 0.5\\ \\mathrm{kg}$ ball is dropped from rest and falls $5\\ \\mathrm{m}$ with no air resistance. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its speed at the bottom?","stemZh":"$0.5\\ \\mathrm{kg}$ 的球由静止下落 $5\\ \\mathrm{m}，无空气阻力，取 g=10\\ \\mathrm{m/s^{2}}，落地时速度多大？$","options":[{"id":"A","text":"$10\\ \\mathrm{m/s}$"},{"id":"B","text":"$5\\ \\mathrm{m/s}$"},{"id":"C","text":"$25\\ \\mathrm{m/s}$"},{"id":"D","text":"$7.1\\ \\mathrm{m/s}$"}],"answer":"A","explanation":"By energy conservation, mgh $= \\dfrac{1}{2}mv^{2}$, so $v = \\sqrt{2gh} = \\sqrt{2 \\times 10 \\times 5} = 10\\ \\mathrm{m/s}$.","explanationZh":"$由机械能守恒，mgh=\\dfrac{1}{2}mv^{2}，故 v=\\sqrt{2gh}=\\sqrt{2\\times10\\times5}=10\\ \\mathrm{m/s}。$","tags":["applied","energy","conservation","fall"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-energy-theorem-009","images":[{"src":"/physics/diagrams/incline-smooth-30.svg","alt":"Block sliding 4 m down a smooth 30-degree incline; vertical drop is $4 \\sin 30 degrees$, giving the final speed by energy conservation.","altZh":"物块沿光滑 30° 斜面下滑 4 m；竖直下降为 4 \\sin30°，由机械能守恒得末速度。"}],"moduleId":"mechanics","unitId":"work-energy","topicId":"energy-theorem","difficulty":3,"stem":"$A 1\\ \\mathrm{kg}$ block slides $4\\ \\mathrm{m}$ down a smooth $30^{\\circ}$ incline from rest. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its speed at the bottom?","stemZh":"质量 $1\\ \\mathrm{kg}$ 的物块由静止沿光滑 $30^{\\circ}$ 斜面下滑 $4\\ \\mathrm{m}，取 g=10\\ \\mathrm{m/s^{2}}，到达底端时速度多大？$","options":[{"id":"A","text":"$4\\ \\mathrm{m/s}$"},{"id":"B","text":"$6.3\\ \\mathrm{m/s}$"},{"id":"C","text":"$8\\ \\mathrm{m/s}$"},{"id":"D","text":"$2\\ \\mathrm{m/s}$"}],"answer":"B","explanation":"Vertical drop $h = 4\\sin 30^{\\circ} = 2\\ \\mathrm{m}$. Energy conservation gives $v = \\sqrt{2gh} = \\sqrt{2 \\times 10 \\times 2} \\approx 6.3\\ \\mathrm{m/s}$.","explanationZh":"竖直下降 $h=4\\sin30^{\\circ}=2\\ \\mathrm{m}。机械能守恒得 v=\\sqrt{2gh}=\\sqrt{2\\times10\\times2}\\approx6.3\\ \\mathrm{m/s}。$","tags":["applied","energy","conservation","incline"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-energy-theorem-010","moduleId":"mechanics","unitId":"work-energy","topicId":"energy-theorem","difficulty":3,"stem":"$A 2\\ \\mathrm{kg}$ object is pushed from rest by a constant net force of $10\\ \\mathrm{N}$ over $5\\ \\mathrm{m}$. What is its final speed?","stemZh":"$2\\ \\mathrm{kg}$ 的物体在恒定合外力 $10\\ \\mathrm{N}$ 作用下由静止移动 $5\\ \\mathrm{m}，末速度多大？$","options":[{"id":"A","text":"$5\\ \\mathrm{m/s}$"},{"id":"B","text":"$10\\ \\mathrm{m/s}$"},{"id":"C","text":"$7.1\\ \\mathrm{m/s}$"},{"id":"D","text":"$50\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"Net work $=$ Fs $= 10 \\times 5 = 50\\ \\mathrm{J} = \\Delta E_k = \\dfrac{1}{2} \\times 2 \\times v^{2}$, so $v^{2} = 50$ and $v \\approx 7.1\\ \\mathrm{m/s}$.","explanationZh":"合外力做功 $=Fs=10\\times5=50\\ \\mathrm{J}=\\Delta E_k=\\dfrac{1}{2}\\times2\\times v^{2}，故 v^{2}=50，v\\approx7.1\\ \\mathrm{m/s}。$","tags":["applied","energy","work-energy-theorem"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-energy-theorem-011","moduleId":"mechanics","unitId":"work-energy","topicId":"energy-theorem","difficulty":4,"stem":"$A 2\\ \\mathrm{kg}$ block slides down a frictionless incline, dropping $3\\ \\mathrm{m}$ vertically. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its speed at the bottom?","stemZh":"$2\\ \\mathrm{kg}$ 的物块沿光滑斜面下滑，竖直下降 $3\\ \\mathrm{m}，取 g=10\\ \\mathrm{m/s^{2}}，底端速度多大？$","options":[{"id":"A","text":"$6\\ \\mathrm{m/s}$"},{"id":"B","text":"$3\\ \\mathrm{m/s}$"},{"id":"C","text":"$10\\ \\mathrm{m/s}$"},{"id":"D","text":"$7.7\\ \\mathrm{m/s}$"}],"answer":"D","explanation":"Mechanical energy is $conserved: v = \\sqrt{2gh} = \\sqrt{2 \\times 10 \\times 3} = \\sqrt{60} \\approx 7.7\\ \\mathrm{m/s}$. Mass does not matter; using gh alone gives the wrong 3.","explanationZh":"$机械能守恒：v=\\sqrt{2gh}=\\sqrt{2\\times10\\times3}=\\sqrt{60}\\approx7.7\\ \\mathrm{m/s}。与质量无关；只用$ gh 会错得 3。","tags":["trap","energy","conservation","sqrt"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-energy-theorem-012","moduleId":"mechanics","unitId":"work-energy","topicId":"energy-theorem","difficulty":4,"stem":"A ball is thrown vertically upward at $10\\ \\mathrm{m/s}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what maximum height does it reach?","stemZh":"小球以 $10\\ \\mathrm{m/s}$ 竖直上抛，取 $g=10\\ \\mathrm{m/s^{2}}，上升的最大高度是多少？$","options":[{"id":"A","text":"$5\\ \\mathrm{m}$"},{"id":"B","text":"$10\\ \\mathrm{m}$"},{"id":"C","text":"$20\\ \\mathrm{m}$"},{"id":"D","text":"$2.5\\ \\mathrm{m}$"}],"answer":"A","explanation":"At the top all kinetic energy becomes $potential: \\dfrac{1}{2}mv^{2} = mgh$, so $h = \\dfrac{v^{2}}{2g} = \\dfrac{100}{20} = 5\\ \\mathrm{m}$.","explanationZh":"$最高点动能全部转化为势能：\\dfrac{1}{2}mv^{2}=mgh，故 h=\\dfrac{v^{2}}{2g}=\\dfrac{100}{20}=5\\ \\mathrm{m}。$","tags":["trap","energy","projectile","height"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-energy-theorem-013","moduleId":"mechanics","unitId":"work-energy","topicId":"energy-theorem","difficulty":4,"stem":"$A 1\\ \\mathrm{kg}$ block slides $5\\ \\mathrm{m}$ on a rough floor with $\\mu = 0.2$, starting at $6\\ \\mathrm{m/s}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its final speed?","stemZh":"$1\\ \\mathrm{kg}$ 的物块在粗糙水平面上滑行 $5\\ \\mathrm{m}，\\mu=0.2，初速 6\\ \\mathrm{m/s}，取 g=10\\ \\mathrm{m/s^{2}}，末速度多大？$","options":[{"id":"A","text":"$6\\ \\mathrm{m/s}$"},{"id":"B","text":"$5\\ \\mathrm{m/s}$"},{"id":"C","text":"$4\\ \\mathrm{m/s}$"},{"id":"D","text":"$2\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"Work by friction $= -\\mu$ mg $s = -0.2 \\times 1 \\times 10 \\times 5 = -10\\ \\mathrm{J}$. By the theorem, final KE $= \\dfrac{1}{2}\\times1\\times6^{2} - 10 = 18 - 10 = 8\\ \\mathrm{J}$, so $v = \\sqrt{16} = 4\\ \\mathrm{m/s}$. Ignoring friction wrongly keeps $6\\ \\mathrm{m/s}$.","explanationZh":"摩擦力做功 $=-\\mu mgs=-0.2\\times1\\times10\\times5=-10\\ \\mathrm{J}。由动能定理，末动能 =\\dfrac{1}{2}\\times1\\times6^{2}-10=18-10=8\\ \\mathrm{J}，故 v=\\sqrt{16}=4\\ \\mathrm{m/s}。忽略摩擦会错保留 6\\ \\mathrm{m/s}。$","tags":["trap","energy","work-energy-theorem","friction"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-energy-theorem-014","moduleId":"mechanics","unitId":"work-energy","topicId":"energy-theorem","difficulty":5,"stem":"For a pendulum swinging with no air resistance, what is true of its mechanical energy?","stemZh":"无空气阻力时单摆摆动，其机械能如何？","options":[{"id":"A","text":"Potential energy is maximum at the lowest point"},{"id":"B","text":"Kinetic energy is maximum at the lowest point"},{"id":"C","text":"Speed is zero throughout"},{"id":"D","text":"Mechanical energy is zero"}],"optionsZh":[{"id":"A","text":"最低点势能最大"},{"id":"B","text":"最低点动能最大"},{"id":"C","text":"全程速度为零"},{"id":"D","text":"机械能为零"}],"answer":"B","explanation":"With no non-conservative forces, mechanical energy is conserved. At the lowest point all the stored potential energy has become kinetic energy, so KE is maximum there.","explanationZh":"无非保守力时机械能守恒。最低点处储存的势能全部转化为动能，故此处动能最大。","tags":["trap","energy","conservation","pendulum"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-energy-theorem-015","moduleId":"mechanics","unitId":"work-energy","topicId":"energy-theorem","difficulty":5,"stem":"$A 0.5\\ \\mathrm{kg}$ ball is dropped from $20\\ \\mathrm{m}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what is its kinetic energy just before impact?","stemZh":"$0.5\\ \\mathrm{kg}$ 的球从 $20\\ \\mathrm{m}$ 高处自由下落，取 $g=10\\ \\mathrm{m/s^{2}}，落地前瞬间的动能多大？$","options":[{"id":"A","text":"$10\\ \\mathrm{J}$"},{"id":"B","text":"$50\\ \\mathrm{J}$"},{"id":"C","text":"$200\\ \\mathrm{J}$"},{"id":"D","text":"$100\\ \\mathrm{J}$"}],"answer":"D","explanation":"By $conservation, E_k =$ mgh $= 0.5 \\times 10 \\times 20 = 100\\ \\mathrm{J}$ at impact. Forgetting the $\\dfrac{1}{2}$ in $E_k = \\dfrac{1}{2}mv^{2}$ would give the wrong $200\\ \\mathrm{J}$.","explanationZh":"由守恒，落地前 $E_k=mgh=0.5\\times10\\times20=100\\ \\mathrm{J}。若忘记动能公式中的 \\dfrac{1}{2}$ 会错得 $200\\ \\mathrm{J}。$","tags":["trap","energy","conservation","impact"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-circular-motion-001","moduleId":"mechanics","unitId":"circular-motion","topicId":"circular-motion","difficulty":1,"stem":"A wheel rotates at $120\\ \\mathrm{rpm} (revolutions\\text{ per }minute)$. What is its angular velocity $\\omega$?","stemZh":"车轮以 $120\\ \\mathrm{rpm}（转/分）旋转，其角速度 \\omega$ 是多少？","options":[{"id":"A","text":"$4\\pi\\ \\mathrm{rad/s}$"},{"id":"B","text":"$2\\pi\\ \\mathrm{rad/s}$"},{"id":"C","text":"$\\pi\\ \\mathrm{rad/s}$"},{"id":"D","text":"$8\\pi\\ \\mathrm{rad/s}$"}],"answer":"A","explanation":"$\\omega = 120 \\times \\dfrac{2\\pi}{60} = 4\\pi\\ \\mathrm{rad/s}$.","explanationZh":"$\\omega=120\\times\\dfrac{2\\pi}{60}=4\\pi\\ \\mathrm{rad/s}。$","tags":["basic","circular-motion","angular-velocity"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-circular-motion-002","moduleId":"mechanics","unitId":"circular-motion","topicId":"circular-motion","difficulty":1,"stem":"A point on a rotating body completes one cycle every $4\\ \\mathrm{s}$. What is its frequency f?","stemZh":"旋转物体上某点每 $4\\ \\mathrm{s}$ 完成一次转动，其频率 f 是多少？","options":[{"id":"A","text":"$8\\ \\mathrm{Hz}$"},{"id":"B","text":"$0.25\\ \\mathrm{Hz}$"},{"id":"C","text":"$4\\ \\mathrm{Hz}$"},{"id":"D","text":"$2\\ \\mathrm{Hz}$"}],"answer":"B","explanation":"$f = \\dfrac{1}{T} = \\dfrac{1}{4} = 0.25\\ \\mathrm{Hz}$.","explanationZh":"$f=\\dfrac{1}{T}=\\dfrac{1}{4}=0.25\\ \\mathrm{Hz}。$","tags":["basic","circular-motion","frequency"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-circular-motion-003","moduleId":"mechanics","unitId":"circular-motion","topicId":"circular-motion","difficulty":1,"stem":"A point is $3\\ \\mathrm{m}$ from the axis and rotates with angular velocity $\\omega = 4\\ \\mathrm{rad/s}$. What is its linear speed v?","stemZh":"某点距转轴 $3\\ \\mathrm{m}，以角速度 \\omega=4\\ \\mathrm{rad/s}$ 旋转，其线速度 v 是多少？","options":[{"id":"A","text":"$0.75\\ \\mathrm{m/s}$"},{"id":"B","text":"$7\\ \\mathrm{m/s}$"},{"id":"C","text":"$12\\ \\mathrm{m/s}$"},{"id":"D","text":"$1.33\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"$v = \\omega r = 4 \\times 3 = 12\\ \\mathrm{m/s}$.","explanationZh":"$v=\\omega r=4\\times3=12\\ \\mathrm{m/s}。$","tags":["basic","circular-motion","linear-speed"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-circular-motion-004","moduleId":"mechanics","unitId":"circular-motion","topicId":"circular-motion","difficulty":2,"stem":"An object moves in a circle of radius $50\\ \\mathrm{m}$ at a constant speed $20\\ \\mathrm{m/s}$. What is its centripetal acceleration?","stemZh":"物体以恒定速度 $20\\ \\mathrm{m/s}$ 在半径 $50\\ \\mathrm{m}$ 的圆周上运动，向心加速度多大？","options":[{"id":"A","text":"$0.4\\ \\mathrm{m/s^{2}}$"},{"id":"B","text":"$1000\\ \\mathrm{m/s^{2}}$"},{"id":"C","text":"$2.5\\ \\mathrm{m/s^{2}}$"},{"id":"D","text":"$8\\ \\mathrm{m/s^{2}}$"}],"answer":"D","explanation":"$a = \\dfrac{v^{2}}{r} = \\dfrac{400}{50} = 8\\ \\mathrm{m/s^{2}}$.","explanationZh":"$a=\\dfrac{v^{2}}{r}=\\dfrac{400}{50}=8\\ \\mathrm{m/s^{2}}。$","tags":["basic","circular-motion","centripetal-acceleration"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-circular-motion-005","moduleId":"mechanics","unitId":"circular-motion","topicId":"circular-motion","difficulty":2,"stem":"A rotating body has period $T = 2\\ \\mathrm{s}$. What is its angular velocity $\\omega$?","stemZh":"旋转物体的周期 $T=2\\ \\mathrm{s}，其角速度 \\omega$ 是多少？","options":[{"id":"A","text":"$\\pi\\ \\mathrm{rad/s}$"},{"id":"B","text":"$2\\pi\\ \\mathrm{rad/s}$"},{"id":"C","text":"$\\dfrac{\\pi}{2}\\ \\mathrm{rad/s}$"},{"id":"D","text":"$4\\pi\\ \\mathrm{rad/s}$"}],"answer":"A","explanation":"$\\omega = \\dfrac{2\\pi}{T} = \\dfrac{2\\pi}{2} = \\pi\\ \\mathrm{rad/s}$.","explanationZh":"$\\omega=\\dfrac{2\\pi}{T}=\\dfrac{2\\pi}{2}=\\pi\\ \\mathrm{rad/s}。$","tags":["basic","circular-motion","angular-velocity"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-circular-motion-006","moduleId":"mechanics","unitId":"circular-motion","topicId":"circular-motion","difficulty":3,"stem":"$A 2\\ \\mathrm{kg}$ object moves in a circle of radius $5\\ \\mathrm{m}$ at speed $10\\ \\mathrm{m/s}$. What centripetal force is required?","stemZh":"质量 $2\\ \\mathrm{kg}$ 的物体在半径 $5\\ \\mathrm{m}$ 的圆周上以 $10\\ \\mathrm{m/s}$ 运动，所需向心力多大？","options":[{"id":"A","text":"$20\\ \\mathrm{N}$"},{"id":"B","text":"$40\\ \\mathrm{N}$"},{"id":"C","text":"$4\\ \\mathrm{N}$"},{"id":"D","text":"$100\\ \\mathrm{N}$"}],"answer":"B","explanation":"$F = \\dfrac{mv^{2}}{r} = \\dfrac{2 \\times 100}{5} = 40\\ \\mathrm{N}$.","explanationZh":"$F=\\dfrac{mv^{2}}{r}=\\dfrac{2\\times100}{5}=40\\ \\mathrm{N}。