[{"id":"math-sets-rep-001","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":1,"stem":"Which notation represents the set {2, 4, 6, 8}?","options":[{"id":"A","text":"${x | x < 8}$"},{"id":"B","text":"${x ∈ ℕ⁺ | x\\text{ is even and }x ≤ 8}$"},{"id":"C","text":"$(2, 8)$"},{"id":"D","text":"$[2, 8]$"}],"answer":"B","explanation":"It lists the positive even integers up to 8.","knowledgePoint":"set","tags":["set","notation"],"status":"published"},{"id":"math-sets-rel-001","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":1,"stem":"If $A = {1, 2}$ and $B = {1, 2, 3}$, which is true?","options":[{"id":"A","text":"$A ⊂ B$"},{"id":"B","text":"$B ⊂ A$"},{"id":"C","text":"$A = B$"},{"id":"D","text":"$A ∩ B = ∅$"}],"answer":"A","explanation":"Every element of A is in B, and B has an extra element, so A is a proper subset of B.","knowledgePoint":"subset","tags":["set","subset"],"status":"published"},{"id":"math-sets-op-001","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":2,"stem":"If $A = {1, 2, 3}$ and $B = {3, 4}$, then $|A ∪ B| =$ ?","options":[{"id":"A","text":"3"},{"id":"B","text":"5"},{"id":"C","text":"4"},{"id":"D","text":"6"}],"answer":"C","explanation":"$A ∪ B = {1, 2, 3, 4}$, which has 4 elements.","knowledgePoint":"union","tags":["set","union"],"status":"published"},{"id":"math-ineq-quad-001","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":2,"stem":"Solve $x^{2} − 4 < 0$.","options":[{"id":"A","text":"$x < 2$"},{"id":"B","text":"$−2 < x < 2$"},{"id":"C","text":"$x > 2$ or $x < −2$"},{"id":"D","text":"$x ≠ 2$"}],"answer":"B","explanation":"$(x − 2)(x + 2) < 0$, so x lies strictly between −2 and 2.","tags":["inequality","quadratic"],"status":"published"},{"id":"math-ineq-quad-002","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"The solution set of $x^{2} − 5x + 6 ≤ 0$ is the interval $[a, b]$ with $a <$ b. Find a + b.","options":[{"id":"A","text":"1"},{"id":"B","text":"5"},{"id":"C","text":"6"},{"id":"D","text":"−5"}],"answer":"B","explanation":"$(x − 2)(x − 3) ≤ 0$ gives $[2, 3]$, so $a + b = 5$.","tags":["inequality","quadratic"],"status":"published"},{"id":"math-ineq-rational-001","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $1 / (x − 1) > 0$.","options":[{"id":"A","text":"$x > 1$"},{"id":"B","text":"$x < 1$"},{"id":"C","text":"$x ≠ 1$"},{"id":"D","text":"$x > 0$"}],"answer":"A","explanation":"A positive fraction needs a positive denominator, so $x − 1 > 0$.","tags":["inequality","rational"],"status":"published"},{"id":"math-ineq-abs-001","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":2,"stem":"Solve $|x| < 3$.","options":[{"id":"A","text":"$−3 < x < 3$"},{"id":"B","text":"$x < 3$"},{"id":"C","text":"$x > −3$"},{"id":"D","text":"$x ≤ 3$"}],"answer":"A","explanation":"$|x| < 3$ means x is within 3 units of 0.","tags":["inequality","absolute"],"status":"published"},{"id":"math-ineq-abs-002","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"The solution set of $|x − 2| ≤ 5$ is $[a, 7]$. Find a.","options":[{"id":"A","text":"−3"},{"id":"B","text":"3"},{"id":"C","text":"−7"},{"id":"D","text":"7"}],"answer":"A","explanation":"$−5 ≤ x − 2 ≤ 5$ gives $−3 ≤ x ≤ 7$, so $a = −3$.","tags":["inequality","absolute"],"status":"published"},{"id":"math-set-representation-001","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":1,"stem":"Which of the following lists the set of odd positive integers less than 8?","stemZh":"下列哪个选项用列举法表示了小于8的正奇数集？","options":[{"id":"A","text":"{1, 3, 5, 7}"},{"id":"B","text":"{1, 3, 5, 7, 9}"},{"id":"C","text":"{0, 2, 4, 6}"},{"id":"D","text":"{2, 4, 6}"}],"answer":"A","explanation":"The odd positive integers below 8 are 1, 3, 5 and 7. 9 is not less than 8, and the even numbers do not qualify.","explanationZh":"小于8的正奇数为1、3、5、7。9不小于8，偶数也不符合要求。","tags":["easy","set","listing"],"status":"published"},{"id":"math-set-representation-002","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":1,"stem":"Write the set of real numbers x with $−1 < x ≤ 3$ as an interval.","stemZh":"用区间表示满足 $−1 < x ≤ 3$ 的实数集合。","options":[{"id":"A","text":"$(−1, 3)$"},{"id":"B","text":"$[−1, 3]$"},{"id":"C","text":"$(−1, 3]$"},{"id":"D","text":"$[−1, 3)$"}],"answer":"C","explanation":"The strict inequality $−1 < x$ excludes −1, so the left end is $open; x ≤ 3$ includes 3, so the right end is closed. Hence $(−1, 3]$.","explanationZh":"$−1 < x$ 不含 $−1，左端为开；x ≤ 3 含 3，右端为闭。故为 (−1, 3]。