[{"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-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"}]

Functions & Introductory Calculus