$","tags":["applied","circular-motion","centripetal-force"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-circular-motion-007","moduleId":"mechanics","unitId":"circular-motion","topicId":"circular-motion","difficulty":3,"stem":"A point on a circle of radius $2\\ \\mathrm{m}$ completes one revolution in $4\\ \\mathrm{s}$. What is its linear speed?","stemZh":"半径 $2\\ \\mathrm{m}$ 圆周上的点每 $4\\ \\mathrm{s}$ 转一圈，其线速度多大？","options":[{"id":"A","text":"$4\\pi\\ \\mathrm{m/s}$"},{"id":"B","text":"$0.5\\pi\\ \\mathrm{m/s}$"},{"id":"C","text":"$\\pi\\ \\mathrm{m/s}$"},{"id":"D","text":"$8\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"$v = \\dfrac{2\\pi r}{T} = \\dfrac{2\\pi \\times 2}{4} = \\pi\\ \\mathrm{m/s}$.","explanationZh":"$v=\\dfrac{2\\pi r}{T}=\\dfrac{2\\pi\\times2}{4}=\\pi\\ \\mathrm{m/s}。$","tags":["applied","circular-motion","linear-speed"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-circular-motion-008","moduleId":"mechanics","unitId":"circular-motion","topicId":"circular-motion","difficulty":3,"stem":"A point is $3\\ \\mathrm{m}$ from the axis and rotates with angular velocity $\\omega = 2\\ \\mathrm{rad/s}$. What is its centripetal acceleration?","stemZh":"距转轴 $3\\ \\mathrm{m}$ 的点以角速度 $\\omega=2\\ \\mathrm{rad/s}$ 旋转，其向心加速度多大？","options":[{"id":"A","text":"$1.5\\ \\mathrm{m/s^{2}}$"},{"id":"B","text":"$6\\ \\mathrm{m/s^{2}}$"},{"id":"C","text":"$18\\ \\mathrm{m/s^{2}}$"},{"id":"D","text":"$12\\ \\mathrm{m/s^{2}}$"}],"answer":"D","explanation":"$a = \\omega^{2}r = 2^{2} \\times 3 = 12\\ \\mathrm{m/s^{2}}$.","explanationZh":"$a=\\omega^{2}r=2^{2}\\times3=12\\ \\mathrm{m/s^{2}}。$","tags":["applied","circular-motion","centripetal-acceleration"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-circular-motion-009","moduleId":"mechanics","unitId":"circular-motion","topicId":"circular-motion","difficulty":3,"stem":"A rotating body has period $T = 0.02\\ \\mathrm{s}$. What is its frequency?","stemZh":"旋转物体的周期 $T=0.02\\ \\mathrm{s}，其频率是多少？$","options":[{"id":"A","text":"$50\\ \\mathrm{Hz}$"},{"id":"B","text":"$0.02\\ \\mathrm{Hz}$"},{"id":"C","text":"$20\\ \\mathrm{Hz}$"},{"id":"D","text":"$2\\ \\mathrm{Hz}$"}],"answer":"A","explanation":"$f = \\dfrac{1}{T} = \\dfrac{1}{0.02} = 50\\ \\mathrm{Hz}$.","explanationZh":"$f=\\dfrac{1}{T}=\\dfrac{1}{0.02}=50\\ \\mathrm{Hz}。$","tags":["applied","circular-motion","frequency"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-circular-motion-010","moduleId":"mechanics","unitId":"circular-motion","topicId":"circular-motion","difficulty":3,"stem":"$A 1\\ \\mathrm{kg}$ mass moves in a circle of radius $0.5\\ \\mathrm{m}$ with angular velocity $\\omega = 4\\ \\mathrm{rad/s}$. What centripetal force acts on it?","stemZh":"质量 $1\\ \\mathrm{kg}$ 的物体在半径 $0.5\\ \\mathrm{m}$ 的圆周上以角速度 $\\omega=4\\ \\mathrm{rad/s}$ 运动，向心力多大？","options":[{"id":"A","text":"$32\\ \\mathrm{N}$"},{"id":"B","text":"$8\\ \\mathrm{N}$"},{"id":"C","text":"$2\\ \\mathrm{N}$"},{"id":"D","text":"$16\\ \\mathrm{N}$"}],"answer":"B","explanation":"$F = m\\omega^{2}r = 1 \\times 4^{2} \\times 0.5 = 8\\ \\mathrm{N}$.","explanationZh":"$F=m\\omega^{2}r=1\\times4^{2}\\times0.5=8\\ \\mathrm{N}。$","tags":["applied","circular-motion","centripetal-force"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-circular-motion-011","images":[{"src":"/physics/diagrams/circular-motion.svg","alt":"A ball tied to a string moves in a horizontal circle; the string tension provides the inward centripetal force.","altZh":"小球系在绳端做水平圆周运动，绳的拉力提供指向圆心的向心力。"}],"moduleId":"mechanics","unitId":"circular-motion","topicId":"circular-motion","difficulty":4,"stem":"A ball on a string moves in a horizontal circle at constant speed. Which statement about the centripetal force is correct?","stemZh":"用绳系住的小球以恒定速率做水平圆周运动，关于向心力的说法哪项正确？","options":[{"id":"A","text":"It is a new outward force pushing the ball away from the centre."},{"id":"B","text":"It is a separate kind of force, independent of tension and gravity."},{"id":"C","text":"It is the net inward force, supplied here by the string tension."},{"id":"D","text":"It is balanced by an equal outward centrifugal force on the ball."}],"optionsZh":[{"id":"A","text":"是一种新的向外的力，把小球推离圆心。"},{"id":"B","text":"是一种独立的力，与拉力、重力无关。"},{"id":"C","text":"是合外力中指向圆心的部分，此处由绳的拉力提供。"},{"id":"D","text":"与小球受到的一个等大向外的离心力平衡。"}],"answer":"C","explanation":"Centripetal force is not a new force; it is the name for the net force directed toward the centre. Here the string tension supplies it. In an inertial frame there is no outward centrifugal force on the ball.","explanationZh":"向心力不是一种新力，而是所有指向圆心的合力成分的统称；此处由绳的拉力提供。在惯性系中小球并不受向外的离心力。","tags":["trap","circular-motion","centripetal-force","concept"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-circular-motion-012","moduleId":"mechanics","unitId":"circular-motion","topicId":"circular-motion","difficulty":4,"stem":"Two points A and B on a rigid rotating disc have radii $r_A = 1\\ \\mathrm{m}$ and $r_B = 3\\ \\mathrm{m}$. With the same angular velocity, how does the linear speed at B compare with A?","stemZh":"刚性转盘上 A、B 两点半径分别为 $r_A=1\\ \\mathrm{m}、r_B=3\\ \\mathrm{m}，角速度相同，B$ 点线速度与 A 点相比如何？","options":[{"id":"A","text":"$v_B = v_A$"},{"id":"B","text":"$v_B = \\dfrac{1}{3}v_A$"},{"id":"C","text":"$v_B = 9v_A$"},{"id":"D","text":"$v_B = 3v_A$"}],"answer":"D","explanation":"Since $v = \\omega r$ and $\\omega$ is the $same, v \\propto$ r. With $r_B = 3r_A, v_B = 3v_A$.","explanationZh":"$由 v=\\omega r 且 \\omega$ 相同，得 $v\\propto$ r。因 $r_B=3r_A，故 v_B=3v_A。$","tags":["trap","circular-motion","linear-speed","radius"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-circular-motion-013","moduleId":"mechanics","unitId":"circular-motion","topicId":"circular-motion","difficulty":4,"stem":"An object moves at constant speed v in a circle of radius r. If the radius is doubled but the speed stays the same, what happens to the centripetal acceleration?","stemZh":"物体以恒定速率 v 在半径 r 的圆周上运动。若半径加倍而速率不变，向心加速度如何变化？","options":[{"id":"A","text":"It is halved."},{"id":"B","text":"It doubles."},{"id":"C","text":"It quadruples."},{"id":"D","text":"It stays the same."}],"answer":"A","explanation":"$a = \\dfrac{v^{2}}{r}$. With $v fixed, a \\propto \\dfrac{1}{r}$, so doubling r halves the acceleration.","explanationZh":"$a=\\dfrac{v^{2}}{r}。v$ 不变时 $a\\propto\\dfrac{1}{r}，半径加倍则加速度减半。$","tags":["trap","circular-motion","centripetal-acceleration","inverse"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-circular-motion-014","images":[{"src":"/physics/diagrams/circular-motion.svg","alt":"A ball on a string moves in a vertical circle; at the top both weight and tension point toward the centre.","altZh":"小球在竖直平面内做圆周运动，在最高点重力与拉力均指向圆心。"}],"moduleId":"mechanics","unitId":"circular-motion","topicId":"circular-motion","difficulty":5,"stem":"$A 0.5\\ \\mathrm{kg}$ ball on $a 0.4\\ \\mathrm{m}$ string just barely completes a vertical circle. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what minimum speed must it have at the top?","stemZh":"质量 $0.5\\ \\mathrm{kg}$ 的小球系于 $0.4\\ \\mathrm{m}$ 长绳，恰好能完成竖直圆周运动。取 $g=10\\ \\mathrm{m/s^{2}}，最高点最小速率是多少？$","options":[{"id":"A","text":"$\\sqrt{10}\\ \\mathrm{m/s}$"},{"id":"B","text":"$4\\ \\mathrm{m/s}$"},{"id":"C","text":"$2\\ \\mathrm{m/s}$"},{"id":"D","text":"$0.2\\ \\mathrm{m/s}$"}],"answer":"C","explanation":"At the top the minimum condition is mg $= \\dfrac{mv^{2}}{r}$, so $v = \\sqrt{gr} = \\sqrt{10 \\times 0.4} = \\sqrt{4} = 2\\ \\mathrm{m/s}$.","explanationZh":"最高点临界条件为 $mg=\\dfrac{mv^{2}}{r}，故 v=\\sqrt{gr}=\\sqrt{10\\times0.4}=\\sqrt{4}=2\\ \\mathrm{m/s}。$","tags":["trap","circular-motion","vertical-circle","minimum-speed"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-circular-motion-015","moduleId":"mechanics","unitId":"circular-motion","topicId":"circular-motion","difficulty":5,"stem":"$A 1000\\ \\mathrm{kg}$ car rounds a flat curve of radius $50\\ \\mathrm{m}$ at $20\\ \\mathrm{m/s}$. Taking $g = 10\\ \\mathrm{m/s^{2}}$, what minimum coefficient of friction is needed to avoid skidding?","stemZh":"质量 $1000\\ \\mathrm{kg}$ 的汽车以 $20\\ \\mathrm{m/s}$ 绕过半径 $50\\ \\mathrm{m}$ 的水平弯道。取 $g=10\\ \\mathrm{m/s^{2}}，避免侧滑所需的最小摩擦因数多大？$","options":[{"id":"A","text":"0.4"},{"id":"B","text":"0.8"},{"id":"C","text":"1.25"},{"id":"D","text":"0.08"}],"answer":"B","explanation":"Friction supplies the centripetal $force: \\mu$ mg $= \\dfrac{mv^{2}}{r}$, so $\\mu = \\dfrac{v^{2}}{rg} = \\dfrac{400}{50 \\times 10} = 0.8$.","explanationZh":"$摩擦力提供向心力：\\mu mg=\\dfrac{mv^{2}}{r}，故 \\mu=\\dfrac{v^{2}}{rg}=\\dfrac{400}{50\\times10}=0.8。$","tags":["trap","circular-motion","friction","banked-curve"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-oscillations-waves-001","moduleId":"mechanics","unitId":"oscillations-waves","topicId":"oscillations-waves","difficulty":1,"stem":"An oscillator has frequency $f = 2\\ \\mathrm{Hz}$. What is its period T?","stemZh":"振子的频率 $f=2\\ \\mathrm{Hz}，其周期 T$ 是多少？","options":[{"id":"A","text":"$0.5\\ \\mathrm{s}$"},{"id":"B","text":"$2\\ \\mathrm{s}$"},{"id":"C","text":"$4\\ \\mathrm{s}$"},{"id":"D","text":"$1\\ \\mathrm{s}$"}],"answer":"A","explanation":"$T = \\dfrac{1}{f} = \\dfrac{1}{2} = 0.5\\ \\mathrm{s}$.","explanationZh":"$T=\\dfrac{1}{f}=\\dfrac{1}{2}=0.5\\ \\mathrm{s}。$","tags":["basic","oscillations-waves","period"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-oscillations-waves-002","images":[{"src":"/physics/diagrams/wave-sine.svg","alt":"A sine curve of displacement versus time; the peak height above the axis is the amplitude A.","altZh":"位移—时间正弦曲线，曲线相对坐标轴的最高高度为振幅 A。"}],"moduleId":"mechanics","unitId":"oscillations-waves","topicId":"oscillations-waves","difficulty":1,"stem":"A particle's displacement is described by $x = 0.1\\sin(5t)$ in metres, where t is in seconds. What is the amplitude?","stemZh":"某质点的位移为 $x=0.1\\sin(5t)（单位：米，t$ 单位：秒），其振幅是多少？","options":[{"id":"A","text":"$5\\ \\mathrm{m}$"},{"id":"B","text":"$0.1\\ \\mathrm{m}$"},{"id":"C","text":"$0.2\\ \\mathrm{m}$"},{"id":"D","text":"$5\\ \\mathrm{rad/s}$"}],"answer":"B","explanation":"In $x = A\\sin(\\omega t)$, the coefficient of the sine is the amplitude, so $A = 0.1\\ \\mathrm{m}$.","explanationZh":"$在 x=A\\sin(\\omega t)$ 中，正弦前的系数即振幅，故 $A=0.1\\ \\mathrm{m}。$","tags":["basic","oscillations-waves","amplitude","shm"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-oscillations-waves-003","moduleId":"mechanics","unitId":"oscillations-waves","topicId":"oscillations-waves","difficulty":1,"stem":"An oscillator has frequency $f = 1\\ \\mathrm{Hz}$. What is its angular frequency $\\omega$?","stemZh":"振子的频率 $f=1\\ \\mathrm{Hz}，其角频率 \\omega$ 是多少？","options":[{"id":"A","text":"$\\pi\\ \\mathrm{rad/s}$"},{"id":"B","text":"$1\\ \\mathrm{rad/s}$"},{"id":"C","text":"$2\\pi\\ \\mathrm{rad/s}$"},{"id":"D","text":"$0.5\\ \\mathrm{rad/s}$"}],"answer":"C","explanation":"$\\omega = 2\\pi f = 2\\pi \\times 1 = 2\\pi\\ \\mathrm{rad/s}$.","explanationZh":"$\\omega=2\\pi f=2\\pi\\times1=2\\pi\\ \\mathrm{rad/s}。$","tags":["basic","oscillations-waves","angular-frequency"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-oscillations-waves-004","moduleId":"mechanics","unitId":"oscillations-waves","topicId":"oscillations-waves","difficulty":2,"stem":"A sound wave has frequency $f = 200\\ \\mathrm{Hz}$ and wavelength $\\lambda = 1.5\\ \\mathrm{m}$. What is its speed?","stemZh":"声波的频率 $f=200\\ \\mathrm{Hz}，波长 \\lambda=1.5\\ \\mathrm{m}，其波速是多少？$","options":[{"id":"A","text":"$133\\ \\mathrm{m/s}$"},{"id":"B","text":"$400\\ \\mathrm{m/s}$"},{"id":"C","text":"$0.0075\\ \\mathrm{m/s}$"},{"id":"D","text":"$300\\ \\mathrm{m/s}$"}],"answer":"D","explanation":"$v = f\\lambda = 200 \\times 1.5 = 300\\ \\mathrm{m/s}$.","explanationZh":"$v=f\\lambda=200\\times1.5=300\\ \\mathrm{m/s}。$","tags":["basic","oscillations-waves","wave-speed"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-oscillations-waves-005","moduleId":"mechanics","unitId":"oscillations-waves","topicId":"oscillations-waves","difficulty":2,"stem":"In simple harmonic motion, the restoring force on the particle is proportional to its displacement and directed toward equilibrium. Which statement captures this?","stemZh":"简谐运动中，回复力与位移成正比且指向平衡位置。下列哪项正确表述了这一关系？","options":[{"id":"A","text":"It is proportional to the displacement and directed toward equilibrium."},{"id":"B","text":"It is proportional to the velocity."},{"id":"C","text":"It is proportional to the square of the displacement."},{"id":"D","text":"It is proportional to the period."}],"optionsZh":[{"id":"A","text":"与位移成正比，且指向平衡位置。"},{"id":"B","text":"与速度成正比。"},{"id":"C","text":"与位移的平方成正比。"},{"id":"D","text":"与周期成正比。"}],"answer":"A","explanation":"SHM is defined by $F = -kx$: the restoring force is proportional to displacement and opposes it (toward equilibrium).","explanationZh":"简谐运动满足 $F=-kx：回复力与位移成正比且方向与之相反（指向平衡位置）。$","tags":["basic","oscillations-waves","restoring-force","shm"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-oscillations-waves-006","moduleId":"mechanics","unitId":"oscillations-waves","topicId":"oscillations-waves","difficulty":3,"stem":"A wave travels at $340\\ \\mathrm{m/s}$ with frequency $170\\ \\mathrm{Hz}$. What is its wavelength?","stemZh":"波以 $340\\ \\mathrm{m/s}$ 传播，频率 $170\\ \\mathrm{Hz}，其波长是多少？$","options":[{"id":"A","text":"$1\\ \\mathrm{m}$"},{"id":"B","text":"$2\\ \\mathrm{m}$"},{"id":"C","text":"$57800\\ \\mathrm{m}$"},{"id":"D","text":"$0.5\\ \\mathrm{m}$"}],"answer":"B","explanation":"$\\lambda = \\dfrac{v}{f} = \\dfrac{340}{170} = 2\\ \\mathrm{m}$.","explanationZh":"$\\lambda=\\dfrac{v}{f}=\\dfrac{340}{170}=2\\ \\mathrm{m}。