$","tags":["easy","set","interval"],"status":"published"},{"id":"math-set-representation-003","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":1,"stem":"Which symbol is used to say that an object is an element of a set?","stemZh":"表示“某个对象是一个集合的元素”使用哪个符号？","options":[{"id":"A","text":"$⊆$"},{"id":"B","text":"$∈$"},{"id":"C","text":"$∪$"},{"id":"D","text":"$∩$"}],"answer":"B","explanation":"$∈$ relates an element to a set. $⊆$ relates a set to another $set, ∪$ is union and $∩$ is intersection.","explanationZh":"$∈ 表示元素与集合的关系；⊆ 表示集合与集合的包含关系，∪ 是并集，∩$ 是交集。","tags":["easy","set","notation"],"status":"published"},{"id":"math-set-representation-004","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":2,"stem":"Which description represents the set {2, 4, 6, 8}?","stemZh":"下列哪个描述法表示了集合 {2, 4, 6, 8}？","options":[{"id":"A","text":"${x | x\\text{ is }even}$"},{"id":"B","text":"${x | x < 9}$"},{"id":"C","text":"${x | x = 2n, n ∈ ℝ}$"},{"id":"D","text":"${x | x = 2n, n ∈ ℕ⁺, n ≤ 4}$"}],"answer":"D","explanation":"The set contains exactly 2, 4, 6, 8, which is 2n for the positive integers $n = 1, 2, 3, 4$. Option A is all even numbers and option C is all real numbers of the form 2n.","explanationZh":"$集合恰好含2、4、6、8，即 n = 1、2、3、4$ 时的 2n。A 是全体偶数，C 是形如 2n 的一切实数。","tags":["easy","set","set-builder"],"status":"published"},{"id":"math-set-representation-005","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":2,"stem":"Which of the following is NOT the empty set?","stemZh":"下列哪个集合不是空集？","options":[{"id":"A","text":"${x ∈ ℝ | x^{2} + 1 = 0}$"},{"id":"B","text":"${x ∈ ℝ | x > 0\\text{ and }x < 0}$"},{"id":"C","text":"{0}"},{"id":"D","text":"$∅$"}],"answer":"C","explanation":"{0} contains one element, namely the number 0, so it is not empty. $x^{2} + 1 = 0$ has no real solution, and no number is both positive and negative, so A, B and D are all empty.","explanationZh":"{0} 含有一个元素 $0，因此不是空集。x^{2} + 1 = 0 无实数解，也没有数既大于0又小于0，故$ A、B、D 都是空集。","tags":["easy","set","empty-set"],"status":"published"},{"id":"math-set-representation-006","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":2,"stem":"List the set ${x ∈ ℤ | −2 ≤ x < 2}$.","stemZh":"用列举法表示集合 ${x ∈ ℤ | −2 ≤ x < 2}。$","options":[{"id":"A","text":"{−2, −1, 0, 1, 2}"},{"id":"B","text":"{−2, −1, 0, 1}"},{"id":"C","text":"{−1, 0, 1}"},{"id":"D","text":"{−2, −1, 1}"}],"answer":"B","explanation":"$−2 ≤ x$ includes −2, while $x < 2$ excludes 2, so the integers are −2, −1, 0, 1.","explanationZh":"$−2 ≤ x 含 −2，而 x < 2$ 不含 2，故整数为 −2、−1、0、1。","tags":["easy","set","listing"],"status":"published"},{"id":"math-set-representation-007","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":1,"stem":"Write the solution set of $x ≥ 5$ as an interval.","stemZh":"用区间表示 $x ≥ 5$ 的解集。","options":[{"id":"A","text":"$(5, +∞)$"},{"id":"B","text":"$(−∞, 5]$"},{"id":"C","text":"$(−∞, 5)$"},{"id":"D","text":"$[5, +∞)$"}],"answer":"D","explanation":"$x ≥ 5$ includes 5 and extends without bound to the right, so the interval is $[5, +∞)$.","explanationZh":"$x ≥ 5$ 包含 $5，且向右无限延伸，故为 [5, +∞)。$","tags":["easy","set","interval"],"status":"published"},{"id":"math-set-representation-008","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":2,"stem":"Which of the following sets is infinite?","stemZh":"下列哪个集合是无限集？","options":[{"id":"A","text":"${x ∈ ℝ | 0 < x < 1}$"},{"id":"B","text":"${x ∈ ℤ | −3 ≤ x ≤ 3}$"},{"id":"C","text":"{1, 2, 3}"},{"id":"D","text":"${x ∈ ℕ⁺ | x < 10}$"}],"answer":"A","explanation":"Between any two real numbers there are infinitely many reals, so $0 < x < 1$ is infinite. The other three contain 7, 3 and 9 elements respectively.","explanationZh":"任意两个实数之间都有无穷多个实数，故 $0 < x < 1$ 是无限集。其余三个分别含 7、3、9 个元素。","tags":["easy","set","finite-infinite"],"status":"published"},{"id":"math-set-representation-009","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":1,"stem":"Solve the system $x + y = 3, x − y = 1$ and write its solution set by listing.","stemZh":"解方程组 $x + y = 3，x − y = 1，并用列举法写出其解集。$","options":[{"id":"A","text":"{2, 1}"},{"id":"B","text":"${(2, 1)}$"},{"id":"C","text":"$(2, 1)$"},{"id":"D","text":"${x = 2, y = 1}$"}],"answer":"B","explanation":"Adding gives $2x = 4$, so $x = 2$ and $y = 1. A$ solution of a system in two unknowns is an ordered pair, so the solution set is the singleton ${(2, 1)}$.","explanationZh":"两式相加得 $2x = 4，故 x = 2，y = 1。二元方程组的解是有序数对，所以解集为单元素集 {(2, 1)}。$","tags":["easy","set","listing"],"status":"published"},{"id":"math-set-representation-010","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":2,"stem":"List the set $A = {x | x^{2} − 4 = 0}$.","stemZh":"用列举法表示集合 $A = {x | x^{2} − 4 = 0}。$","options":[{"id":"A","text":"{2}"},{"id":"B","text":"{4}"},{"id":"C","text":"{−2, 2}"},{"id":"D","text":"{−2}"}],"answer":"C","explanation":"$x^{2} − 4 = 0$ gives $x^{2} = 4$, so $x = 2$ or $x = −2$. Both roots must be listed.","explanationZh":"$由 x^{2} − 4 = 0 得 x^{2} = 4，故 x = 2 或 x = −2，两个根都要列出。$","tags":["easy","set","listing"],"status":"published"},{"id":"math-set-representation-011","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"List the set $A = {x ∈ ℝ | x^{2} − 3x + 2 = 0}$.","stemZh":"用列举法表示集合 $A = {x ∈ ℝ | x^{2} − 3x + 2 = 0}。$","options":[{"id":"A","text":"{1}"},{"id":"B","text":"{1, 2}"},{"id":"C","text":"{2}"},{"id":"D","text":"{−1, −2}"}],"answer":"B","explanation":"$x^{2} − 3x + 2 = (x − 1)(x − 2) = 0$, so $x = 1$ or $x = 2$.","explanationZh":"$x^{2} − 3x + 2 = (x − 1)(x − 2) = 0，故 x = 1 或 x = 2。$","tags":["medium","set","listing"],"status":"published"},{"id":"math-set-representation-012","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"The set ${x | x = 2k + 1, k ∈ ℤ}$ is","stemZh":"集合 ${x | x = 2k + 1, k ∈ ℤ} 是$","options":[{"id":"A","text":"the set of odd numbers"},{"id":"B","text":"the set of even numbers"},{"id":"C","text":"the set of positive integers"},{"id":"D","text":"the set of integers"}],"answer":"A","explanation":"2k is even for every integer k, so 2k + 1 is always odd, and every odd integer can be written this way.","explanationZh":"对任意整数 k，2k 是偶数，因此 2k + 1 恒为奇数，且每个奇数都能写成这种形式。","tags":["medium","set","set-builder"],"status":"published"},{"id":"math-set-representation-013","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"Which set-builder notation describes the points in the second quadrant?","stemZh":"下列哪个描述法表示第二象限内的点集？","options":[{"id":"A","text":"${(x, y) | x > 0, y > 0}$"},{"id":"B","text":"${(x, y) | x < 0, y < 0}$"},{"id":"C","text":"${(x, y) | x < 0, y > 0}$"},{"id":"D","text":"${(x, y) | x > 0, y < 0}$"}],"answer":"C","explanation":"In the second quadrant the x-coordinate is negative and the y-coordinate is positive, so $x < 0$ and $y > 0$.","explanationZh":"第二象限内点的横坐标为负、纵坐标为正，即 $x < 0 且 y > 0。$","tags":["medium","set","coordinates"],"status":"published"},{"id":"math-set-representation-014","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"How many subsets does the set $A = {1, 2}$ have?","stemZh":"集合 $A = {1, 2}$ 共有多少个子集？","options":[{"id":"A","text":"2"},{"id":"B","text":"3"},{"id":"C","text":"8"},{"id":"D","text":"4"}],"answer":"D","explanation":"A set with n elements has $2^{n}$ subsets. Here $n = 2$, so there are $2^{2} = 4 subsets: ∅, {1}, {2}, {1, 2}$.","explanationZh":"含 n 个元素的集合有 $2^{n}$ 个子集。此处 $n = 2，故有 2^{2} = 4 个：∅、{1}、{2}、{1, 2}。$","tags":["medium","set","subsets"],"status":"published"},{"id":"math-set-representation-015","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"If the set $A = {x |\\text{ ax }+ 1 = 0}$ is empty, then $a =$","stemZh":"若集合 $A = {x |$ ax $+ 1 = 0}$ 是空集，则 $a =$","options":[{"id":"A","text":"0"},{"id":"B","text":"1"},{"id":"C","text":"−1"},{"id":"D","text":"any real number"}],"answer":"A","explanation":"When $a = 0$ the equation becomes $1 = 0$, which has no solution, so the set is empty. For $a ≠ 0$ the equation has the solution $x = −1/a$.","explanationZh":"$当 a = 0$ 时方程变为 $1 = 0，无解，集合为空集。当 a ≠ 0$ 时方程有解 $x = −1/a。$","tags":["medium","set","empty-set"],"status":"published"},{"id":"math-set-representation-016","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"List the set ${x ∈ ℕ | |x| < 3}$.","stemZh":"用列举法表示集合 ${x ∈ ℕ | |x| < 3}。$","options":[{"id":"A","text":"{1, 2}"},{"id":"B","text":"{0, 1, 2}"},{"id":"C","text":"{−2, −1, 0, 1, 2}"},{"id":"D","text":"{0, 1, 2, 3}"}],"answer":"B","explanation":"$|x| < 3$ gives $−3 < x < 3$; restricting to the natural numbers (which include $0)$ leaves 0, 1, 2.","explanationZh":"$由 |x| < 3 得 −3 < x < 3；限制在自然数集（含 0）内，剩下 0、1、2。$","tags":["medium","set","listing"],"status":"published"},{"id":"math-set-representation-017","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"The set $A = {x ∈ ℝ | x^{2} + 2x + a = 0}$ has exactly one element. Find a.","stemZh":"集合 $A = {x ∈ ℝ | x^{2} + 2x + a = 0}$ 恰有一个元素，求 a。","options":[{"id":"A","text":"−1"},{"id":"B","text":"0"},{"id":"C","text":"1"},{"id":"D","text":"2"}],"answer":"C","explanation":"Exactly one element means the quadratic has a double root, so the discriminant is $0: Δ = 4 − 4a = 0$, giving $a = 1$.","explanationZh":"恰有一个元素说明二次方程有两个相等实根，判别式为 $0：Δ = 4 − 4a = 0，得 a = 1。$","tags":["medium","set","discriminant"],"status":"published"},{"id":"math-set-representation-018","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"List the set $A = {(x, y) | x + y = 4, x ∈ ℕ, y ∈ ℕ}$.","stemZh":"用列举法表示集合 $A = {(x, y) | x + y = 4, x ∈ ℕ, y ∈ ℕ}。$","options":[{"id":"A","text":"${(1, 3), (2, 2), (3, 1)}$"},{"id":"B","text":"${(0, 4), (1, 3), (2, 2), (3, 1), (4, 0)}$"},{"id":"C","text":"{0, 1, 2, 3, 4}"},{"id":"D","text":"${(4, 0), (3, 1), (2, 2), (1, 3)}$"}],"answer":"B","explanation":"The natural numbers include 0, so the pairs are $(0, 4), (1, 3), (2, 2), (3, 1), (4, 0)$. Option A forgets the two pairs involving 0.","explanationZh":"自然数集含 $0，故数对为 (0, 4)、(1, 3)、(2, 2)、(3, 1)、(4, 0)。