$","tags":["applied","oscillations-waves","wavelength"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-oscillations-waves-007","moduleId":"mechanics","unitId":"oscillations-waves","topicId":"oscillations-waves","difficulty":3,"stem":"An oscillator has angular frequency $\\omega = 4\\pi\\ \\mathrm{rad/s}$. What is its period?","stemZh":"振子的角频率 $\\omega=4\\pi\\ \\mathrm{rad/s}，其周期是多少？$","options":[{"id":"A","text":"$4\\pi\\ \\mathrm{s}$"},{"id":"B","text":"$2\\pi\\ \\mathrm{s}$"},{"id":"C","text":"$0.5\\ \\mathrm{s}$"},{"id":"D","text":"$8\\pi\\ \\mathrm{s}$"}],"answer":"C","explanation":"$T = \\dfrac{2\\pi}{\\omega} = \\dfrac{2\\pi}{4\\pi} = 0.5\\ \\mathrm{s}$.","explanationZh":"$T=\\dfrac{2\\pi}{\\omega}=\\dfrac{2\\pi}{4\\pi}=0.5\\ \\mathrm{s}。$","tags":["applied","oscillations-waves","period","angular-frequency"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-oscillations-waves-008","moduleId":"mechanics","unitId":"oscillations-waves","topicId":"oscillations-waves","difficulty":3,"stem":"A particle in SHM has displacement $x = 0.2\\sin(10t)$ in metres. What is its maximum speed?","stemZh":"某质点做简谐运动，位移 $x=0.2\\sin(10t)（单位：米），其最大速率是多少？$","options":[{"id":"A","text":"$10\\ \\mathrm{m/s}$"},{"id":"B","text":"$0.02\\ \\mathrm{m/s}$"},{"id":"C","text":"$0.2\\ \\mathrm{m/s}$"},{"id":"D","text":"$2\\ \\mathrm{m/s}$"}],"answer":"D","explanation":"$v_{\\max} = A\\omega = 0.2 \\times 10 = 2\\ \\mathrm{m/s}$.","explanationZh":"$v_{\\max}=A\\omega=0.2\\times10=2\\ \\mathrm{m/s}。$","tags":["applied","oscillations-waves","maximum-speed","shm"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-oscillations-waves-009","moduleId":"mechanics","unitId":"oscillations-waves","topicId":"oscillations-waves","difficulty":3,"stem":"A periodic wave has period $T = 0.01\\ \\mathrm{s}$. What is its frequency?","stemZh":"周期性波的周期 $T=0.01\\ \\mathrm{s}，其频率是多少？$","options":[{"id":"A","text":"$100\\ \\mathrm{Hz}$"},{"id":"B","text":"$0.01\\ \\mathrm{Hz}$"},{"id":"C","text":"$10\\ \\mathrm{Hz}$"},{"id":"D","text":"$50\\ \\mathrm{Hz}$"}],"answer":"A","explanation":"$f = \\dfrac{1}{T} = \\dfrac{1}{0.01} = 100\\ \\mathrm{Hz}$.","explanationZh":"$f=\\dfrac{1}{T}=\\dfrac{1}{0.01}=100\\ \\mathrm{Hz}。$","tags":["applied","oscillations-waves","frequency"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-oscillations-waves-010","images":[{"src":"/physics/diagrams/wave-sine.svg","alt":"A transverse sine wave on a string; the particles move perpendicular to the direction the wave travels.","altZh":"绳上的横波正弦曲线；质点振动方向垂直于波的传播方向。"}],"moduleId":"mechanics","unitId":"oscillations-waves","topicId":"oscillations-waves","difficulty":3,"stem":"A wave on a string is transverse. How do the string particles move relative to the direction the wave travels?","stemZh":"绳上的波是横波，绳上质点的振动方向与波的传播方向关系如何？","options":[{"id":"A","text":"Parallel to the direction of travel"},{"id":"B","text":"Perpendicular to the direction of travel"},{"id":"C","text":"They do not move at all"},{"id":"D","text":"In circles"}],"optionsZh":[{"id":"A","text":"平行于传播方向"},{"id":"B","text":"垂直于传播方向"},{"id":"C","text":"完全不动"},{"id":"D","text":"做圆周运动"}],"answer":"B","explanation":"In a transverse wave the particle displacement is perpendicular to the propagation direction; in a longitudinal wave it is parallel.","explanationZh":"横波中质点振动方向垂直于传播方向；纵波中则平行。","tags":["applied","oscillations-waves","transverse","longitudinal"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-oscillations-waves-011","moduleId":"mechanics","unitId":"oscillations-waves","topicId":"oscillations-waves","difficulty":4,"stem":"A wave has period $T = 0.5\\ \\mathrm{s}. A$ student writes its frequency as $0.5\\ \\mathrm{Hz}$. What is the correct frequency?","stemZh":"某波的周期 $T=0.5\\ \\mathrm{s}。某同学误将其频率写成 0.5\\ \\mathrm{Hz}。正确频率是多少？$","options":[{"id":"A","text":"$0.5\\ \\mathrm{Hz}$"},{"id":"B","text":"$0.25\\ \\mathrm{Hz}$"},{"id":"C","text":"$2\\ \\mathrm{Hz}$"},{"id":"D","text":"$1\\ \\mathrm{Hz}$"}],"answer":"C","explanation":"Frequency is the reciprocal of $period: f = \\dfrac{1}{T} = \\dfrac{1}{0.5} = 2\\ \\mathrm{Hz}$, not equal to T.","explanationZh":"$频率是周期的倒数：f=\\dfrac{1}{T}=\\dfrac{1}{0.5}=2\\ \\mathrm{Hz}，并不等于$ T。","tags":["trap","oscillations-waves","frequency","period-confusion"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-oscillations-waves-012","moduleId":"mechanics","unitId":"oscillations-waves","topicId":"oscillations-waves","difficulty":4,"stem":"A wave on a string has speed $v = 20\\ \\mathrm{m/s}$. If the frequency is doubled but the speed stays the same, what happens to the wavelength?","stemZh":"绳上波速 $v=20\\ \\mathrm{m/s}。若频率加倍而波速不变，波长如何变化？$","options":[{"id":"A","text":"It doubles."},{"id":"B","text":"It stays the same."},{"id":"C","text":"It quadruples."},{"id":"D","text":"It is halved."}],"answer":"D","explanation":"From $v = f\\lambda$ with $v fixed, \\lambda = \\dfrac{v}{f}$, so doubling f halves $\\lambda$.","explanationZh":"$由 v=f\\lambda，波速不变时 \\lambda=\\dfrac{v}{f}，频率加倍则波长减半。$","tags":["trap","oscillations-waves","wavelength","v-f-lambda"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-oscillations-waves-013","moduleId":"mechanics","unitId":"oscillations-waves","topicId":"oscillations-waves","difficulty":4,"stem":"For a particle in simple harmonic motion, which statement is correct?","stemZh":"关于做简谐运动的质点，下列哪项正确？","options":[{"id":"A","text":"At the equilibrium position the restoring force is zero and the speed is maximum."},{"id":"B","text":"At the equilibrium position the restoring force is maximum."},{"id":"C","text":"The acceleration is zero at the extreme positions."},{"id":"D","text":"The speed is zero at the equilibrium position."}],"optionsZh":[{"id":"A","text":"在平衡位置回复力为零、速率最大。"},{"id":"B","text":"在平衡位置回复力最大。"},{"id":"C","text":"在最大位移处加速度为零。"},{"id":"D","text":"在平衡位置速率为零。"}],"answer":"A","explanation":"At equilibrium $x = 0$, so $F =$ -kx $= 0$ and acceleration is zero, while speed reaches its maximum. At the extremes the force and acceleration are maximum and speed is zero.","explanationZh":"平衡位置 $x=0，故 F=-kx=0、加速度为零，而速率达到最大。在最大位移处力与加速度最大、速率为零。$","tags":["trap","oscillations-waves","shm","equilibrium","concept"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-oscillations-waves-014","moduleId":"mechanics","unitId":"oscillations-waves","topicId":"oscillations-waves","difficulty":5,"stem":"A string wave has speed $v = 30\\ \\mathrm{m/s}$ and frequency $f = 6\\ \\mathrm{Hz}. A$ student writes $\\lambda =$ vf $= 180\\ \\mathrm{m}$. What is the correct wavelength?","stemZh":"绳上波速 $v=30\\ \\mathrm{m/s}，频率 f=6\\ \\mathrm{Hz}。某同学写成 \\lambda=vf=180\\ \\mathrm{m}。正确波长是多少？$","options":[{"id":"A","text":"$180\\ \\mathrm{m}$"},{"id":"B","text":"$36\\ \\mathrm{m}$"},{"id":"C","text":"$5\\ \\mathrm{m}$"},{"id":"D","text":"$24\\ \\mathrm{m}$"}],"answer":"C","explanation":"The correct relation is $v = f\\lambda$, so $\\lambda = \\dfrac{v}{f} = \\dfrac{30}{6} = 5\\ \\mathrm{m}$. Multiplying v and f is wrong.","explanationZh":"正确关系是 $v=f\\lambda，故 \\lambda=\\dfrac{v}{f}=\\dfrac{30}{6}=5\\ \\mathrm{m}。将 v 与 f$ 相乘是错误的。","tags":["trap","oscillations-waves","wavelength","formula-misuse"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-oscillations-waves-015","moduleId":"mechanics","unitId":"oscillations-waves","topicId":"oscillations-waves","difficulty":5,"stem":"For a simple pendulum in small oscillations, which change doubles its period?","stemZh":"对于做小角度振动的单摆，下列哪一种改变会使周期加倍？","options":[{"id":"A","text":"Doubling the bob mass"},{"id":"B","text":"Quadrupling the length"},{"id":"C","text":"Doubling the amplitude"},{"id":"D","text":"Halving the length"}],"optionsZh":[{"id":"A","text":"摆球质量加倍"},{"id":"B","text":"摆长变为四倍"},{"id":"C","text":"振幅加倍"},{"id":"D","text":"摆长减半"}],"answer":"B","explanation":"The period is $T = 2\\pi\\sqrt{\\dfrac{L}{g}}$, so $T \\propto \\sqrt{L}$. To double T, the length must be quadrupled $(4\\times)$. It does not depend on mass or (for small angles) amplitude.","explanationZh":"周期 $T=2\\pi\\sqrt{\\dfrac{L}{g}}，故 T\\propto\\sqrt{L}。要使 T 加倍，摆长须变为四倍（4\\times）。周期与质量无关，小角度下也与振幅无关。$","tags":["trap","oscillations-waves","pendulum","period-dependence"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-gas-laws-001","moduleId":"thermal-physics","unitId":"thermal","topicId":"gas-laws","difficulty":1,"stem":"A fixed mass of gas at constant temperature has $V_1 = 2.0\\ \\mathrm{L}$ at $p_1 = 100\\ \\mathrm{kPa}$. It is compressed to $V_2 = 1.0\\ \\mathrm{L}$. Find $p_2$.","stemZh":"一定质量的气体在恒温下，初态 $V_1=2.0\\ \\mathrm{L}、p_1=100\\ \\mathrm{kPa}，被压缩到 V_2=1.0\\ \\mathrm{L}，求 p_2。$","options":[{"id":"A","text":"$50\\ \\mathrm{kPa}$"},{"id":"B","text":"$100\\ \\mathrm{kPa}$"},{"id":"C","text":"$200\\ \\mathrm{kPa}$"},{"id":"D","text":"$400\\ \\mathrm{kPa}$"}],"answer":"C","explanation":"Boyle's $law: p_1 V_1 = p_2 V_2$, so $p_2 = \\dfrac{p_1 V_1}{V_2} = \\dfrac{100 \\times 2.0}{1.0} = 200\\ \\mathrm{kPa}$.","explanationZh":"$玻意耳定律：p_1 V_1=p_2 V_2，故 p_2=\\dfrac{p_1 V_1}{V_2}=\\dfrac{100\\times2.0}{1.0}=200\\ \\mathrm{kPa}。$","tags":["basic","gas-laws","boyle"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-gas-laws-002","moduleId":"thermal-physics","unitId":"thermal","topicId":"gas-laws","difficulty":1,"stem":"A gas at constant pressure has $V_1 = 3.0\\ \\mathrm{L}$ at $T_1 = 300\\ \\mathrm{K}$. Its temperature rises to $T_2 = 400\\ \\mathrm{K}$. Find $V_2$.","stemZh":"气体在恒压下，初态 $V_1=3.0\\ \\mathrm{L}、T_1=300\\ \\mathrm{K}，温度升高到 T_2=400\\ \\mathrm{K}，求 V_2。$","options":[{"id":"A","text":"$4.0\\ \\mathrm{L}$"},{"id":"B","text":"$2.25\\ \\mathrm{L}$"},{"id":"C","text":"$3.0\\ \\mathrm{L}$"},{"id":"D","text":"$5.0\\ \\mathrm{L}$"}],"answer":"A","explanation":"Charles's $law: \\dfrac{V}{T} = constant$, so $V_2 = V_1\\dfrac{T_2}{T_1} = 3.0 \\times \\dfrac{400}{300} = 4.0\\ \\mathrm{L}$.","explanationZh":"$查理定律：\\dfrac{V}{T}=\\text{常量}，故 V_2=V_1\\dfrac{T_2}{T_1}=3.0\\times\\dfrac{400}{300}=4.0\\ \\mathrm{L}。$","tags":["basic","gas-laws","charles"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-gas-laws-003","moduleId":"thermal-physics","unitId":"thermal","topicId":"gas-laws","difficulty":1,"stem":"At constant $volume, p_1 = 200\\ \\mathrm{kPa}$ at $T_1 = 400\\ \\mathrm{K}$. The gas is cooled to $T_2 = 300\\ \\mathrm{K}$. Find $p_2$.","stemZh":"在恒定体积下，初态 $p_1=200\\ \\mathrm{kPa}、T_1=400\\ \\mathrm{K}，冷却到 T_2=300\\ \\mathrm{K}，求 p_2。$","options":[{"id":"A","text":"$266.7\\ \\mathrm{kPa}$"},{"id":"B","text":"$150\\ \\mathrm{kPa}$"},{"id":"C","text":"$100\\ \\mathrm{kPa}$"},{"id":"D","text":"$600\\ \\mathrm{kPa}$"}],"answer":"B","explanation":"Gay-Lussac's $law: \\dfrac{p}{T} = constant$, so $p_2 = p_1\\dfrac{T_2}{T_1} = 200 \\times \\dfrac{300}{400} = 150\\ \\mathrm{kPa}$.","explanationZh":"$盖-吕萨克定律：\\dfrac{p}{T}=\\text{常量}，故 p_2=p_1\\dfrac{T_2}{T_1}=200\\times\\dfrac{300}{400}=150\\ \\mathrm{kPa}。$","tags":["basic","gas-laws","gay-lussac"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-gas-laws-004","moduleId":"thermal-physics","unitId":"thermal","topicId":"gas-laws","difficulty":2,"stem":"Room temperature is $27^{\\circ}\\mathrm{C}$. What is this temperature on the absolute (Kelvin) scale?","stemZh":"室温为 $27^{\\circ}\\mathrm{C}，换算成热力学温度（开尔文）是多少？$","options":[{"id":"A","text":"$27\\ \\mathrm{K}$"},{"id":"B","text":"$246\\ \\mathrm{K}$"},{"id":"C","text":"$300^{\\circ}\\mathrm{C}$"},{"id":"D","text":"$300\\ \\mathrm{K}$"}],"answer":"D","explanation":"Absolute temperature is $T = t + 273$, so $T = 27 + 273 = 300\\ \\mathrm{K}$.","explanationZh":"热力学温度 $T=t+273，故 T=27+273=300\\ \\mathrm{K}。$","tags":["basic","gas-laws","absolute-temperature"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-gas-laws-005","moduleId":"thermal-physics","unitId":"thermal","topicId":"gas-laws","difficulty":2,"stem":"A gas has $p_1 = 100\\ \\mathrm{kPa}, V_1 = 4.0\\ \\mathrm{L}, T_1 = 300\\ \\mathrm{K}$. It changes to $p_2 = 200\\ \\mathrm{kPa}, T_2 = 600\\ \\mathrm{K}$. Find $V_2$.","stemZh":"某气体初态 $p_1=100\\ \\mathrm{kPa}、V_1=4.0\\ \\mathrm{L}、T_1=300\\ \\mathrm{K}，变为 p_2=200\\ \\mathrm{kPa}、T_2=600\\ \\mathrm{K}，求 V_2。$","options":[{"id":"A","text":"$1.0\\ \\mathrm{L}$"},{"id":"B","text":"$2.0\\ \\mathrm{L}$"},{"id":"C","text":"$4.0\\ \\mathrm{L}$"},{"id":"D","text":"$8.0\\ \\mathrm{L}$"}],"answer":"C","explanation":"Combined gas $law: \\dfrac{p_1 V_1}{T_1} = \\dfrac{p_2 V_2}{T_2}$, so $V_2 = \\dfrac{p_1 V_1 T_2}{p_2 T_1} = \\dfrac{100 \\times 4.0 \\times 600}{200 \\times 300} = 4.0\\ \\mathrm{L}$.","explanationZh":"$理想气体状态方程：\\dfrac{p_1 V_1}{T_1}=\\dfrac{p_2 V_2}{T_2}，故 V_2=\\dfrac{p_1 V_1 T_2}{p_2 T_1}=\\dfrac{100\\times4.0\\times600}{200\\times300}=4.0\\ \\mathrm{L}。$","tags":["basic","gas-laws","combined"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-gas-laws-006","moduleId":"thermal-physics","unitId":"thermal","topicId":"gas-laws","difficulty":3,"stem":"For an ideal gas, pV $=$ nRT with $R = 8.31\\ \\mathrm{J/(mol\\cdot K)}$. If $n = 2\\ \\mathrm{mol}, V = 0.050\\ \\mathrm{m^{3}}, T = 300\\ \\mathrm{K}$, what is p?","stemZh":"理想气体满足 $pV=nRT，取 R=8.31\\ \\mathrm{J/(mol\\cdot K)}。