A$ 漏掉了含 0 的两组。","tags":["medium","set","listing"],"status":"published"},{"id":"math-set-representation-019","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"Write ${x | x ≠ 2}$ as a union of intervals.","stemZh":"用区间的并表示集合 ${x | x ≠ 2}。$","options":[{"id":"A","text":"$(−∞, 2]$"},{"id":"B","text":"$[2, +∞)$"},{"id":"C","text":"$(−∞, +∞)$"},{"id":"D","text":"$(−∞, 2) ∪ (2, +∞)$"}],"answer":"D","explanation":"Excluding the single point $x = 2$ splits the real line into two open intervals: everything below 2 and everything above 2.","explanationZh":"去掉唯一点 $x = 2$ 后，实数轴被分成两个开区间：小于 2 的部分和大于 2 的部分。","tags":["medium","set","interval"],"status":"published"},{"id":"math-set-representation-020","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":3,"stem":"How many sets A satisfy ${1, 2} ⊆ A ⊆ {1, 2, 3, 4, 5}$?","stemZh":"满足 ${1, 2} ⊆ A ⊆ {1, 2, 3, 4, 5}$ 的集合 A 有多少个？","options":[{"id":"A","text":"4"},{"id":"B","text":"6"},{"id":"C","text":"8"},{"id":"D","text":"7"}],"answer":"C","explanation":"A must contain 1 and 2, and each of 3, 4, 5 may be either included or excluded, giving $2^{3} = 8$ possibilities.","explanationZh":"A 必须含 1 和 2，而 3、4、5 每个可选可不选，共 $2^{3} = 8$ 种。","tags":["medium","set","counting"],"status":"published"},{"id":"math-set-representation-021","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":4,"stem":"The set $A = {x | x^{2} + (a − 1)x + 1 = 0}$ has exactly one element. Find a.","stemZh":"集合 $A = {x | x^{2} + (a − 1)x + 1 = 0}$ 恰有一个元素，求 a。","options":[{"id":"A","text":"3"},{"id":"B","text":"−1"},{"id":"C","text":"3 or −1"},{"id":"D","text":"1"}],"answer":"C","explanation":"A single element requires $Δ = 0: (a − 1)^{2} − 4 = 0$, so $(a − 1)^{2} = 4$ and $a − 1 = \\pm2$, giving $a = 3$ or $a = −1$. Both values must be given.","explanationZh":"恰有一个元素需 $Δ = 0：(a − 1)^{2} − 4 = 0，即 (a − 1)^{2} = 4，a − 1 = \\pm2，得 a = 3 或 a = −1，两个值都要给出。$","tags":["hard","set","discriminant"],"status":"published"},{"id":"math-set-representation-022","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":5,"stem":"Let $A = {x | x^{2} − 5x + 6 = 0}$ and $B = {x |\\text{ mx }− 1 = 0}$. If $B ⊆ A$, then the set of possible values of m is","stemZh":"$设 A = {x | x^{2} − 5x + 6 = 0}，B = {x |$ mx $− 1 = 0}。若 B ⊆$ A，则 m 的取值集合为","options":[{"id":"A","text":"{1/2, 1/3}"},{"id":"B","text":"{0, 1/2, 1/3}"},{"id":"C","text":"{1/2}"},{"id":"D","text":"{0, 1/2}"}],"answer":"B","explanation":"$A = {2, 3}$. For $m = 0$ the equation mx $= 1$ has no solution, so $B = ∅$ and $∅ ⊆ A$ holds. For $m ≠ 0, B = {1/m}$, so $1/m = 2$ gives $m = 1/2$ and $1/m = 3$ gives $m = 1/3$. Hence $m ∈ {0, 1/2, 1/3}$; the empty-set case is the one most often missed.","explanationZh":"$A = {2, 3}。当 m = 0$ 时方程 mx $= 1$ 无解，B $= ∅，满足 ∅ ⊆$ A；当 $m ≠ 0 时 B = {1/m}，由 1/m = 2 得 m = 1/2，由 1/m = 3 得 m = 1/3。故 m ∈ {0, 1/2, 1/3}，其中空集情形最易漏掉。$","tags":["hard","set","empty-set"],"status":"published"},{"id":"math-set-representation-023","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":5,"stem":"If the set $A = {x | x^{2} + (a + 2)x + 1 = 0, x ∈ ℝ}$ is empty, find the range of a.","stemZh":"若集合 $A = {x | x^{2} + (a + 2)x + 1 = 0, x ∈ ℝ}$ 为空集，求 a 的取值范围。","options":[{"id":"A","text":"$(−∞, −4) ∪ (0, +∞)$"},{"id":"B","text":"$[−4, 0]$"},{"id":"C","text":"$(0, 4)$"},{"id":"D","text":"$(−4, 0)$"}],"answer":"D","explanation":"The set is empty when the quadratic has no real root, so $Δ < 0: (a + 2)^{2} − 4 < 0$, i.e. $(a + 2)^{2} < 4$, giving $−2 < a + 2 < 2$ and therefore $−4 < a < 0$.","explanationZh":"集合为空说明二次方程无实根，即 $Δ < 0：(a + 2)^{2} − 4 < 0，亦即 (a + 2)^{2} < 4，得 −2 < a + 2 < 2，故 −4 < a < 0。$","tags":["hard","set","discriminant"],"status":"published"},{"id":"math-set-representation-024","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":4,"stem":"For the set $A = {a, a^{2} − a}$ to be well defined, what must a satisfy?","stemZh":"要使集合 $A = {a, a^{2} − a}$ 有意义，a 应满足什么条件？","options":[{"id":"A","text":"$a ≠ 0$"},{"id":"B","text":"$a ≠ 2$"},{"id":"C","text":"$a ≠ 0$ and $a ≠ 2$"},{"id":"D","text":"a can be any real number"}],"answer":"C","explanation":"Elements of a set must be distinct, so $a ≠ a^{2} − a$, i.e. $a^{2} − 2a ≠ 0$, i.e. $a(a − 2) ≠ 0$, giving $a ≠ 0$ and $a ≠ 2$. Both exclusions are needed.","explanationZh":"集合中的元素必须互异，故 $a ≠ a^{2} −$ a，即 $a^{2} − 2a ≠ 0，亦即 a(a − 2) ≠ 0，得 a ≠ 0 且 a ≠ 2，两个条件都要排除。$","tags":["hard","set","distinctness"],"status":"published"},{"id":"math-set-representation-025","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":5,"stem":"Let $A = {x | x^{2} − 3x + 2 ≤ 0}$ and $B = {x | x > a}$. If $A ⊆ B$, find the range of a.","stemZh":"$设 A = {x | x^{2} − 3x + 2 ≤ 0}，B = {x | x > a}。