若 n=2\\ \\mathrm{mol}、V=0.050\\ \\mathrm{m^{3}}、T=300\\ \\mathrm{K}，求$ p。","options":[{"id":"A","text":"$1.0 \\times 10^{5}\\ \\mathrm{Pa}$"},{"id":"B","text":"$5.0 \\times 10^{4}\\ \\mathrm{Pa}$"},{"id":"C","text":"$2.0 \\times 10^{5}\\ \\mathrm{Pa}$"},{"id":"D","text":"$9.97 \\times 10^{3}\\ \\mathrm{Pa}$"}],"answer":"A","explanation":"$p = \\dfrac{nRT}{V} = \\dfrac{2 \\times 8.31 \\times 300}{0.050} = \\dfrac{4986}{0.050} \\approx 1.0 \\times 10^{5}\\ \\mathrm{Pa}$.","explanationZh":"$p=\\dfrac{nRT}{V}=\\dfrac{2\\times8.31\\times300}{0.050}=\\dfrac{4986}{0.050}\\approx1.0\\times10^{5}\\ \\mathrm{Pa}。$","tags":["applied","gas-laws","ideal-gas"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-gas-laws-007","moduleId":"thermal-physics","unitId":"thermal","topicId":"gas-laws","difficulty":3,"stem":"A gas at $p = 1.0 \\times 10^{5}\\ \\mathrm{Pa}$ occupies $V = 0.0246\\ \\mathrm{m^{3}}$ at $T = 300\\ \\mathrm{K}$ with $R = 8.31\\ \\mathrm{J/(mol\\cdot K)}$. How many moles n?","stemZh":"气体在 $p=1.0\\times10^{5}\\ \\mathrm{Pa}、V=0.0246\\ \\mathrm{m^{3}}、T=300\\ \\mathrm{K}$ 下，取 $R=8.31\\ \\mathrm{J/(mol\\cdot K)}，物质的量 n$ 为多少？","options":[{"id":"A","text":"$0.5\\ \\mathrm{mol}$"},{"id":"B","text":"$1.0\\ \\mathrm{mol}$"},{"id":"C","text":"$2.0\\ \\mathrm{mol}$"},{"id":"D","text":"$0.25\\ \\mathrm{mol}$"}],"answer":"B","explanation":"$n = \\dfrac{pV}{RT} = \\dfrac{1.0 \\times 10^{5} \\times 0.0246}{8.31 \\times 300} = \\dfrac{2460}{2493} \\approx 1.0\\ \\mathrm{mol}$.","explanationZh":"$n=\\dfrac{pV}{RT}=\\dfrac{1.0\\times10^{5}\\times0.0246}{8.31\\times300}=\\dfrac{2460}{2493}\\approx1.0\\ \\mathrm{mol}。$","tags":["applied","gas-laws","ideal-gas"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-gas-laws-008","images":[{"src":"/physics/diagrams/pv-diagram.svg","alt":"p–V diagram of an ideal gas: an isotherm is a curve where pressure falls as volume rises $(pV = constant)$.","altZh":"理想气体的 p–V 图：等温线为压强随体积增大而下降的曲线（pV = 常量）。"}],"moduleId":"thermal-physics","unitId":"thermal","topicId":"gas-laws","difficulty":3,"stem":"On a p–V diagram, how is an isothermal $(constant-temperature)$ change of a fixed mass of ideal gas drawn?","stemZh":"在 p–V 图上，一定质量理想气体的等温过程应如何表示？","options":[{"id":"A","text":"A straight line through the origin"},{"id":"B","text":"A horizontal line"},{"id":"C","text":"A vertical line"},{"id":"D","text":"A curve with pV $=$ constant (a hyperbola)"}],"answer":"D","explanation":"At constant T, pV $=$ nRT $= constant$, so $p \\propto \\dfrac{1}{V} — a hyperbola$, not a straight line.","explanationZh":"恒温时 $pV=nRT=\\text{常量}，故 p\\propto\\dfrac{1}{V}，是双曲线而非直线。$","tags":["applied","gas-laws","pv-diagram"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-gas-laws-009","moduleId":"thermal-physics","unitId":"thermal","topicId":"gas-laws","difficulty":3,"stem":"A gas has $V_1 = 3.0\\ \\mathrm{L}$ at $p_1 = 120\\ \\mathrm{kPa}, T_1 = 360\\ \\mathrm{K}$. It changes to $p_2 = 80\\ \\mathrm{kPa}, T_2 = 240\\ \\mathrm{K}$. Find $V_2$.","stemZh":"气体初态 $V_1=3.0\\ \\mathrm{L}、p_1=120\\ \\mathrm{kPa}、T_1=360\\ \\mathrm{K}，变为 p_2=80\\ \\mathrm{kPa}、T_2=240\\ \\mathrm{K}，求 V_2。$","options":[{"id":"A","text":"$1.5\\ \\mathrm{L}$"},{"id":"B","text":"$6.0\\ \\mathrm{L}$"},{"id":"C","text":"$3.0\\ \\mathrm{L}$"},{"id":"D","text":"$4.5\\ \\mathrm{L}$"}],"answer":"C","explanation":"$V_2 = \\dfrac{p_1 V_1 T_2}{p_2 T_1} = \\dfrac{120 \\times 3.0 \\times 240}{80 \\times 360} = \\dfrac{86400}{28800} = 3.0\\ \\mathrm{L}$.","explanationZh":"$V_2=\\dfrac{p_1 V_1 T_2}{p_2 T_1}=\\dfrac{120\\times3.0\\times240}{80\\times360}=\\dfrac{86400}{28800}=3.0\\ \\mathrm{L}。$","tags":["applied","gas-laws","combined"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-gas-laws-010","moduleId":"thermal-physics","unitId":"thermal","topicId":"gas-laws","difficulty":3,"stem":"Two moles of an ideal gas are at $p = 2.0 \\times 10^{5}\\ \\mathrm{Pa}, V = 0.025\\ \\mathrm{m^{3}}$ with $R = 8.31\\ \\mathrm{J/(mol\\cdot K)}$. Find T.","stemZh":"2 mol 理想气体处于 $p=2.0\\times10^{5}\\ \\mathrm{Pa}、V=0.025\\ \\mathrm{m^{3}}，取 R=8.31\\ \\mathrm{J/(mol\\cdot K)}，求$ T。","options":[{"id":"A","text":"$300\\ \\mathrm{K}$"},{"id":"B","text":"$150\\ \\mathrm{K}$"},{"id":"C","text":"$600\\ \\mathrm{K}$"},{"id":"D","text":"$30\\ \\mathrm{K}$"}],"answer":"A","explanation":"$T = \\dfrac{pV}{nR} = \\dfrac{2.0 \\times 10^{5} \\times 0.025}{2 \\times 8.31} = \\dfrac{5000}{16.62} \\approx 300\\ \\mathrm{K}$.","explanationZh":"$T=\\dfrac{pV}{nR}=\\dfrac{2.0\\times10^{5}\\times0.025}{2\\times8.31}=\\dfrac{5000}{16.62}\\approx300\\ \\mathrm{K}。$","tags":["applied","gas-laws","ideal-gas"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-gas-laws-011","moduleId":"thermal-physics","unitId":"thermal","topicId":"gas-laws","difficulty":4,"stem":"A gas at constant pressure has $V_1 = 1.0\\ \\mathrm{L}$ at $27^{\\circ}\\mathrm{C}$. It is heated to $127^{\\circ}\\mathrm{C}$. What is $V_2$?","stemZh":"恒压下气体初态 $V_1=1.0\\ \\mathrm{L}、27^{\\circ}\\mathrm{C}，加热到 127^{\\circ}\\mathrm{C}，求 V_2。$","options":[{"id":"A","text":"$4.70\\ \\mathrm{L}$"},{"id":"B","text":"$1.33\\ \\mathrm{L}$"},{"id":"C","text":"$0.75\\ \\mathrm{L}$"},{"id":"D","text":"$2.0\\ \\mathrm{L}$"}],"answer":"B","explanation":"Charles's law needs absolute $temperature: T_1 = 27 + 273 = 300\\ \\mathrm{K}, T_2 = 127 + 273 = 400\\ \\mathrm{K}$. So $V_2 = 1.0 \\times \\dfrac{400}{300} = 1.33\\ \\mathrm{L}$. Using Celsius $(127/27)$ gives the wrong $4.70\\ \\mathrm{L}$.","explanationZh":"$查理定律必须用热力学温度：T_1=27+273=300\\ \\mathrm{K}，T_2=127+273=400\\ \\mathrm{K}，故 V_2=1.0\\times\\dfrac{400}{300}=1.33\\ \\mathrm{L}。直接代入摄氏度 127/27$ 会得到错误的 $4.70\\ \\mathrm{L}。$","tags":["trap","gas-laws","absolute-temperature","celsius-trap"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-gas-laws-012","moduleId":"thermal-physics","unitId":"thermal","topicId":"gas-laws","difficulty":4,"stem":"A rigid sealed container holds gas at $p_1 = 100\\ \\mathrm{kPa}, T_1 = 300\\ \\mathrm{K}$. It is heated to $T_2 = 450\\ \\mathrm{K}$ at constant volume. Find $p_2$.","stemZh":"刚性密闭容器中气体初态 $p_1=100\\ \\mathrm{kPa}、T_1=300\\ \\mathrm{K}，在恒容下加热到 T_2=450\\ \\mathrm{K}，求 p_2。$","options":[{"id":"A","text":"$200\\ \\mathrm{kPa}$"},{"id":"B","text":"$66.7\\ \\mathrm{kPa}$"},{"id":"C","text":"$100\\ \\mathrm{kPa}$"},{"id":"D","text":"$150\\ \\mathrm{kPa}$"}],"answer":"D","explanation":"Constant volume means Gay-Lussac's law $\\dfrac{p}{T} =$ constant $(not V/T): p_2 = p_1\\dfrac{T_2}{T_1} = 100 \\times \\dfrac{450}{300} = 150\\ \\mathrm{kPa}$.","explanationZh":"恒容用盖-吕萨克定律 $\\dfrac{p}{T}=\\text{常量}（而非 V/T）：p_2=p_1\\dfrac{T_2}{T_1}=100\\times\\dfrac{450}{300}=150\\ \\mathrm{kPa}。$","tags":["trap","gas-laws","gay-lussac","law-choice"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-gas-laws-013","images":[{"src":"/physics/diagrams/pv-diagram.svg","alt":"p–V diagram of an ideal gas: an isotherm is the curve pV $= constant$, so pressure falls as volume rises.","altZh":"理想气体的 p–V 图：等温线满足 pV = 常量，故压强随体积增大而减小。"}],"moduleId":"thermal-physics","unitId":"thermal","topicId":"gas-laws","difficulty":4,"stem":"A fixed mass of ideal gas undergoes a change shown on a p–V diagram. If the curve is an isotherm $(constant T)$, which relation holds between any two points on it?","stemZh":"一定质量理想气体的变化画在 p–V 图上，若该曲线为等温线（温度恒定），其上任意两点满足什么关系？","options":[{"id":"A","text":"$p \\propto V$"},{"id":"B","text":"$p \\propto T$"},{"id":"C","text":"pV $=$ constant"},{"id":"D","text":"$V \\propto T$"}],"answer":"C","explanation":"For an isotherm of a fixed mass, pV $=$ nRT $= constant; V \\propto T$ only holds at constant pressure, not on this curve.","explanationZh":"固定质量的等温过程满足 $pV=nRT=\\text{常量}；V\\propto T$ 只在恒压时成立，不适用于该曲线。","tags":["trap","gas-laws","pv-diagram","isotherm"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-gas-laws-014","moduleId":"thermal-physics","unitId":"thermal","topicId":"gas-laws","difficulty":5,"stem":"A fixed mass of gas is compressed to half its volume while its absolute temperature doubles. By what factor does its pressure change?","stemZh":"一定质量的气体被压缩到原来体积的一半，同时热力学温度加倍，压强变为原来的几倍？","options":[{"id":"A","text":"$4\\times$"},{"id":"B","text":"$2\\times$"},{"id":"C","text":"$1\\times (unchanged)$"},{"id":"D","text":"$0.5\\times$"}],"answer":"A","explanation":"From $\\dfrac{pV}{T} = constant, \\dfrac{p_2}{p_1} = \\dfrac{T_2}{T_1} \\div \\dfrac{V_2}{V_1} = \\dfrac{2}{1} \\div \\dfrac{1}{2} = 4$. Pressure becomes $4\\times —$ forgetting the volume factor gives the wrong $2\\times$.","explanationZh":"$由 \\dfrac{pV}{T}=\\text{常量}，\\dfrac{p_2}{p_1}=\\dfrac{T_2}{T_1}\\div\\dfrac{V_2}{V_1}=\\dfrac{2}{1}\\div\\dfrac{1}{2}=4，压强变为原来的 4$ 倍；漏掉体积因子会错得 $2\\times。$","tags":["trap","gas-laws","combined","two-factors"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-gas-laws-015","moduleId":"thermal-physics","unitId":"thermal","topicId":"gas-laws","difficulty":5,"stem":"Statement: 'If the temperature of a gas is doubled $(in °C)$, its volume doubles at constant pressure.' Which is correct?","stemZh":"命题：“若气体温度（以 °C 计）加倍，则在恒压下其体积加倍。” 正确的是？","options":[{"id":"A","text":"True, because of Charles's law"},{"id":"B","text":"False — Charles's law requires absolute temperature $(K)$, not °C"},{"id":"C","text":"True only for ideal gases"},{"id":"D","text":"False, volume is inversely related to temperature"}],"optionsZh":[{"id":"A","text":"正确，依据查理定律"},{"id":"B","text":"错误——查理定律要求用热力学温度 K，而非 °C"},{"id":"C","text":"仅对理想气体正确"},{"id":"D","text":"错误，体积与温度成反比"}],"answer":"B","explanation":"Charles's law uses absolute temperature T in K. Doubling a Celsius value does not double T in K, so the volume does not double.","explanationZh":"查理定律使用热力学温度 T（K）。摄氏度加倍并不等于开尔文温度加倍，因此体积不会加倍。","tags":["trap","gas-laws","charles","celsius-trap"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-first-law-001","moduleId":"thermal-physics","unitId":"thermal","topicId":"first-law","difficulty":1,"stem":"A gas is heated so it absorbs $Q = 200\\ \\mathrm{J}$. It sits in a rigid container so no work is done $(W = 0)$. What is $\\Delta$ U?","stemZh":"气体吸热 $Q=200\\ \\mathrm{J}，置于刚性容器中不做功（W=0），求 \\Delta$ U。","options":[{"id":"A","text":"$+200\\ \\mathrm{J}$"},{"id":"B","text":"$+400\\ \\mathrm{J}$"},{"id":"C","text":"$0\\ \\mathrm{J}$"},{"id":"D","text":"$-200\\ \\mathrm{J}$"}],"answer":"A","explanation":"Using $\\Delta U = Q + W (W =\\text{ work done ON the }gas): \\Delta U = 200 + 0 = +200\\ \\mathrm{J}$.","explanationZh":"采用 $\\Delta U=Q+W（W 为外界对气体做的功）：\\Delta U=200+0=+200\\ \\mathrm{J}。$","tags":["basic","first-law","delta-u"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-first-law-002","moduleId":"thermal-physics","unitId":"thermal","topicId":"first-law","difficulty":1,"stem":"A gas expands and does $150\\ \\mathrm{J}$ of work on its surroundings, with no heat exchange $(Q = 0, adiabatic)$. What is $\\Delta$ U?","stemZh":"气体膨胀对外界做功 $150\\ \\mathrm{J}，且无热交换（Q=0，绝热），求 \\Delta$ U。","options":[{"id":"A","text":"$+150\\ \\mathrm{J}$"},{"id":"B","text":"$-150\\ \\mathrm{J}$"},{"id":"C","text":"$0\\ \\mathrm{J}$"},{"id":"D","text":"$+300\\ \\mathrm{J}$"}],"answer":"B","explanation":"Work done ON the gas is negative when it $expands: W = -150\\ \\mathrm{J}$. So $\\Delta U = Q + W = 0 - 150 = -150\\ \\mathrm{J}$.","explanationZh":"$气体膨胀时外界对气体做功为负：W=-150\\ \\mathrm{J}，故 \\Delta U=Q+W=0-150=-150\\ \\mathrm{J}。$","tags":["basic","first-law","sign","adiabatic"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-first-law-003","moduleId":"thermal-physics","unitId":"thermal","topicId":"first-law","difficulty":1,"stem":"An ideal gas expands isothermally (constant T) and absorbs $Q = 100\\ \\mathrm{J}$. What is W (work done ON the gas)?","stemZh":"理想气体等温膨胀（T 不变）并吸热 $Q=100\\ \\mathrm{J}，求$ W（外界对气体做的功）。","options":[{"id":"A","text":"$+100\\ \\mathrm{J}$"},{"id":"B","text":"$0\\ \\mathrm{J}$"},{"id":"C","text":"$-100\\ \\mathrm{J}$"},{"id":"D","text":"$+200\\ \\mathrm{J}$"}],"answer":"C","explanation":"For an ideal gas, internal energy depends only on T, so isothermal means $\\Delta U = 0$. Then $W = \\Delta U - Q = 0 - 100 = -100\\ \\mathrm{J} (the\\text{ gas does }+100\\ \\mathrm{J}\\text{ on the }surroundings)$.","explanationZh":"理想气体内能只与温度有关，等温时 $\\Delta U=0，故 W=\\Delta U-Q=0-100=-100\\ \\mathrm{J}（气体对外界做功 +100\\ \\mathrm{J}）。$","tags":["basic","first-law","isothermal","sign"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-first-law-004","moduleId":"thermal-physics","unitId":"thermal","topicId":"first-law","difficulty":2,"stem":"An isochoric $(constant-volume) process: W = 0$. The gas receives $Q = 80\\ \\mathrm{J}$. What is $\\Delta$ U?","stemZh":"$等容过程（W=0），气体吸热 Q=80\\ \\mathrm{J}，求 \\Delta$ U。","options":[{"id":"A","text":"$0\\ \\mathrm{J}$"},{"id":"B","text":"$-80\\ \\mathrm{J}$"},{"id":"C","text":"$160\\ \\mathrm{J}$"},{"id":"D","text":"$+80\\ \\mathrm{J}$"}],"answer":"D","explanation":"At constant volume no work is done $(W = 0)$, so $\\Delta U = Q + W = 80 + 0 = +80\\ \\mathrm{J}$.","explanationZh":"$等容过程不做功（W=0），故 \\Delta U=Q+W=80+0=+80\\ \\mathrm{J}。$","tags":["basic","first-law","isochoric"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-first-law-005","moduleId":"thermal-physics","unitId":"thermal","topicId":"first-law","difficulty":2,"stem":"The internal energy of a gas increases by $\\Delta U = +50\\ \\mathrm{J}$, and $W = +30\\ \\mathrm{J}$ of work is done ON it (compression). Find Q.","stemZh":"气体内能增加 $\\Delta U=+50\\ \\mathrm{J}，外界对其做功 W=+30\\ \\mathrm{J}（压缩），求$ Q。","options":[{"id":"A","text":"$+20\\ \\mathrm{J}$"},{"id":"B","text":"$+80\\ \\mathrm{J}$"},{"id":"C","text":"$-20\\ \\mathrm{J}$"},{"id":"D","text":"$+30\\ \\mathrm{J}$"}],"answer":"A","explanation":"$\\Delta U = Q + W \\Rightarrow Q = \\Delta U - W = 50 - 30 = +20\\ \\mathrm{J} (the\\text{ gas absorbs }20\\ \\mathrm{J})$.","explanationZh":"$\\Delta U=Q+W\\Rightarrow Q=\\Delta U-W=50-30=+20\\ \\mathrm{J}（气体吸热 20\\ \\mathrm{J}）。$","tags":["basic","first-law","delta-u"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-first-law-006","moduleId":"thermal-physics","unitId":"thermal","topicId":"first-law","difficulty":3,"stem":"An adiabatic compression does $W = +200\\ \\mathrm{J}$ of work ON the gas $(Q = 0)$. What is $\\Delta$ U?","stemZh":"绝热压缩中外界对气体做功 $W=+200\\ \\mathrm{J}（Q=0），求 \\Delta$ U。","options":[{"id":"A","text":"$0\\ \\mathrm{J}$"},{"id":"B","text":"$+200\\ \\mathrm{J}$"},{"id":"C","text":"$-200\\ \\mathrm{J}$"},{"id":"D","text":"$+400\\ \\mathrm{J}$"}],"answer":"B","explanation":"Adiabatic means $Q = 0$; compression means work is done ON the gas $(W = +200\\ \\mathrm{J}). \\Delta U = 0 + 200 = +200\\ \\mathrm{J}$ and temperature rises.","explanationZh":"绝热即 $Q=0；压缩表示外界对气体做功（W=+200\\ \\mathrm{J}），故 \\Delta U=0+200=+200\\ \\mathrm{J}，温度升高。$","tags":["applied","first-law","adiabatic","compression"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-first-law-007","moduleId":"thermal-physics","unitId":"thermal","topicId":"first-law","difficulty":3,"stem":"A gas absorbs $Q = 500\\ \\mathrm{J}$ and expands, doing $200\\ \\mathrm{J}$ of work on the surroundings. What is $\\Delta$ U?","stemZh":"气体吸热 $Q=500\\ \\mathrm{J}$ 并膨胀，对外界做功 $200\\ \\mathrm{J}，求 \\Delta$ U。","options":[{"id":"A","text":"$+700\\ \\mathrm{J}$"},{"id":"B","text":"$+500\\ \\mathrm{J}$"},{"id":"C","text":"$+300\\ \\mathrm{J}$"},{"id":"D","text":"$+200\\ \\mathrm{J}$"}],"answer":"C","explanation":"Work done ON the gas is negative when it $expands: W = -200\\ \\mathrm{J}. \\Delta U = Q + W = 500 - 200 = +300\\ \\mathrm{J}$.","explanationZh":"$膨胀时外界对气体做功为负：W=-200\\ \\mathrm{J}，故 \\Delta U=Q+W=500-200=+300\\ \\mathrm{J}。$","tags":["applied","first-law","sign","expansion"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-first-law-008","moduleId":"thermal-physics","unitId":"thermal","topicId":"first-law","difficulty":3,"stem":"In an isobaric $(constant-pressure) expansion, p = 2.0 \\times 10^{5}\\ \\mathrm{Pa}$, from $V_1 = 1.0 \\times 10^{-3}\\ \\mathrm{m^{3}}$ to $V_2 = 4.0 \\times 10^{-3}\\ \\mathrm{m^{3}}$. What is the work done ON the gas?","stemZh":"等压膨胀中 $p=2.0\\times10^{5}\\ \\mathrm{Pa}，体积从 V_1=1.0\\times10^{-3}\\ \\mathrm{m^{3}} 到 V_2=4.0\\times10^{-3}\\ \\mathrm{m^{3}}，外界对气体做功为多少？$","options":[{"id":"A","text":"$+600\\ \\mathrm{J}$"},{"id":"B","text":"$0\\ \\mathrm{J}$"},{"id":"C","text":"$+300\\ \\mathrm{J}$"},{"id":"D","text":"$-600\\ \\mathrm{J}$"}],"answer":"D","explanation":"Work done BY the gas is $p\\Delta V = 2.0\\times10^{5} \\times 3.0\\times10^{-3} = 600\\ \\mathrm{J}$. Work done ON the gas is the $negative: W = -600\\ \\mathrm{J}$.","explanationZh":"气体对外界做功为 $p\\Delta V=2.0\\times10^{5}\\times3.0\\times10^{-3}=600\\ \\mathrm{J}，故外界对气体做功为其相反数 W=-600\\ \\mathrm{J}。$","tags":["applied","first-law","isobaric","work-sign"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-first-law-009","moduleId":"thermal-physics","unitId":"thermal","topicId":"first-law","difficulty":3,"stem":"A gas in a piston is compressed isobarically from $V_1 = 3.0 \\times 10^{-3}\\ \\mathrm{m^{3}}$ to $V_2 = 1.0 \\times 10^{-3}\\ \\mathrm{m^{3}}$ at $p = 1.0 \\times 10^{5}\\ \\mathrm{Pa}$, and loses $Q = -100\\ \\mathrm{J}$ of heat. What is $\\Delta$ U?","stemZh":"活塞中气体在 $p=1.0\\times10^{5}\\ \\mathrm{Pa}$ 下等压压缩，体积从 $V_1=3.0\\times10^{-3}\\ \\mathrm{m^{3}} 到 V_2=1.0\\times10^{-3}\\ \\mathrm{m^{3}}，并放热 Q=-100\\ \\mathrm{J}，求 \\Delta$ U。","options":[{"id":"A","text":"$+100\\ \\mathrm{J}$"},{"id":"B","text":"$+300\\ \\mathrm{J}$"},{"id":"C","text":"$-100\\ \\mathrm{J}$"},{"id":"D","text":"$0\\ \\mathrm{J}$"}],"answer":"A","explanation":"Compression means work done ON the gas is $positive: W = -p\\Delta V = -1.0\\times10^{5}(1.0\\times10^{-3} - 3.0\\times10^{-3}) = +200\\ \\mathrm{J}. \\Delta U = Q + W = -100 + 200 = +100\\ \\mathrm{J}$.","explanationZh":"$压缩时外界对气体做功为正：W=-p\\Delta V=-1.0\\times10^{5}(1.0\\times10^{-3}-3.0\\times10^{-3})=+200\\ \\mathrm{J}。\\Delta U=Q+W=-100+200=+100\\ \\mathrm{J}。$","tags":["applied","first-law","isobaric","compression"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-first-law-010","moduleId":"thermal-physics","unitId":"thermal","topicId":"first-law","difficulty":3,"stem":"Over one complete cycle, a gas net absorbs $Q_{net} = +250\\ \\mathrm{J}$. What is the net work done ON the $gas, W_{net}$?","stemZh":"在一个完整循环中，气体净吸热 $Q_{\\text{net}}=+250\\ \\mathrm{J}，求外界对气体的净功 W_{\\text{net}}。$","options":[{"id":"A","text":"$+250\\ \\mathrm{J}$"},{"id":"B","text":"$-250\\ \\mathrm{J}$"},{"id":"C","text":"$0\\ \\mathrm{J}$"},{"id":"D","text":"$+500\\ \\mathrm{J}$"}],"answer":"B","explanation":"For a complete cycle $\\Delta U = 0$, so $0 = Q_{net} + W_{net} \\Rightarrow W_{net} = -Q_{net} = -250\\ \\mathrm{J} (the\\text{ gas does }+250\\ \\mathrm{J}\\text{ on the }surroundings)$.","explanationZh":"完成一个循环时 $\\Delta U=0，故 0=Q_{\\text{net}}+W_{\\text{net}}，得 W_{\\text{net}}=-Q_{\\text{net}}=-250\\ \\mathrm{J}（气体对外界做功 +250\\ \\mathrm{J}）。$","tags":["applied","first-law","cycle"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-first-law-011","moduleId":"thermal-physics","unitId":"thermal","topicId":"first-law","difficulty":4,"stem":"A gas expands and does $100\\ \\mathrm{J}$ of work on the surroundings while absorbing $Q = 100\\ \\mathrm{J}$ of heat. A student claims $\\Delta U = Q + W = 100 + 100 = 200\\ \\mathrm{J}$. What is wrong?","stemZh":"气体膨胀对外界做功 $100\\ \\mathrm{J}，同时吸热 Q=100\\ \\mathrm{J}。有学生认为 \\Delta U=Q+W=100+100=200\\ \\mathrm{J}，错在哪里？$","options":[{"id":"A","text":"$Nothing, 200\\ \\mathrm{J}$ is correct"},{"id":"B","text":"$\\Delta U$ should be $-200\\ \\mathrm{J}$"},{"id":"C","text":"W must be $-100\\ \\mathrm{J}$, so $\\Delta U = 0$"},{"id":"D","text":"Q should be $-100\\ \\mathrm{J}$"}],"answer":"C","explanation":"Here W is work done ON the gas. Since the gas does work on the $surroundings, W = -100\\ \\mathrm{J}$. Then $\\Delta U = 100 + (-100) = 0$. The student used the wrong sign for W.","explanationZh":"此处 W 是外界对气体做的功。气体对外界做功，故 $W=-100\\ \\mathrm{J}，\\Delta U=100+(-100)=0。学生把 W$ 的符号弄错了。","tags":["trap","first-law","sign","work-on-gas"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-first-law-012","moduleId":"thermal-physics","unitId":"thermal","topicId":"first-law","difficulty":4,"stem":"A gas in a rigid sealed container is heated, gaining $Q = 120\\ \\mathrm{J}$. What is $\\Delta$ U?","stemZh":"刚性密闭容器中的气体被加热，吸热 $Q=120\\ \\mathrm{J}，求 \\Delta$ U。","options":[{"id":"A","text":"$0\\ \\mathrm{J}$"},{"id":"B","text":"$-120\\ \\mathrm{J}$"},{"id":"C","text":"$+240\\ \\mathrm{J}$"},{"id":"D","text":"$+120\\ \\mathrm{J}$"}],"answer":"D","explanation":"Rigid means constant volume, so $W = 0 (there\\text{ is no }p\\Delta V work). \\Delta U = Q + W = 120 + 0 = +120\\ \\mathrm{J}$.","explanationZh":"刚性即体积不变，故 $W=0（没有 p\\Delta V 功），\\Delta U=Q+W=120+0=+120\\ \\mathrm{J}。$","tags":["trap","first-law","isochoric","no-work"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-first-law-013","moduleId":"thermal-physics","unitId":"thermal","topicId":"first-law","difficulty":4,"stem":"An ideal gas is compressed isothermally while $50\\ \\mathrm{J}$ of work is done ON it. What is the heat exchange Q?","stemZh":"理想气体被等温压缩，外界对其做功 $50\\ \\mathrm{J}，热交换 Q$ 为多少？","options":[{"id":"A","text":"$-50\\ \\mathrm{J}$"},{"id":"B","text":"$+50\\ \\mathrm{J}$"},{"id":"C","text":"$0\\ \\mathrm{J}$"},{"id":"D","text":"$+100\\ \\mathrm{J}$"}],"answer":"A","explanation":"For an ideal gas, isothermal means $\\Delta U = 0$. Then $Q = \\Delta U - W = 0 - 50 = -50\\ \\mathrm{J}: 50\\ \\mathrm{J}$ of heat must leave the gas. Forgetting $\\Delta U = 0$ is the common trap.","explanationZh":"理想气体等温时 $\\Delta U=0，故 Q=\\Delta U-W=0-50=-50\\ \\mathrm{J}，即有 50\\ \\mathrm{J}$ 热量放出。常见错误是忽略等温时 $\\Delta U=0。$","tags":["trap","first-law","isothermal","heat-out"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-first-law-014","moduleId":"thermal-physics","unitId":"thermal","topicId":"first-law","difficulty":5,"stem":"A gas expands adiabatically, doing $300\\ \\mathrm{J}$ of work on the surroundings $(Q = 0)$. What is $\\Delta$ U?","stemZh":"气体绝热膨胀，对外界做功 $300\\ \\mathrm{J}（Q=0），求 \\Delta$ U。","options":[{"id":"A","text":"$0\\ \\mathrm{J}$"},{"id":"B","text":"$-300\\ \\mathrm{J}$"},{"id":"C","text":"$+300\\ \\mathrm{J}$"},{"id":"D","text":"$-600\\ \\mathrm{J}$"}],"answer":"B","explanation":"Adiabatic means $Q = 0$. The gas does work on the surroundings, so work done ON it is $W = -300\\ \\mathrm{J}. \\Delta U = 0 + (-300) = -300\\ \\mathrm{J}$; temperature falls. $(Trap:\\text{ thinking adiabatic }\\Rightarrow \\Delta U = 0.)$","explanationZh":"绝热即 $Q=0。气体对外做功，故 W=-300\\ \\mathrm{J}，\\Delta U=0+(-300)=-300\\ \\mathrm{J}，温度下降。（陷阱：误以为绝热即 \\Delta U=0。）$","tags":["trap","first-law","adiabatic","expansion"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-first-law-015","moduleId":"thermal-physics","unitId":"thermal","topicId":"first-law","difficulty":5,"stem":"A gas undergoes two $steps: (1)$ isochoric heating with $Q = +100\\ \\mathrm{J}; (2)$ isobaric expansion absorbing $Q = +200\\ \\mathrm{J}$ while doing $100\\ \\mathrm{J}$ of work on the surroundings. What is the total $\\Delta$ U?","stemZh":"$气体经历两步：(1)$ 等容加热 $Q=+100\\ \\mathrm{J}；(2)$ 等压膨胀吸热 $Q=+200\\ \\mathrm{J}$ 且对外界做功 $100\\ \\mathrm{J}。总 \\Delta U$ 是多少？","options":[{"id":"A","text":"$+100\\ \\mathrm{J}$"},{"id":"B","text":"$0\\ \\mathrm{J}$"},{"id":"C","text":"$+200\\ \\mathrm{J}$"},{"id":"D","text":"$+400\\ \\mathrm{J}$"}],"answer":"C","explanation":"Step $1: W_1 = 0$, so $\\Delta U_1 = +100\\ \\mathrm{J}$. Step 2: work ON gas $W_2 = -100\\ \\mathrm{J}$, so $\\Delta U_2 = 200 + (-100) = +100\\ \\mathrm{J}$. Total $\\Delta U = 100 + 100 = +200\\ \\mathrm{J}$.","explanationZh":"$第一步：W_1=0，故 \\Delta U_1=+100\\ \\mathrm{J}。第二步：外界对气体做功 W_2=-100\\ \\mathrm{J}，\\Delta U_2=200+(-100)=+100\\ \\mathrm{J}。合计 \\Delta U=100+100=+200\\ \\mathrm{J}。$","tags":["trap","first-law","multi-step","delta-u"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-reflection-refraction-001","moduleId":"optics","unitId":"geometric-optics","topicId":"reflection-refraction","difficulty":1,"stem":"A light ray strikes a mirror with an angle of incidence $i = 30^{\\circ}$. By the law of reflection, what is the angle of reflection?","stemZh":"一束光以入射角 $i=30^{\\circ}$ 射到镜面，根据反射定律，反射角是多少？","options":[{"id":"A","text":"$60^{\\circ}$"},{"id":"B","text":"$90^{\\circ}$"},{"id":"C","text":"$30^{\\circ}$"},{"id":"D","text":"$0^{\\circ}$"}],"answer":"C","explanation":"The law of reflection states the angle of reflection equals the angle of incidence: i' $= i = 30^{\\circ}$.","explanationZh":"反射定律：反射角等于入射角，即 $i'=i=30^{\\circ}。$","tags":["basic","reflection-refraction","reflection"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-reflection-refraction-002","images":[{"src":"/physics/diagrams/refraction-ray.svg","alt":"Ray of light entering glass from air, bending toward the $normal; n_air \\sin i = n_glass \\sin$ r.","altZh":"光从空气进入玻璃，向法线偏折；满足 n_空气 \\sin i = n_玻璃 \\sin r。"}],"moduleId":"optics","unitId":"geometric-optics","topicId":"reflection-refraction","difficulty":1,"stem":"Light in air $(n_1 = 1.0)$ enters glass $(n_2 = 1.5)$ at an angle of incidence $i = 30^{\\circ}$. Find the angle of refraction r.","stemZh":"$光从空气（n_1=1.0）射入玻璃（n_2=1.5），入射角 i=30^{\\circ}，求折射角$ r。","options":[{"id":"A","text":"$30^{\\circ}$"},{"id":"B","text":"$45^{\\circ}$"},{"id":"C","text":"$15^{\\circ}$"},{"id":"D","text":"$20^{\\circ}$"}],"answer":"D","explanation":"Snell's $law: n_1\\sin i = n_2\\sin r \\Rightarrow \\sin r = \\dfrac{1.0}{1.5}\\sin 30^{\\circ} = 0.333$, so $r \\approx 19.5^{\\circ} \\approx 20^{\\circ}$.","explanationZh":"$折射定律：n_1\\sin i=n_2\\sin r\\Rightarrow\\sin r=\\dfrac{1.0}{1.5}\\sin30^{\\circ}=0.333，故 r\\approx19.5^{\\circ}\\approx20^{\\circ}。$","tags":["basic","reflection-refraction","snell"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-reflection-refraction-003","moduleId":"optics","unitId":"geometric-optics","topicId":"reflection-refraction","difficulty":1,"stem":"The speed of light in a medium is $v = 2.0 \\times 10^{8}\\ \\mathrm{m/s} (c = 3.0 \\times 10^{8}\\ \\mathrm{m/s})$. What is the refractive index n?","stemZh":"光在介质中的速度 $v=2.0\\times10^{8}\\ \\mathrm{m/s}（c=3.0\\times10^{8}\\ \\mathrm{m/s}），折射率 n$ 为多少？","options":[{"id":"A","text":"1.5"},{"id":"B","text":"0.67"},{"id":"C","text":"2.0"},{"id":"D","text":"1.0"}],"answer":"A","explanation":"Refractive index $n = \\dfrac{c}{v} = \\dfrac{3.0 \\times 10^{8}}{2.0 \\times 10^{8}} = 1.5$.","explanationZh":"折射率 $n=\\dfrac{c}{v}=\\dfrac{3.0\\times10^{8}}{2.0\\times10^{8}}=1.5。