若 A ⊆$ B，求 a 的取值范围。","options":[{"id":"A","text":"$a < 1$"},{"id":"B","text":"$a ≤ 1$"},{"id":"C","text":"$a < 2$"},{"id":"D","text":"$a ≤ 2$"}],"answer":"A","explanation":"$A = [1, 2]$. For $[1, 2] ⊆ (a, +∞)$ every element of A must exceed a, in particular the smallest one, 1, so $a < 1$. Equality $a = 1$ fails because $1 ∈ A$ but $1 ∉ (1, +∞)$.","explanationZh":"$A = [1, 2]。要使 [1, 2] ⊆ (a, +∞)，A$ 中每个元素都要大于 a，特别是最小的元素 $1，故 a < 1。取 a = 1$ 不行，因为 $1 ∈ A 但 1 ∉ (1, +∞)。$","tags":["hard","set","interval"],"status":"published"},{"id":"math-set-representation-026","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":4,"stem":"List the set ${x ∈ ℤ | 0 ≤ 2x − 1 < 5}$.","stemZh":"用列举法表示集合 ${x ∈ ℤ | 0 ≤ 2x − 1 < 5}。$","options":[{"id":"A","text":"{0, 1, 2}"},{"id":"B","text":"{1, 2, 3}"},{"id":"C","text":"{0, 1, 2, 3}"},{"id":"D","text":"{1, 2}"}],"answer":"D","explanation":"$0 ≤ 2x − 1$ gives $x ≥ 1/2$ and $2x − 1 < 5$ gives $x < 3$, so $1/2 ≤ x < 3$. The integers in that range are 1 and 2.","explanationZh":"$由 0 ≤ 2x − 1 得 x ≥ 1/2，由 2x − 1 < 5 得 x < 3，故 1/2 ≤ x < 3，其中的整数为 1 和 2。$","tags":["hard","set","listing"],"status":"published"},{"id":"math-set-representation-027","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":4,"stem":"Let the universal set be $U = ℝ$ and $A = {x | x ≥ 2}$. Write the complement ∁ᵤA as an interval.","stemZh":"设全集 $U = ℝ，A = {x | x ≥ 2}，用区间表示补集$ ∁ᵤA。","options":[{"id":"A","text":"$(−∞, 2]$"},{"id":"B","text":"$(−∞, 2)$"},{"id":"C","text":"$(2, +∞)$"},{"id":"D","text":"$[2, +∞)$"}],"answer":"B","explanation":"$A = [2, +∞)$ contains 2, so its complement contains everything strictly less than 2, namely $(−∞, 2)$. The point 2 belongs to A, so it must be excluded from the complement.","explanationZh":"$A = [2, +∞) 含 2，故其补集为严格小于 2$ 的全体实数，即 $(−∞, 2)。点 2$ 属于 A，必须从补集中排除。","tags":["hard","set","complement"],"status":"published"},{"id":"math-set-representation-028","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":5,"stem":"Let $A = {x | x = k/2 + 1/4, k ∈ ℤ}$ and $B = {x | x = k/4 + 1/2, k ∈ ℤ}$. Which relation holds?","stemZh":"$设 A = {x | x = k/2 + 1/4, k ∈ ℤ}，B = {x | x = k/4 + 1/2, k ∈ ℤ}，二者的关系是？$","options":[{"id":"A","text":"$A = B$"},{"id":"B","text":"$B ⊆ A$"},{"id":"C","text":"$A ⊂ B$"},{"id":"D","text":"cannot be determined"}],"answer":"C","explanation":"$A = {(2k + 1)/4 : k ∈ ℤ}$, the quarters whose numerator is odd. $B = {(k + 2)/4 : k ∈ ℤ}$, all multiples of 1/4. So every element of A lies in B, but B also contains e.g. $1/2 (take k = 0)$, which is not in A. Hence A is a proper subset of B.","explanationZh":"$A = {(2k + 1)/4 : k ∈ ℤ}，即分子为奇数的四分之一；B = {(k + 2)/4 : k ∈ ℤ}，即一切 1/4$ 的整数倍。故 A 中元素都在 B 中，而 B 还含 $1/2（取 k = 0），它不在 A$ 中，所以 A 是 B 的真子集。","tags":["hard","set","set-builder"],"status":"published"},{"id":"math-set-representation-029","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":5,"stem":"Let $A = {1, 2, 3}$. How many sets B satisfy $A ∪ B =$ A?","stemZh":"$设 A = {1, 2, 3}，满足 A ∪ B = A$ 的集合 B 有多少个？","options":[{"id":"A","text":"8"},{"id":"B","text":"7"},{"id":"C","text":"4"},{"id":"D","text":"3"}],"answer":"A","explanation":"$A ∪ B = A$ holds exactly when $B ⊆$ A. A has 3 elements, so it has $2^{3} = 8 subsets$, and each of them works as B.","explanationZh":"$A ∪ B = A$ 当且仅当 $B ⊆$ A。A 有 3 个元素，共有 $2^{3} = 8$ 个子集，每一个都可作为 B。","tags":["hard","set","counting"],"status":"published"},{"id":"math-set-representation-030","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-representation","type":"single-choice","difficulty":4,"stem":"Which set-builder notation describes the points of the first quadrant, excluding the axes?","stemZh":"下列哪个描述法表示第一象限内（不含坐标轴）的点集？","options":[{"id":"A","text":"${(x, y) | x ≥ 0, y ≥ 0}$"},{"id":"B","text":"${(x, y) | x > 0, y < 0}$"},{"id":"C","text":"${(x, y) |\\text{ xy }> 0}$"},{"id":"D","text":"${(x, y) | x > 0, y > 0}$"}],"answer":"D","explanation":"Excluding the axes means both coordinates are strictly positive, so $x > 0$ and $y > 0$. Option C describes quadrants I and III together, since xy $> 0$ also holds when both coordinates are negative.","explanationZh":"不含坐标轴意味着两个坐标都严格为正，即 $x > 0 且 y > 0。C$ 描述的是第一和第三象限，因为两坐标同负时也有 xy $> 0。$","tags":["hard","set","coordinates"],"status":"published"},{"id":"math-set-relations-001","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":1,"stem":"Let $A = {2, 4, 6}$ and $B = {2, 4, 6, 8}$. Which statement is true?","stemZh":"已知 $A = {2, 4, 6}，B = {2, 4, 6, 8}，下列关系正确的是？$","options":[{"id":"A","text":"$A ∈ B$"},{"id":"B","text":"$B ⊆ A$"},{"id":"C","text":"$A = B$"},{"id":"D","text":"$A ⊆ B$"}],"answer":"D","explanation":"Every element of $A (2, 4, 6)$ also belongs to B, so $A ⊆ B (in\\text{ fact }A\\text{ is }a\\text{ proper subset of }B). A$ itself is a set, not one of the elements listed inside B, so $A ∈ B$ is false.","explanationZh":"A 中的元素 2、4、6 都属于 B，所以 $A ⊆$ B（且为真子集）。