$","tags":["basic","reflection-refraction","refractive-index"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-reflection-refraction-004","moduleId":"optics","unitId":"geometric-optics","topicId":"reflection-refraction","difficulty":2,"stem":"Glass has refractive index $n = 1.5$. What is the speed of light inside it?","stemZh":"玻璃折射率 $n=1.5，光在其中的传播速度是多少？$","options":[{"id":"A","text":"$4.5 \\times 10^{8}\\ \\mathrm{m/s}$"},{"id":"B","text":"$2.0 \\times 10^{8}\\ \\mathrm{m/s}$"},{"id":"C","text":"$1.5 \\times 10^{8}\\ \\mathrm{m/s}$"},{"id":"D","text":"$3.0 \\times 10^{8}\\ \\mathrm{m/s}$"}],"answer":"B","explanation":"$v = \\dfrac{c}{n} = \\dfrac{3.0 \\times 10^{8}}{1.5} = 2.0 \\times 10^{8}\\ \\mathrm{m/s}$.","explanationZh":"$v=\\dfrac{c}{n}=\\dfrac{3.0\\times10^{8}}{1.5}=2.0\\times10^{8}\\ \\mathrm{m/s}。$","tags":["basic","reflection-refraction","speed-in-medium"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-reflection-refraction-005","moduleId":"optics","unitId":"geometric-optics","topicId":"reflection-refraction","difficulty":2,"stem":"Light goes from water $(n_1 = 1.33)$ into air $(n_2 = 1.0)$ at $i = 30^{\\circ}$. Find the angle of refraction r.","stemZh":"$光从水（n_1=1.33）射入空气（n_2=1.0），入射角 i=30^{\\circ}，求折射角$ r。","options":[{"id":"A","text":"$22^{\\circ}$"},{"id":"B","text":"$30^{\\circ}$"},{"id":"C","text":"$42^{\\circ}$"},{"id":"D","text":"$48^{\\circ}$"}],"answer":"C","explanation":"$\\sin r = \\dfrac{n_1}{n_2}\\sin i = \\dfrac{1.33}{1.0}\\sin 30^{\\circ} = 0.665$, so $r \\approx 41.7^{\\circ} \\approx 42^{\\circ}$.","explanationZh":"$\\sin r=\\dfrac{n_1}{n_2}\\sin i=\\dfrac{1.33}{1.0}\\sin30^{\\circ}=0.665，故 r\\approx41.7^{\\circ}\\approx42^{\\circ}。$","tags":["basic","reflection-refraction","snell"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-reflection-refraction-006","moduleId":"optics","unitId":"geometric-optics","topicId":"reflection-refraction","difficulty":3,"stem":"Light in air $(n_1 = 1.0)$ enters glass $(n_2 = 1.5)$ at $i = 60^{\\circ}$. Find r.","stemZh":"$光从空气（n_1=1.0）射入玻璃（n_2=1.5），入射角 i=60^{\\circ}，求$ r。","options":[{"id":"A","text":"$60^{\\circ}$"},{"id":"B","text":"$40^{\\circ}$"},{"id":"C","text":"$30^{\\circ}$"},{"id":"D","text":"$35^{\\circ}$"}],"answer":"D","explanation":"$\\sin r = \\dfrac{1.0}{1.5}\\sin 60^{\\circ} = 0.6667 \\times 0.8660 = 0.577$, so $r \\approx 35.3^{\\circ} \\approx 35^{\\circ}$.","explanationZh":"$\\sin r=\\dfrac{1.0}{1.5}\\sin60^{\\circ}=0.6667\\times0.8660=0.577，故 r\\approx35.3^{\\circ}\\approx35^{\\circ}。$","tags":["applied","reflection-refraction","snell"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-reflection-refraction-007","moduleId":"optics","unitId":"geometric-optics","topicId":"reflection-refraction","difficulty":3,"stem":"Light takes $t = 2.0 \\times 10^{-11}\\ \\mathrm{s}$ to cross a slab of thickness $d = 3.0 \\times 10^{-3}\\ \\mathrm{m}$. What is the refractive index $(c = 3.0 \\times 10^{8}\\ \\mathrm{m/s})$?","stemZh":"光穿过厚度 $d=3.0\\times10^{-3}\\ \\mathrm{m}$ 的平板耗时 $t=2.0\\times10^{-11}\\ \\mathrm{s}，折射率是多少（c=3.0\\times10^{8}\\ \\mathrm{m/s}）？$","options":[{"id":"A","text":"2.0"},{"id":"B","text":"1.0"},{"id":"C","text":"1.5"},{"id":"D","text":"0.5"}],"answer":"A","explanation":"$v = \\dfrac{d}{t} = \\dfrac{3.0 \\times 10^{-3}}{2.0 \\times 10^{-11}} = 1.5 \\times 10^{8}\\ \\mathrm{m/s}$, so $n = \\dfrac{c}{v} = \\dfrac{3.0 \\times 10^{8}}{1.5 \\times 10^{8}} = 2.0$.","explanationZh":"$v=\\dfrac{d}{t}=\\dfrac{3.0\\times10^{-3}}{2.0\\times10^{-11}}=1.5\\times10^{8}\\ \\mathrm{m/s}，故 n=\\dfrac{c}{v}=\\dfrac{3.0\\times10^{8}}{1.5\\times10^{8}}=2.0。$","tags":["applied","reflection-refraction","refractive-index","speed"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-reflection-refraction-008","moduleId":"optics","unitId":"geometric-optics","topicId":"reflection-refraction","difficulty":3,"stem":"A ray strikes a mirror at an angle of incidence of $40^{\\circ}$. What is the angle between the incident ray and the reflected ray?","stemZh":"光线以入射角 $40^{\\circ}$ 射到镜面，入射光线与反射光线之间的夹角是多少？","options":[{"id":"A","text":"$40^{\\circ}$"},{"id":"B","text":"$80^{\\circ}$"},{"id":"C","text":"$50^{\\circ}$"},{"id":"D","text":"$140^{\\circ}$"}],"answer":"B","explanation":"Both rays make $40^{\\circ}$ with the normal, on opposite sides, so the angle between them is $40^{\\circ} + 40^{\\circ} = 80^{\\circ}$.","explanationZh":"两光线分别在法线两侧与法线成 $40^{\\circ}，故二者夹角为 40^{\\circ}+40^{\\circ}=80^{\\circ}。$","tags":["applied","reflection-refraction","reflection","angle-between"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-reflection-refraction-009","images":[{"src":"/physics/diagrams/refraction-ray.svg","alt":"Ray leaving glass into air, bending away from the $normal; n_glass \\sin i = n_air \\sin$ r.","altZh":"光从玻璃射向空气，偏离法线；满足 n_玻璃 \\sin i = n_空气 \\sin r。"}],"moduleId":"optics","unitId":"geometric-optics","topicId":"reflection-refraction","difficulty":3,"stem":"Light in glass $(n_1 = 1.5)$ strikes a glass–air boundary at $i = 37^{\\circ} (the\\text{ critical angle for glass is about }42^{\\circ})$. Find r in air $(n_2 = 1.0)$.","stemZh":"$光在玻璃（n_1=1.5）中以 i=37^{\\circ}$ 射向玻璃–空气界面（玻璃临界角约 $42^{\\circ}），求在空气中的折射角 r（n_2=1.0）。$","options":[{"id":"A","text":"$37^{\\circ}$"},{"id":"B","text":"$45^{\\circ}$"},{"id":"C","text":"$64^{\\circ}$"},{"id":"D","text":"$90^{\\circ}$"}],"answer":"C","explanation":"$\\sin r = \\dfrac{1.5}{1.0}\\sin 37^{\\circ} \\approx 1.5 \\times 0.60 = 0.90$, so $r \\approx 64^{\\circ} (since 37^{\\circ} < 42^{\\circ},\\text{ refraction }occurs)$.","explanationZh":"$\\sin r=\\dfrac{1.5}{1.0}\\sin37^{\\circ}\\approx1.5\\times0.60=0.90，故 r\\approx64^{\\circ}（因 37^{\\circ}<42^{\\circ}，发生折射）。$","tags":["applied","reflection-refraction","snell"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-reflection-refraction-010","moduleId":"optics","unitId":"geometric-optics","topicId":"reflection-refraction","difficulty":3,"stem":"Monochromatic light of wavelength $\\lambda_0 = 600\\ \\mathrm{nm}$ in air enters glass with $n = 1.5$. What is its wavelength in the glass?","stemZh":"真空中波长 $\\lambda_0=600\\ \\mathrm{nm}$ 的单色光进入折射率 $n=1.5$ 的玻璃，在玻璃中的波长是多少？","options":[{"id":"A","text":"$900\\ \\mathrm{nm}$"},{"id":"B","text":"$600\\ \\mathrm{nm}$"},{"id":"C","text":"$300\\ \\mathrm{nm}$"},{"id":"D","text":"$400\\ \\mathrm{nm}$"}],"answer":"D","explanation":"Frequency is unchanged; wavelength shortens by the refractive $index: \\lambda = \\dfrac{\\lambda_0}{n} = \\dfrac{600}{1.5} = 400\\ \\mathrm{nm}$.","explanationZh":"$频率不变，波长按折射率缩短：\\lambda=\\dfrac{\\lambda_0}{n}=\\dfrac{600}{1.5}=400\\ \\mathrm{nm}。$","tags":["applied","reflection-refraction","wavelength","medium"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-reflection-refraction-011","moduleId":"optics","unitId":"geometric-optics","topicId":"reflection-refraction","difficulty":4,"stem":"A student writes Snell's law as $n_1\\sin i = n_2\\sin r and$, for light going from air into glass, uses $n_1 = 1.5, n_2 = 1.0$. What is the error?","stemZh":"某学生把折射定律写成 $n_1\\sin i=n_2\\sin r，并用于空气→玻璃时取 n_1=1.5、n_2=1.0，错在哪里？$","options":[{"id":"A","text":"They swapped $n_1$ and $n_2$; air is the incident medium so $n_1 = 1.0$, glass is $n_2 = 1.5$"},{"id":"B","text":"They used the wrong formula"},{"id":"C","text":"Angles must be measured from the surface, not the normal"},{"id":"D","text":"There is nothing wrong"}],"optionsZh":[{"id":"A","text":"他们把 $n_1 与 n_2$ 颠倒了；空气是入射介质，应 $n_1=1.0、玻璃 n_2=1.5$"},{"id":"B","text":"公式用错了"},{"id":"C","text":"角度应从界面而非法线起算"},{"id":"D","text":"没有问题"}],"answer":"A","explanation":"In Snell's law $n_1$ is the refractive index of the incident medium. $Air→glass$ means $n_1 = 1.0 (air), n_2 = 1.5 (glass)$. Swapping them gives the wrong angle.","explanationZh":"折射定律中 $n_1 是入射介质的折射率。空气→玻璃应 n_1=1.0（空气）、n_2=1.5（玻璃）。颠倒两者会得到错误角度。$","tags":["trap","reflection-refraction","snell","n-order"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-reflection-refraction-012","moduleId":"optics","unitId":"geometric-optics","topicId":"reflection-refraction","difficulty":4,"stem":"A ray meets a mirror making an angle of $60^{\\circ}$ with the mirror surface. What is the angle of reflection (measured from the normal)?","stemZh":"光线与镜面成 $60^{\\circ}$ 夹角入射，反射角（相对于法线）是多少？","options":[{"id":"A","text":"$60^{\\circ}$"},{"id":"B","text":"$30^{\\circ}$"},{"id":"C","text":"$120^{\\circ}$"},{"id":"D","text":"$90^{\\circ}$"}],"answer":"B","explanation":"The angle with the surface is $60^{\\circ}$, so the angle of incidence from the normal is $90^{\\circ} - 60^{\\circ} = 30^{\\circ}$. Reflection equals $incidence: 30^{\\circ}$ from the normal.","explanationZh":"与镜面夹角为 $60^{\\circ}，故相对法线的入射角为 90^{\\circ}-60^{\\circ}=30^{\\circ}。反射角等于入射角：相对法线 30^{\\circ}。$","tags":["trap","reflection-refraction","reflection","surface-vs-normal"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-reflection-refraction-013","moduleId":"optics","unitId":"geometric-optics","topicId":"reflection-refraction","difficulty":4,"stem":"Light travels from air $(n = 1.0)$ into water $(n = 1.33)$. How does the refracted ray behave relative to the normal?","stemZh":"$光从空气（n=1.0）进入水（n=1.33），折射光线相对法线如何偏折？$","options":[{"id":"A","text":"Away from the normal"},{"id":"B","text":"It does not refract"},{"id":"C","text":"Toward the normal"},{"id":"D","text":"Back along the incident ray"}],"optionsZh":[{"id":"A","text":"偏离法线"},{"id":"B","text":"不发生折射"},{"id":"C","text":"偏向法线"},{"id":"D","text":"沿原入射光线返回"}],"answer":"C","explanation":"Entering a denser $(higher-n) medium$, the ray slows and bends toward the normal, so the angle of refraction is smaller than the angle of incidence.","explanationZh":"进入折射率更大的介质时，光速变慢，光线偏向法线，折射角小于入射角。","tags":["trap","reflection-refraction","bending","denser"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-reflection-refraction-014","moduleId":"optics","unitId":"geometric-optics","topicId":"reflection-refraction","difficulty":5,"stem":"A ray enters a glass block $(n = 1.5) normally$, hitting the surface at $i = 0^{\\circ} (along\\text{ the }normal)$. What happens to its direction?","stemZh":"$光线垂直射入玻璃块（n=1.5），即沿法线以 i=0^{\\circ}$ 入射，其方向如何变化？","options":[{"id":"A","text":"It bends toward the normal"},{"id":"B","text":"$30^{\\circ}$"},{"id":"C","text":"It reflects back"},{"id":"D","text":"$0^{\\circ} —$ no bending"}],"answer":"D","explanation":"Normal incidence means $i = 0^{\\circ}$, so $\\sin r = \\dfrac{n_1}{n_2}\\sin 0^{\\circ} = 0$ and $r = 0^{\\circ}$. The ray passes straight through without bending (it does slow down, but direction is unchanged).","explanationZh":"正入射即 $i=0^{\\circ}，故 \\sin r=\\dfrac{n_1}{n_2}\\sin0^{\\circ}=0，r=0^{\\circ}。光线直线穿过不偏折（速度会变慢，但方向不变）。$","tags":["trap","reflection-refraction","normal-incidence","no-bend"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-reflection-refraction-015","moduleId":"optics","unitId":"geometric-optics","topicId":"reflection-refraction","difficulty":5,"stem":"Light in water $(n_1 = 1.33)$ strikes a water–air boundary at $i = 30^{\\circ} (water's\\text{ critical angle is about }49^{\\circ})$. Find r in air $(n_2 = 1.0)$.","stemZh":"$光在水（n_1=1.33）中以 i=30^{\\circ}$ 射向水–空气界面（水的临界角约 $49^{\\circ}），求在空气中的 r（n_2=1.0）。$","options":[{"id":"A","text":"$42^{\\circ}$"},{"id":"B","text":"$48^{\\circ}$"},{"id":"C","text":"Total internal reflection"},{"id":"D","text":"$30^{\\circ}$"}],"answer":"A","explanation":"$\\sin r = \\dfrac{1.33}{1.0}\\sin 30^{\\circ} = 0.665$, so $r \\approx 41.7^{\\circ} \\approx 42^{\\circ}$. Since $i = 30^{\\circ} < 49^{\\circ}$, refraction occurs — total internal reflection does NOT apply here.","explanationZh":"$\\sin r=\\dfrac{1.33}{1.0}\\sin30^{\\circ}=0.665，故 r\\approx41.7^{\\circ}\\approx42^{\\circ}。因 i=30^{\\circ}<49^{\\circ}$ 发生折射，此处不适用全反射。","tags":["trap","reflection-refraction","snell","no-tir-here"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-total-internal-reflection-001","moduleId":"optics","unitId":"geometric-optics","topicId":"total-internal-reflection","difficulty":1,"stem":"For total internal reflection to occur, light must travel from the optically denser medium to the less dense one. Which setup allows it?","stemZh":"要发生全反射，光必须从光密介质射向光疏介质。下列哪种情形可以发生？","options":[{"id":"A","text":"air $(n = 1.0) →$ glass $(n = 1.5)$"},{"id":"B","text":"glass $(n = 1.5) →$ air $(n = 1.0)$"},{"id":"C","text":"air $→$ water $(n = 1.33)$"},{"id":"D","text":"water $→$ glass"}],"optionsZh":[{"id":"A","text":"$空气（n=1.0）→ 玻璃（n=1.5）$"},{"id":"B","text":"$玻璃（n=1.5）→ 空气（n=1.0）$"},{"id":"C","text":"空气 $→ 水（n=1.33）$"},{"id":"D","text":"$水 →$ 玻璃"}],"answer":"B","explanation":"Total internal reflection requires starting in the higher-index (denser) medium: glass $(1.5)$ to air $(1.0)$, with the incidence angle above the critical angle.","explanationZh":"全反射要求从折射率较大的介质（光密）出发：玻璃（1.5）到空气（1.0），且入射角大于临界角。","tags":["basic","total-internal-reflection","condition"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-total-internal-reflection-002","images":[{"src":"/physics/diagrams/critical-angle.svg","alt":"Critical angle $θc$ at a glass–air $boundary: \\sin θc = n_air / n_glass$; beyond $θc$ the ray is totally reflected.","altZh":"玻璃–空气界面的临界角 θc：\\sin θc = n_空气 / n_玻璃；超过 θc 光线发生全反射。"}],"moduleId":"optics","unitId":"geometric-optics","topicId":"total-internal-reflection","difficulty":1,"stem":"For light going from glass $(n_1 = 1.5)$ to air $(n_2 = 1.0)$, the critical angle satisfies $\\sin\\theta_c = \\dfrac{n_2}{n_1}$. Find $\\theta_c$.","stemZh":"$光从玻璃（n_1=1.5）到空气（n_2=1.0），临界角满足 \\sin\\theta_c=\\dfrac{n_2}{n_1}，求 \\theta_c。$","options":[{"id":"A","text":"$48^{\\circ}$"},{"id":"B","text":"$60^{\\circ}$"},{"id":"C","text":"$42^{\\circ}$"},{"id":"D","text":"$90^{\\circ}$"}],"answer":"C","explanation":"$\\sin\\theta_c = \\dfrac{1.0}{1.5} = 0.667$, so $\\theta_c = \\arcsin(0.667) \\approx 41.8^{\\circ} \\approx 42^{\\circ}$.","explanationZh":"$\\sin\\theta_c=\\dfrac{1.0}{1.5}=0.667，故 \\theta_c=\\arcsin(0.667)\\approx41.8^{\\circ}\\approx42^{\\circ}。$","tags":["basic","total-internal-reflection","critical-angle"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-total-internal-reflection-003","moduleId":"optics","unitId":"geometric-optics","topicId":"total-internal-reflection","difficulty":1,"stem":"At a glass–air boundary, the critical angle is $42^{\\circ}. A$ ray inside the glass strikes the boundary at $i = 30^{\\circ}$. What happens?","stemZh":"在玻璃–空气界面，临界角为 $42^{\\circ}。玻璃内的光线以 i=30^{\\circ}$ 射向界面，会发生什么？","options":[{"id":"A","text":"Total internal reflection"},{"id":"B","text":"No light emerges"},{"id":"C","text":"It reflects completely"},{"id":"D","text":"It refracts into the air"}],"optionsZh":[{"id":"A","text":"全反射"},{"id":"B","text":"没有光射出"},{"id":"C","text":"完全反射"},{"id":"D","text":"折射进入空气"}],"answer":"D","explanation":"Total internal reflection only occurs when $i > \\theta_c$. Here $30^{\\circ} < 42^{\\circ}$, so the ray refracts into the air.","explanationZh":"全反射只在 $i>\\theta_c$ 时发生。此处 $30^{\\circ}<42^{\\circ}，故光线折射进入空气。$","tags":["basic","total-internal-reflection","below-critical"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-total-internal-reflection-004","moduleId":"optics","unitId":"geometric-optics","topicId":"total-internal-reflection","difficulty":2,"stem":"Find the critical angle for light going from water $(n_1 = 1.33)$ to air $(n_2 = 1.0)$.","stemZh":"$求光从水（n_1=1.33）到空气（n_2=1.0）的临界角。$","options":[{"id":"A","text":"$49^{\\circ}$"},{"id":"B","text":"$41^{\\circ}$"},{"id":"C","text":"$60^{\\circ}$"},{"id":"D","text":"$30^{\\circ}$"}],"answer":"A","explanation":"$\\sin\\theta_c = \\dfrac{1.0}{1.33} = 0.752$, so $\\theta_c = \\arcsin(0.752) \\approx 48.8^{\\circ} \\approx 49^{\\circ}$.","explanationZh":"$\\sin\\theta_c=\\dfrac{1.0}{1.33}=0.752，故 \\theta_c=\\arcsin(0.752)\\approx48.8^{\\circ}\\approx49^{\\circ}。$","tags":["basic","total-internal-reflection","critical-angle"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-total-internal-reflection-005","moduleId":"optics","unitId":"geometric-optics","topicId":"total-internal-reflection","difficulty":2,"stem":"At a glass–air boundary the critical angle is $42^{\\circ}. A$ ray inside the glass strikes at $i = 50^{\\circ}$. What happens?","stemZh":"玻璃–空气界面临界角为 $42^{\\circ}。玻璃内光线以 i=50^{\\circ}$ 入射，会发生什么？","options":[{"id":"A","text":"It refracts"},{"id":"B","text":"Total internal reflection occurs"},{"id":"C","text":"It stops"},{"id":"D","text":"It reflects only partially"}],"optionsZh":[{"id":"A","text":"折射"},{"id":"B","text":"发生全反射"},{"id":"C","text":"停止"},{"id":"D","text":"仅部分反射"}],"answer":"B","explanation":"Since $i = 50^{\\circ} > \\theta_c = 42^{\\circ}$, the conditions for total internal reflection are met and the ray is completely reflected.","explanationZh":"$因 i=50^{\\circ}>\\theta_c=42^{\\circ}，满足全反射条件，光线被完全反射。$","tags":["basic","total-internal-reflection","above-critical"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-total-internal-reflection-006","moduleId":"optics","unitId":"geometric-optics","topicId":"total-internal-reflection","difficulty":3,"stem":"Light goes from a medium of $n_1 = 1.4$ to air $(n_2 = 1.0)$. Find the critical angle $\\theta_c$.","stemZh":"光从折射率 $n_1=1.4 的介质到空气（n_2=1.0），求临界角 \\theta_c。$","options":[{"id":"A","text":"$44^{\\circ}$"},{"id":"B","text":"$60^{\\circ}$"},{"id":"C","text":"$46^{\\circ}$"},{"id":"D","text":"$35^{\\circ}$"}],"answer":"C","explanation":"$\\sin\\theta_c = \\dfrac{1.0}{1.4} = 0.714$, so $\\theta_c = \\arcsin(0.714) \\approx 45.6^{\\circ} \\approx 46^{\\circ}$.","explanationZh":"$\\sin\\theta_c=\\dfrac{1.0}{1.4}=0.714，故 \\theta_c=\\arcsin(0.714)\\approx45.6^{\\circ}\\approx46^{\\circ}。$","tags":["applied","total-internal-reflection","critical-angle"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-total-internal-reflection-007","moduleId":"optics","unitId":"geometric-optics","topicId":"total-internal-reflection","difficulty":3,"stem":"An optical fibre guides light by repeated total internal reflection at the core–cladding boundary. Which statement is correct?","stemZh":"光纤利用纤芯–包层界面上的多次全反射来传导光。下列说法正确的是？","options":[{"id":"A","text":"Light escapes at every bounce"},{"id":"B","text":"The cladding has a higher n than the core"},{"id":"C","text":"TIR requires going from less dense to denser"},{"id":"D","text":"Light is trapped by repeated TIR because the core is denser than the cladding"}],"optionsZh":[{"id":"A","text":"光在每次反射时都会逸出"},{"id":"B","text":"包层的折射率比纤芯高"},{"id":"C","text":"全反射要求从光疏介质射向光密介质"},{"id":"D","text":"由于纤芯比包层更光密，光被多次全反射约束在纤芯内"}],"answer":"D","explanation":"TIR needs the ray to start in the denser medium $(core$, higher n) striking the less dense cladding (lower n) above the critical angle, so light stays trapped in the core.","explanationZh":"全反射要求光线从光密介质（纤芯，较大 n）射向光疏介质（包层，较小 n）且入射角大于临界角，因此光被约束在纤芯内。","tags":["applied","total-internal-reflection","optical-fibre"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-total-internal-reflection-008","moduleId":"optics","unitId":"geometric-optics","topicId":"total-internal-reflection","difficulty":3,"stem":"In a fibre, light in the core $(n_2 = 1.5,\\text{ the }cladding)$ has a critical angle of $60^{\\circ}$ with the cladding. Find the core's refractive index $n_1 (n_2 = 1.5)$.","stemZh":"某光纤中，纤芯到包层的临界角为 $60^{\\circ}（包层 n_2=1.5），求纤芯折射率 n_1。$","options":[{"id":"A","text":"1.73"},{"id":"B","text":"1.50"},{"id":"C","text":"1.30"},{"id":"D","text":"2.00"}],"answer":"A","explanation":"$\\sin\\theta_c = \\dfrac{n_2}{n_1} \\Rightarrow n_1 = \\dfrac{n_2}{\\sin\\theta_c} = \\dfrac{1.5}{\\sin 60^{\\circ}} = \\dfrac{1.5}{0.866} \\approx 1.73$.","explanationZh":"$\\sin\\theta_c=\\dfrac{n_2}{n_1}\\Rightarrow n_1=\\dfrac{n_2}{\\sin\\theta_c}=\\dfrac{1.5}{\\sin60^{\\circ}}=\\dfrac{1.5}{0.866}\\approx1.73。$","tags":["applied","total-internal-reflection","critical-angle","fibre"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-total-internal-reflection-009","moduleId":"optics","unitId":"geometric-optics","topicId":"total-internal-reflection","difficulty":3,"stem":"For glass $(n_1 = 1.5)$ to air $(n_2 = 1.0)$, the critical angle is $42^{\\circ}$. When does total internal reflection occur?","stemZh":"$玻璃（n_1=1.5）到空气（n_2=1.0）的临界角为 42^{\\circ}。何时发生全反射？$","options":[{"id":"A","text":"$i < 42^{\\circ}$"},{"id":"B","text":"$i > 42^{\\circ}$"},{"id":"C","text":"$i = 42^{\\circ}$ only"},{"id":"D","text":"For all angles"}],"answer":"B","explanation":"TIR occurs only when the incidence angle exceeds the critical $angle: i > \\theta_c = 42^{\\circ}$.","explanationZh":"全反射只在入射角大于临界角时发生，即 $i>\\theta_c=42^{\\circ}。$","tags":["applied","total-internal-reflection","condition"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-total-internal-reflection-010","moduleId":"optics","unitId":"geometric-optics","topicId":"total-internal-reflection","difficulty":3,"stem":"Diamond has $n = 2.42$. Find its critical angle for light going to air $(n = 1.0)$.","stemZh":"钻石折射率 $n=2.42，求光从钻石到空气（n=1.0）的临界角。$","options":[{"id":"A","text":"$48^{\\circ}$"},{"id":"B","text":"$90^{\\circ}$"},{"id":"C","text":"$24^{\\circ}$"},{"id":"D","text":"$66^{\\circ}$"}],"answer":"C","explanation":"$\\sin\\theta_c = \\dfrac{1.0}{2.42} = 0.413$, so $\\theta_c = \\arcsin(0.413) \\approx 24.4^{\\circ} \\approx 24^{\\circ} —$ the small angle gives diamond its sparkle.","explanationZh":"$\\sin\\theta_c=\\dfrac{1.0}{2.42}=0.413，故 \\theta_c=\\arcsin(0.413)\\approx24.4^{\\circ}\\approx24^{\\circ}；小的临界角使钻石更闪耀。$","tags":["applied","total-internal-reflection","critical-angle","diamond"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-total-internal-reflection-011","images":[{"src":"/physics/diagrams/critical-angle.svg","alt":"At a glass–air boundary, TIR only happens for rays starting in the glass with $i > θc; air→glass$ cannot give TIR.","altZh":"玻璃–空气界面，只有从玻璃出发且 i > θc 的光线才会全反射；空气→玻璃不可能发生全反射。"}],"moduleId":"optics","unitId":"geometric-optics","topicId":"total-internal-reflection","difficulty":4,"stem":"A student says total internal reflection happens when light goes from air into water, 'because water has a larger refractive index.' What is wrong?","stemZh":"有学生说，当光从空气进入水中时会全反射，“因为水的折射率更大”。错在哪里？","options":[{"id":"A","text":"TIR has nothing to do with refractive index"},{"id":"B","text":"Water has the lower refractive index"},{"id":"C","text":"TIR needs the ray to start in the lower-n medium"},{"id":"D","text":"TIR needs the ray to start in the higher-n (denser) medium with $i > \\theta_c$"}],"optionsZh":[{"id":"A","text":"全反射与折射率无关"},{"id":"B","text":"水的折射率较小"},{"id":"C","text":"全反射要求光线从较小 n 的介质出发"},{"id":"D","text":"全反射要求光线从较大 n（光密）介质出发，且 $i>\\theta_c$"}],"answer":"D","explanation":"$Air→water$ goes from the LESS dense to the MORE dense medium, so TIR cannot occur. TIR requires starting in the denser medium (higher n) with incidence above the critical angle.","explanationZh":"$空气→水是从光疏到光密，不可能全反射。全反射要求从光密（较大$ n）介质出发且入射角大于临界角。","tags":["trap","total-internal-reflection","condition","direction"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-total-internal-reflection-012","moduleId":"optics","unitId":"geometric-optics","topicId":"total-internal-reflection","difficulty":4,"stem":"A student computes the critical angle as $\\theta_c = \\arcsin\\!\\left(\\dfrac{n_1}{n_2}\\right)$ with $n_1 > n_2$, getting $\\arcsin(1.5)$, which is impossible. What is the correct formula?","stemZh":"某学生把临界角算成 $\\theta_c=\\arcsin\\!\\left(\\dfrac{n_1}{n_2}\\right)（n_1>n_2），得到 \\arcsin(1.5)$ 无意义。正确公式是什么？","options":[{"id":"A","text":"$\\sin\\theta_c = \\dfrac{n_2}{n_1}$ with $n_1 > n_2$"},{"id":"B","text":"$\\sin\\theta_c = \\dfrac{n_1}{n_2}$"},{"id":"C","text":"$\\theta_c = \\dfrac{n_1 - n_2}{n_1}$"},{"id":"D","text":"$\\sin\\theta_c = n_1 n_2$"}],"answer":"A","explanation":"The ratio must be less than 1, so $\\sin\\theta_c = \\dfrac{n_2}{n_1}$ where the incident (denser) medium is $n_1$ and the other is $n_2 < n_1$.","explanationZh":"比值必须小于 $1，故 \\sin\\theta_c=\\dfrac{n_2}{n_1}，其中入射（光密）介质为 n_1，另一介质 n_2 \\theta_c$"},{"id":"C","text":"No light refracts at $i = \\theta_c$"},{"id":"D","text":"The ray is absorbed"}],"optionsZh":[{"id":"A","text":"仅在 $i=\\theta_c$ 时发生全反射"},{"id":"B","text":"$i<\\theta_c 时折射；i>\\theta_c$ 时才全反射"},{"id":"C","text":"$i=\\theta_c$ 时无光折射"},{"id":"D","text":"光线被吸收"}],"answer":"B","explanation":"TIR begins only when i exceeds $\\theta_c$. At and below $\\theta_c$ there is still a refracted ray $(at \\theta_c\\text{ it skims the }surface)$.","explanationZh":"全反射仅在 i 超过 $\\theta_c$ 后才开始。在 $\\theta_c 及以下仍有折射光线（\\theta_c$ 时沿界面掠射）。","tags":["trap","total-internal-reflection","threshold"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-total-internal-reflection-014","moduleId":"optics","unitId":"geometric-optics","topicId":"total-internal-reflection","difficulty":5,"stem":"A fibre has core $n_1 = 1.6$ and cladding $n_2 = 1.4$. At the core–cladding boundary, with incidence $i = 50^{\\circ}$, what happens?","stemZh":"某光纤纤芯 $n_1=1.6、包层 n_2=1.4。在纤芯–包层界面上，入射角 i=50^{\\circ}$ 时会发生什么？","options":[{"id":"A","text":"TIR occurs"},{"id":"B","text":"Light is fully absorbed"},{"id":"C","text":"Light refracts into the cladding (no TIR)"},{"id":"D","text":"The ray travels along the boundary"}],"optionsZh":[{"id":"A","text":"发生全反射"},{"id":"B","text":"光被完全吸收"},{"id":"C","text":"折射进入包层（未发生全反射）"},{"id":"D","text":"光线沿界面传播"}],"answer":"C","explanation":"Critical $angle: \\sin\\theta_c = \\dfrac{1.4}{1.6} = 0.875$, so $\\theta_c \\approx 61^{\\circ}$. Since $i = 50^{\\circ} < 61^{\\circ}$, TIR does NOT occur; the ray refracts into the cladding.","explanationZh":"$临界角：\\sin\\theta_c=\\dfrac{1.4}{1.6}=0.875，故 \\theta_c\\approx61^{\\circ}。因 i=50^{\\circ}<61^{\\circ}，不发生全反射，光线折射进入包层。$","tags":["trap","total-internal-reflection","fibre","angle-check"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-total-internal-reflection-015","moduleId":"optics","unitId":"geometric-optics","topicId":"total-internal-reflection","difficulty":5,"stem":"A glass block $(n_1 = 1.5)$ is immersed in water $(n_2 = 1.33)$. Can total internal reflection occur at the glass–water boundary, and if so when?","stemZh":"$玻璃块（n_1=1.5）浸在水中（n_2=1.33）。