A 是一个集合，不是 B 中的元素，因此 $A ∈ B$ 错误。","tags":["easy","subset","element-relation"],"status":"published"},{"id":"math-set-relations-002","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":1,"stem":"Let $U = {1, 2, 3, 4, 5, 6}$ and $A = {1, 2, 3}$. Find ∁ᵤA.","stemZh":"全集 $U = {1, 2, 3, 4, 5, 6}，A = {1, 2, 3}，求$ ∁ᵤA。","options":[{"id":"A","text":"{4, 5, 6}"},{"id":"B","text":"{1, 2, 3}"},{"id":"C","text":"$∅$"},{"id":"D","text":"{1, 2, 3, 4, 5, 6}"}],"answer":"A","explanation":"The complement of A in U consists of the elements of U that are not in A, namely 4, 5 and 6.","explanationZh":"补集由 U 中不属于 A 的元素组成，即 4、5、6。","tags":["easy","complement","universal-set"],"status":"published"},{"id":"math-set-relations-003","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":2,"stem":"Let $A = {x | x^{2} = 4}$. How many subsets does A have?","stemZh":"已知 $A = {x | x^{2} = 4}，则 A$ 的子集个数为？","options":[{"id":"A","text":"2"},{"id":"B","text":"3"},{"id":"C","text":"4"},{"id":"D","text":"8"}],"answer":"C","explanation":"$x^{2} = 4$ gives $x = −2$ or $x = 2$, so $A = {−2, 2}$ has 2 elements and therefore $2^{2} = 4$ subsets.","explanationZh":"$由 x^{2} = 4 得 x = −2 或 x = 2，故 A = {−2, 2} 有 2$ 个元素，子集个数为 $2^{2} = 4。$","tags":["easy","subsets","counting"],"status":"published"},{"id":"math-set-relations-004","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":1,"stem":"Let $A = {1, 2}$ and $B = {2, 3}$. Find $A ∩$ B.","stemZh":"已知 $A = {1, 2}，B = {2, 3}，求 A ∩$ B。","options":[{"id":"A","text":"{1, 3}"},{"id":"B","text":"{2}"},{"id":"C","text":"{1, 2, 3}"},{"id":"D","text":"$∅$"}],"answer":"B","explanation":"The intersection keeps only the elements common to both sets; 2 is the only one.","explanationZh":"交集只保留两集合的公共元素，只有 2。","tags":["easy","intersection"],"status":"published"},{"id":"math-set-relations-005","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":2,"stem":"Let $A = {1, 2}$ and $B = {2, 3}$. Find $A ∪$ B.","stemZh":"已知 $A = {1, 2}，B = {2, 3}，求 A ∪$ B。","options":[{"id":"A","text":"{2}"},{"id":"B","text":"{1, 3}"},{"id":"C","text":"{1, 2, 3}"},{"id":"D","text":"{1, 2}"}],"answer":"C","explanation":"The union collects every element belonging to at least one of the sets: 1, 2 and 3.","explanationZh":"并集包含属于至少一个集合的所有元素：1、2、3。","tags":["easy","union"],"status":"published"},{"id":"math-set-relations-006","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":1,"stem":"Which relation symbol makes the statement {1, 2} ___ {1, 2} true?","stemZh":"用哪个关系符号可使 {1, 2} ___ {1, 2} 成立？","options":[{"id":"A","text":"$⊆$"},{"id":"B","text":"$⊂$"},{"id":"C","text":"$∈$"},{"id":"D","text":"⊄"}],"answer":"A","explanation":"Every set is a subset of itself, so $⊆$ holds. The proper-subset symbol $⊂$ fails when the two sets are equal, and $∈$ fails because the elements of {1, 2} are the numbers 1 and 2, not the set {1, 2} itself.","explanationZh":"任何集合都是它自己的子集，故 $⊆$ 成立；两集合相等时真子集符号 $⊂ 不成立；{1, 2}$ 的元素是 $1 和 2，而不是集合 {1, 2}$ 本身，故 $∈$ 错误。","tags":["easy","subset","notation"],"status":"published"},{"id":"math-set-relations-007","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":2,"stem":"How many proper subsets does ${a,\\text{ b, }c}$ have?","stemZh":"集合 ${a, b, c}$ 的真子集个数为？","options":[{"id":"A","text":"6"},{"id":"B","text":"8"},{"id":"C","text":"9"},{"id":"D","text":"7"}],"answer":"D","explanation":"A 3-element set has $2^{3} = 8$ subsets. A proper subset excludes the set itself, leaving $8 − 1 = 7$.","explanationZh":"含 3 个元素的集合有 $2^{3} = 8$ 个子集，真子集要去掉集合本身，故为 $8 − 1 = 7。$","tags":["easy","subsets","counting"],"status":"published"},{"id":"math-set-relations-008","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":1,"stem":"Let $A = {x ∈ ℝ | x > 3}$ and $B = {x ∈ ℝ | x > 1}$. Find $A ∩$ B.","stemZh":"已知 $A = {x ∈ ℝ | x > 3}，B = {x ∈ ℝ | x > 1}，求 A ∩$ B。","options":[{"id":"A","text":"${x | x > 1}$"},{"id":"B","text":"${x | x > 3}$"},{"id":"C","text":"$ℝ$"},{"id":"D","text":"$∅$"}],"answer":"B","explanation":"Since $x > 3$ implies $x > 1$, we have $A ⊂ B$, so $A ∩ B = A = {x | x > 3}$.","explanationZh":"$由 x > 3$ 可推出 $x > 1，即 A ⊂$ B，所以 $A ∩ B = A = {x | x > 3}。