在玻璃–水界面能否发生全反射？若能，何时发生？$","options":[{"id":"A","text":"No, glass and water are too similar"},{"id":"B","text":"Only at $i = 0^{\\circ}$"},{"id":"C","text":"Yes, but only if $i > 90^{\\circ}$"},{"id":"D","text":"Yes, for $i > 62.5^{\\circ}$"}],"optionsZh":[{"id":"A","text":"不能，玻璃和水太接近"},{"id":"B","text":"仅在 $i=0^{\\circ} 时$"},{"id":"C","text":"能，但需 $i>90^{\\circ}$"},{"id":"D","text":"能，当 $i>62.5^{\\circ} 时$"}],"answer":"D","explanation":"Glass is denser than water $(1.5 > 1.33)$, so TIR is possible. $\\sin\\theta_c = \\dfrac{1.33}{1.5} = 0.887, \\theta_c \\approx 62.5^{\\circ}$; TIR occurs for $i > 62.5^{\\circ} (not\\text{ only at an air }boundary)$.","explanationZh":"$玻璃比水光密（1.5>1.33），故可发生全反射。\\sin\\theta_c=\\dfrac{1.33}{1.5}=0.887，\\theta_c\\approx62.5^{\\circ}；当 i>62.5^{\\circ}$ 时发生全反射（并非只发生在空气界面）。","tags":["trap","total-internal-reflection","condition","non-air"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-modern-overview-001","moduleId":"modern-physics","unitId":"modern-overview","topicId":"modern-overview","difficulty":1,"stem":"The energy of a photon is $E =$ hf with $h = 6.63 \\times 10^{-34}\\ \\mathrm{J\\cdot s}$. For frequency $f = 6.0 \\times 10^{14}\\ \\mathrm{Hz}$, what is E?","stemZh":"光子能量 $E=hf，取 h=6.63\\times10^{-34}\\ \\mathrm{J\\cdot s}。频率 f=6.0\\times10^{14}\\ \\mathrm{Hz}$ 时，E 为多少？","options":[{"id":"A","text":"$1.1 \\times 10^{-19}\\ \\mathrm{J}$"},{"id":"B","text":"$6.0 \\times 10^{-34}\\ \\mathrm{J}$"},{"id":"C","text":"$3.0 \\times 10^{-8}\\ \\mathrm{J}$"},{"id":"D","text":"$4.0 \\times 10^{-19}\\ \\mathrm{J}$"}],"answer":"D","explanation":"$E =$ hf $= 6.63 \\times 10^{-34} \\times 6.0 \\times 10^{14} = 3.98 \\times 10^{-19} \\approx 4.0 \\times 10^{-19}\\ \\mathrm{J}$.","explanationZh":"$E=hf=6.63\\times10^{-34}\\times6.0\\times10^{14}=3.98\\times10^{-19}\\approx4.0\\times10^{-19}\\ \\mathrm{J}。$","tags":["basic","modern-overview","photon-energy"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-modern-overview-002","moduleId":"modern-physics","unitId":"modern-overview","topicId":"modern-overview","difficulty":1,"stem":"A photon has wavelength $\\lambda = 500\\ \\mathrm{nm}$. Using $E = \\dfrac{hc}{\\lambda}$ with $h = 6.63 \\times 10^{-34}\\ \\mathrm{J\\cdot s}, c = 3.0 \\times 10^{8}\\ \\mathrm{m/s}$, find E.","stemZh":"光子波长 $\\lambda=500\\ \\mathrm{nm}。用 E=\\dfrac{hc}{\\lambda}（h=6.63\\times10^{-34}\\ \\mathrm{J\\cdot s}，c=3.0\\times10^{8}\\ \\mathrm{m/s}）求$ E。","options":[{"id":"A","text":"$4.0 \\times 10^{-19}\\ \\mathrm{J}$"},{"id":"B","text":"$1.0 \\times 10^{-19}\\ \\mathrm{J}$"},{"id":"C","text":"$1.2 \\times 10^{-18}\\ \\mathrm{J}$"},{"id":"D","text":"$3.0 \\times 10^{-8}\\ \\mathrm{J}$"}],"answer":"A","explanation":"$E = \\dfrac{hc}{\\lambda} = \\dfrac{6.63 \\times 10^{-34} \\times 3.0 \\times 10^{8}}{5.0 \\times 10^{-7}} = 3.98 \\times 10^{-19} \\approx 4.0 \\times 10^{-19}\\ \\mathrm{J}$.","explanationZh":"$E=\\dfrac{hc}{\\lambda}=\\dfrac{6.63\\times10^{-34}\\times3.0\\times10^{8}}{5.0\\times10^{-7}}=3.98\\times10^{-19}\\approx4.0\\times10^{-19}\\ \\mathrm{J}。$","tags":["basic","modern-overview","photon-energy","wavelength"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-modern-overview-003","images":[{"src":"/physics/diagrams/photoelectric.svg","alt":"Photoelectric effect: a photon of energy hf strikes a metal; if hf exceeds the work function W, an electron is emitted with kinetic energy Ek $=$ hf - W.","altZh":"光电效应：能量为 hf 的光子撞击金属；若 hf 大于逸出功 W，电子以动能 Ek = hf - W 逸出。"}],"moduleId":"modern-physics","unitId":"modern-overview","topicId":"modern-overview","difficulty":1,"stem":"In the photoelectric effect, electrons are emitted from a metal only when:","stemZh":"在光电效应中，金属发射电子的条件是：","options":[{"id":"A","text":"the light intensity is high enough"},{"id":"B","text":"the photon frequency exceeds the threshold frequency $f_0$"},{"id":"C","text":"the wavelength is long enough"},{"id":"D","text":"the metal is heated"}],"optionsZh":[{"id":"A","text":"光强足够大"},{"id":"B","text":"光子频率超过截止频率 $f_0$"},{"id":"C","text":"波长足够长"},{"id":"D","text":"金属被加热"}],"answer":"B","explanation":"Emission needs one photon's energy hf to exceed the work $function: f > f_0$. Intensity alone cannot cause emission below threshold.","explanationZh":"逸出要求单个光子能量 hf 大于逸出功，即 $f>f_0。低于截止频率时，仅靠增大光强不能产生光电子。$","tags":["basic","modern-overview","photoelectric","threshold"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-modern-overview-004","moduleId":"modern-physics","unitId":"modern-overview","topicId":"modern-overview","difficulty":2,"stem":"Photoelectric equation: hf $= W + E_k. A$ photon of energy hf $= 3.0\\ \\mathrm{eV}$ strikes a metal with work function $W = 2.0\\ \\mathrm{eV}$. What is the maximum kinetic energy $E_k$ of the emitted electron?","stemZh":"$光电方程：hf=W+E_k。能量 hf=3.0\\ \\mathrm{eV}$ 的光子照射逸出功 $W=2.0\\ \\mathrm{eV}$ 的金属，光电子最大动能 $E_k$ 为多少？","options":[{"id":"A","text":"$5.0\\ \\mathrm{eV}$"},{"id":"B","text":"$0\\ \\mathrm{eV}$"},{"id":"C","text":"$1.0\\ \\mathrm{eV}$"},{"id":"D","text":"$2.0\\ \\mathrm{eV}$"}],"answer":"C","explanation":"$E_k =$ hf $- W = 3.0 - 2.0 = 1.0\\ \\mathrm{eV}$.","explanationZh":"$E_k=hf-W=3.0-2.0=1.0\\ \\mathrm{eV}。$","tags":["basic","modern-overview","photoelectric","kinetic-energy"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-modern-overview-005","moduleId":"modern-physics","unitId":"modern-overview","topicId":"modern-overview","difficulty":2,"stem":"The de Broglie wavelength is $\\lambda = \\dfrac{h}{p}$. An electron has momentum $p = 3.3 \\times 10^{-24}\\ \\mathrm{kg\\cdot m/s} (h = 6.63 \\times 10^{-34}\\ \\mathrm{J\\cdot s})$. Find $\\lambda$.","stemZh":"德布罗意波长 $\\lambda=\\dfrac{h}{p}。电子动量 p=3.3\\times10^{-24}\\ \\mathrm{kg\\cdot m/s}（h=6.63\\times10^{-34}\\ \\mathrm{J\\cdot s}），求 \\lambda。$","options":[{"id":"A","text":"$2.0 \\times 10^{-9}\\ \\mathrm{m}$"},{"id":"B","text":"$2.2 \\times 10^{-58}\\ \\mathrm{m}$"},{"id":"C","text":"$2.0 \\times 10^{-8}\\ \\mathrm{m}$"},{"id":"D","text":"$2.0 \\times 10^{-10}\\ \\mathrm{m}$"}],"answer":"D","explanation":"$\\lambda = \\dfrac{h}{p} = \\dfrac{6.63 \\times 10^{-34}}{3.3 \\times 10^{-24}} = 2.0 \\times 10^{-10}\\ \\mathrm{m}$.","explanationZh":"$\\lambda=\\dfrac{h}{p}=\\dfrac{6.63\\times10^{-34}}{3.3\\times10^{-24}}=2.0\\times10^{-10}\\ \\mathrm{m}。$","tags":["basic","modern-overview","de-broglie"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-modern-overview-006","moduleId":"modern-physics","unitId":"modern-overview","topicId":"modern-overview","difficulty":3,"stem":"Light has wavelength $\\lambda = 600\\ \\mathrm{nm} (c = 3.0 \\times 10^{8}\\ \\mathrm{m/s})$. What is its frequency f?","stemZh":"光波长 $\\lambda=600\\ \\mathrm{nm}（c=3.0\\times10^{8}\\ \\mathrm{m/s}），其频率 f$ 是多少？","options":[{"id":"A","text":"$5.0 \\times 10^{14}\\ \\mathrm{Hz}$"},{"id":"B","text":"$2.0 \\times 10^{14}\\ \\mathrm{Hz}$"},{"id":"C","text":"$5.0 \\times 10^{15}\\ \\mathrm{Hz}$"},{"id":"D","text":"$1.8 \\times 10^{6}\\ \\mathrm{Hz}$"}],"answer":"A","explanation":"$f = \\dfrac{c}{\\lambda} = \\dfrac{3.0 \\times 10^{8}}{6.0 \\times 10^{-7}} = 5.0 \\times 10^{14}\\ \\mathrm{Hz}$.","explanationZh":"$f=\\dfrac{c}{\\lambda}=\\dfrac{3.0\\times10^{8}}{6.0\\times10^{-7}}=5.0\\times10^{14}\\ \\mathrm{Hz}。$","tags":["applied","modern-overview","frequency"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-modern-overview-007","moduleId":"modern-physics","unitId":"modern-overview","topicId":"modern-overview","difficulty":3,"stem":"A photon has $\\lambda = 400\\ \\mathrm{nm}$. Using hc $= 1.99 \\times 10^{-25}\\ \\mathrm{J\\cdot m}$ and $1\\ \\mathrm{eV} = 1.6 \\times 10^{-19}\\ \\mathrm{J}$, what is its energy in eV?","stemZh":"光子波长 $\\lambda=400\\ \\mathrm{nm}。取 hc=1.99\\times10^{-25}\\ \\mathrm{J\\cdot m}、1\\ \\mathrm{eV}=1.6\\times10^{-19}\\ \\mathrm{J}，其能量（以$ eV 计）是多少？","options":[{"id":"A","text":"$2.0\\ \\mathrm{eV}$"},{"id":"B","text":"$3.1\\ \\mathrm{eV}$"},{"id":"C","text":"$4.0\\ \\mathrm{eV}$"},{"id":"D","text":"$1.5\\ \\mathrm{eV}$"}],"answer":"B","explanation":"$E = \\dfrac{hc}{\\lambda} = \\dfrac{1.99 \\times 10^{-25}}{4.0 \\times 10^{-7}} = 4.98 \\times 10^{-19}\\ \\mathrm{J}$; in $eV, E = \\dfrac{4.98 \\times 10^{-19}}{1.6 \\times 10^{-19}} \\approx 3.1\\ \\mathrm{eV}$.","explanationZh":"$E=\\dfrac{hc}{\\lambda}=\\dfrac{1.99\\times10^{-25}}{4.0\\times10^{-7}}=4.98\\times10^{-19}\\ \\mathrm{J}；换算为 eV：E=\\dfrac{4.98\\times10^{-19}}{1.6\\times10^{-19}}\\approx3.1\\ \\mathrm{eV}。$","tags":["applied","modern-overview","photon-energy","ev"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-modern-overview-008","moduleId":"modern-physics","unitId":"modern-overview","topicId":"modern-overview","difficulty":3,"stem":"Einstein's mass–energy relation is $E = mc^{2}$ with $c = 3.0 \\times 10^{8}\\ \\mathrm{m/s}$. If a mass $\\Delta m = 1.0 \\times 10^{-6}\\ \\mathrm{kg}$ is converted to energy, what is E?","stemZh":"爱因斯坦质能方程 $E=mc^{2}（c=3.0\\times10^{8}\\ \\mathrm{m/s}）。若质量 \\Delta m=1.0\\times10^{-6}\\ \\mathrm{kg}$ 转化为能量，E 为多少？","options":[{"id":"A","text":"$3.0 \\times 10^{2}\\ \\mathrm{J}$"},{"id":"B","text":"$9.0 \\times 10^{8}\\ \\mathrm{J}$"},{"id":"C","text":"$9.0 \\times 10^{10}\\ \\mathrm{J}$"},{"id":"D","text":"$3.0 \\times 10^{16}\\ \\mathrm{J}$"}],"answer":"C","explanation":"$E = \\Delta m\\, c^{2} = 1.0 \\times 10^{-6} \\times (3.0 \\times 10^{8})^{2} = 1.0 \\times 10^{-6} \\times 9.0 \\times 10^{16} = 9.0 \\times 10^{10}\\ \\mathrm{J}$.","explanationZh":"$E=\\Delta m\\,c^{2}=1.0\\times10^{-6}\\times(3.0\\times10^{8})^{2}=1.0\\times10^{-6}\\times9.0\\times10^{16}=9.0\\times10^{10}\\ \\mathrm{J}。$","tags":["applied","modern-overview","mass-energy"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-modern-overview-009","images":[{"src":"/physics/diagrams/photoelectric.svg","alt":"Photoelectric effect: with hf above the work function W, the emitted electron's max kinetic energy is Ek $=$ hf - W.","altZh":"光电效应：当 hf 大于逸出功 W 时，光电子最大动能为 Ek = hf - W。"}],"moduleId":"modern-physics","unitId":"modern-overview","topicId":"modern-overview","difficulty":3,"stem":"A photon of energy $4.5\\ \\mathrm{eV}$ strikes a metal with work function $W = 2.5\\ \\mathrm{eV}$. What is the maximum kinetic energy of the emitted electron?","stemZh":"能量 $4.5\\ \\mathrm{eV}$ 的光子照射逸出功 $W=2.5\\ \\mathrm{eV}$ 的金属，光电子最大动能是多少？","options":[{"id":"A","text":"$4.5\\ \\mathrm{eV}$"},{"id":"B","text":"$7.0\\ \\mathrm{eV}$"},{"id":"C","text":"$2.5\\ \\mathrm{eV}$"},{"id":"D","text":"$2.0\\ \\mathrm{eV}$"}],"answer":"D","explanation":"$E_k =$ hf $- W = 4.5 - 2.5 = 2.0\\ \\mathrm{eV}$.","explanationZh":"$E_k=hf-W=4.5-2.5=2.0\\ \\mathrm{eV}。$","tags":["applied","modern-overview","photoelectric","kinetic-energy"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-modern-overview-010","moduleId":"modern-physics","unitId":"modern-overview","topicId":"modern-overview","difficulty":3,"stem":"The threshold frequency is $f_0 = \\dfrac{W}{h}$. For a work function $W = 3.3 \\times 10^{-19}\\ \\mathrm{J}$ and $h = 6.63 \\times 10^{-34}\\ \\mathrm{J\\cdot s}$, find $f_0$.","stemZh":"截止频率 $f_0=\\dfrac{W}{h}。逸出功 W=3.3\\times10^{-19}\\ \\mathrm{J}，h=6.63\\times10^{-34}\\ \\mathrm{J\\cdot s}，求 f_0。$","options":[{"id":"A","text":"$5.0 \\times 10^{14}\\ \\mathrm{Hz}$"},{"id":"B","text":"$2.0 \\times 10^{14}\\ \\mathrm{Hz}$"},{"id":"C","text":"$3.3 \\times 10^{-19}\\ \\mathrm{Hz}$"},{"id":"D","text":"$1.0 \\times 10^{15}\\ \\mathrm{Hz}$"}],"answer":"A","explanation":"$f_0 = \\dfrac{W}{h} = \\dfrac{3.3 \\times 10^{-19}}{6.63 \\times 10^{-34}} = 4.98 \\times 10^{14} \\approx 5.0 \\times 10^{14}\\ \\mathrm{Hz}$.","explanationZh":"$f_0=\\dfrac{W}{h}=\\dfrac{3.3\\times10^{-19}}{6.63\\times10^{-34}}=4.98\\times10^{14}\\approx5.0\\times10^{14}\\ \\mathrm{Hz}。$","tags":["applied","modern-overview","threshold-frequency"]},{"subject":"physics","type":"single-choice","status":"published","id":"physics-modern-overview-011","moduleId":"modern-physics","unitId":"modern-overview","topicId":"modern-overview","difficulty":4,"stem":"Statement: 'Increasing the light intensity on a metal that is below its threshold frequency will eventually eject electrons.' Is this true?","stemZh":"命题：“对低于截止频率的金属增大光强，最终也能打出电子。” 这正确吗？","options":[{"id":"A","text":"True, more photons means more total energy"},{"id":"B","text":"False — each photon must exceed $hf_0$; intensity alone cannot"},{"id":"C","text":"True only for low intensity"},{"id":"D","text":"False, it ejects electrons but very slowly"}],"optionsZh":[{"id":"A","text":"正确，光子越多总能量越大"},{"id":"B","text":"错误——每个光子能量必须超过 $hf_0，仅增大光强不行$"},{"id":"C","text":"仅在小光强时正确"},{"id":"D","text":"错误，但会打出电子且很慢"}],"answer":"B","explanation":"The photoelectric effect is one-photon-one-electron. Below threshold no single photon has enough energy $(hf < W)$, no matter how many photons arrive. Intensity does not help.","explanationZh":"$光电效应是一个光子打出一个电子。低于截止频率时，单个光子能量不足（hf

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