$","tags":["easy","intersection","intervals"],"status":"published"},{"id":"math-set-relations-009","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":2,"stem":"Take $ℝ$ as the universal set and let $A = {x | x ≤ 1}$. Find ∁ᵤA.","stemZh":"$以 ℝ$ 为全集，A $= {x | x ≤ 1}，求$ ∁ᵤA。","options":[{"id":"A","text":"${x | x ≥ 1}$"},{"id":"B","text":"${x | x < 1}$"},{"id":"C","text":"${x | x > 1}$"},{"id":"D","text":"${x | x > 0}$"}],"answer":"C","explanation":"Removing the part $x ≤ 1$ from $ℝ$ leaves $x > 1$. The endpoint 1 belongs to A, so it must be excluded from the complement.","explanationZh":"$从 ℝ$ 中去掉 $x ≤ 1$ 的部分，剩下 $x > 1；端点 1$ 属于 A，必须从补集中排除。","tags":["easy","complement","intervals"],"status":"published"},{"id":"math-set-relations-010","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":1,"stem":"Let $A = {1, 2, 3}$ and $B = {3}$. Find A − B.","stemZh":"已知 $A = {1, 2, 3}，B = {3}，求 A −$ B。","options":[{"id":"A","text":"{1, 2}"},{"id":"B","text":"{3}"},{"id":"C","text":"{1, 2, 3}"},{"id":"D","text":"$∅$"}],"answer":"A","explanation":"The difference A − B keeps the elements of A that do not belong to B, namely 1 and 2.","explanationZh":"差集 A − B 保留属于 A 但不属于 B 的元素，即 1 和 2。","tags":["easy","difference"],"status":"published"},{"id":"math-set-relations-011","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"Let $A = {x | x^{2} − x − 2 = 0}$ and $B = {−1, 2}$. Which statement is true?","stemZh":"已知 $A = {x | x^{2} − x − 2 = 0}，B = {−1, 2}，下列关系正确的是？$","options":[{"id":"A","text":"$A ⊂ B$"},{"id":"B","text":"$A = B$"},{"id":"C","text":"$B ⊂ A$"},{"id":"D","text":"$A ∩ B = ∅$"}],"answer":"B","explanation":"$x^{2} − x − 2 = (x − 2)(x + 1) = 0$ gives $x = 2$ or $x = −1$, so $A = {−1, 2}$, which is exactly B.","explanationZh":"$由 x^{2} − x − 2 = (x − 2)(x + 1) = 0 得 x = 2 或 x = −1，故 A = {−1, 2} =$ B。","tags":["medium","equality","subset"],"status":"published"},{"id":"math-set-relations-012","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"A finite set has exactly 16 subsets. How many elements does it have?","stemZh":"某有限集合恰有 16 个子集，则该集合有几个元素？","options":[{"id":"A","text":"8"},{"id":"B","text":"5"},{"id":"C","text":"3"},{"id":"D","text":"4"}],"answer":"D","explanation":"A set with n elements has $2^{n}$ subsets. Solving $2^{n} = 16$ gives $n = 4. (16\\text{ is the number of subsets, not twice the number of }elements.)$","explanationZh":"含 n 个元素的集合有 $2^{n}$ 个子集，由 $2^{n} = 16 得 n = 4（16$ 是子集个数，不是元素个数的两倍）。","tags":["medium","subsets","counting"],"status":"published"},{"id":"math-set-relations-013","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"Let $U = {1, 2, 3, 4, 5}, A = {1, 3}$ and $B = {3, 5}$. Find $∁ᵤ(A ∪ B)$.","stemZh":"全集 $U = {1, 2, 3, 4, 5}，A = {1, 3}，B = {3, 5}，求 ∁ᵤ(A ∪ B)。$","options":[{"id":"A","text":"{2, 4}"},{"id":"B","text":"{2}"},{"id":"C","text":"{4}"},{"id":"D","text":"{1, 5}"}],"answer":"A","explanation":"$A ∪ B = {1, 3, 5}$, so the complement inside $U = {1, 2, 3, 4, 5}$ is {2, 4}.","explanationZh":"$A ∪ B = {1, 3, 5}，故其在 U = {1, 2, 3, 4, 5}$ 中的补集为 {2, 4}。","tags":["medium","complement","union"],"status":"published"},{"id":"math-set-relations-014","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"Let $A = {x | −1 < x < 3}$ and $B = {x | 0 ≤ x ≤ 5}$. Find $A ∩$ B.","stemZh":"已知 $A = {x | −1 < x < 3}，B = {x | 0 ≤ x ≤ 5}，求 A ∩$ B。","options":[{"id":"A","text":"$(0, 3)$"},{"id":"B","text":"$(−1, 5]$"},{"id":"C","text":"$[0, 3)$"},{"id":"D","text":"$[0, 5]$"}],"answer":"C","explanation":"Intersect $(−1, 3)$ with $[0, 5]$: the lower endpoint 0 is included because 0 lies in both sets, while 3 is excluded because A is open there.","explanationZh":"$(−1, 3) 与 [0, 5]$ 取交集：下端点 0 同时属于两个集合故取闭，上端点 3 在 A 中是开的故取开，即 $[0, 3)。$","tags":["medium","intersection","intervals"],"status":"published"},{"id":"math-set-relations-015","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"Which of the following statements is true?","stemZh":"下列命题中正确的一个是？","options":[{"id":"A","text":"$∅ ∈ {0}$"},{"id":"B","text":"$∅ ⊆ {0}$"},{"id":"C","text":"${0} ⊆ ∅$"},{"id":"D","text":"$0 ⊆ {0}$"}],"answer":"B","explanation":"The empty set is a subset of every set, so $∅ ⊆ {0}$. It is not an element of {0}: the only element of {0} is the number 0, and 0 is a number, not a set, so it cannot be a subset of anything.","explanationZh":"空集是任何集合的子集，故 $∅ ⊆ {0} 正确；{0}$ 的唯一元素是数字 0，空集不是它的元素；0 是数不是集合，不能作为子集。","tags":["medium","empty-set","element-relation"],"status":"published"},{"id":"math-set-relations-016","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"In a class of 40 students, 25 study Mathematics, 20 study Physics and 10 study both. How many students study neither?","stemZh":"某班 40 人，25 人学数学，20 人学物理，10 人两科都学，两科都不学的有多少人？","options":[{"id":"A","text":"10"},{"id":"B","text":"0"},{"id":"C","text":"15"},{"id":"D","text":"5"}],"answer":"D","explanation":"By the inclusion–exclusion principle $|M ∪ P| = 25 + 20 − 10 = 35$, so $40 − 35 = 5$ students study neither subject.","explanationZh":"由容斥原理 $|M ∪ P| = 25 + 20 − 10 = 35，故两科都不学的有 40 − 35 = 5$ 人。","tags":["medium","venn","counting"],"status":"published"},{"id":"math-set-relations-017","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"For any sets A and B inside a universal set $U, ∁ᵤ(A ∩ B) equals$:","stemZh":"对全集 U 中的任意集合 $A、B，∁ᵤ(A ∩ B)$ 等于？","options":[{"id":"A","text":"∁ᵤA $∪$ ∁ᵤB"},{"id":"B","text":"∁ᵤA $∩$ ∁ᵤB"},{"id":"C","text":"$A ∪ B$"},{"id":"D","text":"$A ∩ B$"}],"answer":"A","explanation":"De Morgan's law: the complement of an intersection is the union of the complements (and the complement of a union is the intersection of the complements).","explanationZh":"德·摩根定律：交集的补集等于补集的并集（并集的补集等于补集的交集）。","tags":["medium","complement","de-morgan"],"status":"published"},{"id":"math-set-relations-018","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"For sets A and B with universal set U, the difference A − B equals:","stemZh":"在全集 U 中，差集 A − B 等于？","options":[{"id":"A","text":"$A ∪$ ∁ᵤB"},{"id":"B","text":"∁ᵤA $∩ B$"},{"id":"C","text":"$A ∩$ ∁ᵤB"},{"id":"D","text":"A − ∁ᵤB"}],"answer":"C","explanation":"A − B keeps the elements of A that are not in B, which is exactly $A ∩$ ∁ᵤB. $(Note A −\\text{ ∁ᵤB }= A ∩\\text{ B, the opposite }operation.)$","explanationZh":"A − B 保留属于 A 但不属于 B 的元素，即 $A ∩$ ∁ᵤB（而 A − ∁ᵤB $= A ∩$ B，正好相反）。","tags":["medium","difference","complement"],"status":"published"},{"id":"math-set-relations-019","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"If $|A| = 3, |B| = 4$ and $|A ∩ B| = 1$, find $|A ∪ B|$.","stemZh":"已知 $|A| = 3，|B| = 4，|A ∩ B| = 1，求 |A ∪ B|。$","options":[{"id":"A","text":"7"},{"id":"B","text":"6"},{"id":"C","text":"5"},{"id":"D","text":"12"}],"answer":"B","explanation":"$|A ∪ B| = |A| + |B| − |A ∩ B| = 3 + 4 − 1 = 6$. Adding without subtracting would double-count the shared element.","explanationZh":"$|A ∪ B| = |A| + |B| − |A ∩ B| = 3 + 4 − 1 = 6；直接相加会把公共元素重复计算一次。$","tags":["medium","union","counting"],"status":"published"},{"id":"math-set-relations-020","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":3,"stem":"If $A ∪ B = A$, which statement must be true?","stemZh":"$若 A ∪ B =$ A，则一定成立的是？","options":[{"id":"A","text":"$A ⊆ B$"},{"id":"B","text":"$B = ∅$"},{"id":"C","text":"$A ∩ B = ∅$"},{"id":"D","text":"$B ⊆ A$"}],"answer":"D","explanation":"$A ∪ B = A$ means adding B brings in nothing new, i.e. every element of B already lies in A, so $B ⊆$ A. $B = ∅$ is only one possible case, not a necessary consequence.","explanationZh":"$A ∪ B = A$ 说明并上 B 后没有新增元素，即 B 的元素都在 A 中，故 $B ⊆$ A；B $= ∅$ 只是其中一种可能，并非必然。","tags":["medium","union","subset"],"status":"published"},{"id":"math-set-relations-021","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"Let $M = {x | x^{2} − 5x + 6 = 0}$. How many non-empty subsets does M have?","stemZh":"已知 $M = {x | x^{2} − 5x + 6 = 0}，则 M$ 的非空子集个数为？","options":[{"id":"A","text":"4"},{"id":"B","text":"2"},{"id":"C","text":"3"},{"id":"D","text":"1"}],"answer":"C","explanation":"$x^{2} − 5x + 6 = (x − 2)(x − 3) = 0$, so $M = {2, 3}$. It has $2^{2} = 4 subsets$, and removing the empty set leaves 3 non-empty $ones: {2}, {3}, {2, 3}$.","explanationZh":"$由 (x − 2)(x − 3) = 0 得 M = {2, 3}，共有 2^{2} = 4$ 个子集，去掉空集后非空子集为 ${2}、{3}、{2, 3} 共 3$ 个。","tags":["hard","subsets","counting"],"status":"published"},{"id":"math-set-relations-022","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"Let $A = {x |\\text{ ax }+ 1 = 0}$. If $A ⊆ {1}$, then a equals:","stemZh":"已知 $A = {x |$ ax $+ 1 = 0}，若 A ⊆ {1}，则 a$ 等于？","options":[{"id":"A","text":"0 or −1"},{"id":"B","text":"−1"},{"id":"C","text":"0"},{"id":"D","text":"1"}],"answer":"A","explanation":"Two cases. If $a = 0$ the equation becomes $1 = 0$, so $A = ∅$ and $∅ ⊆ {1}$ holds. If $a ≠ 0$ then $A = {−1/a}$, and $−1/a = 1$ gives $a = −1$. Hence $a = 0$ or $a = −1$; forgetting the empty-set case loses $a = 0$.","explanationZh":"分两种情况：当 $a = 0$ 时方程化为 $1 = 0，A = ∅，而 ∅ ⊆ {1}$ 成立；当 $a ≠ 0 时 A = {−1/a}，由 −1/a = 1 得 a = −1。故 a = 0 或 a = −1；漏掉空集情形就会丢掉 a = 0。$","tags":["hard","empty-set","parameter"],"status":"published"},{"id":"math-set-relations-023","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"Let $A = {x | x < a}$ and $B = {x | x^{2} − 3x + 2 ≤ 0}$. If $A ∩ B = ∅, then$:","stemZh":"已知 $A = {x | x < a}，B = {x | x^{2} − 3x + 2 ≤ 0}，若 A ∩ B = ∅，则？$","options":[{"id":"A","text":"$a < 1$"},{"id":"B","text":"$a ≤ 1$"},{"id":"C","text":"$a ≥ 2$"},{"id":"D","text":"$a ≤ 2$"}],"answer":"B","explanation":"$B = [1, 2]$ and $A = (−∞, a)$. They are disjoint exactly when $a ≤ 1$: at $a = 1$ we get $A = (−∞, 1)$, which contains neither 1 nor 2, so the endpoint is included in the range of a.","explanationZh":"$B = [1, 2]，A = (−∞, a)。两者无公共元素当且仅当 a ≤ 1：当 a = 1 时 A = (−∞, 1)$ 不含 1 也不含 2，因此端点可以取到。","tags":["hard","intervals","parameter"],"status":"published"},{"id":"math-set-relations-024","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"Which of the following identities is correct for all sets A, B and C?","stemZh":"下列集合恒等式中对任意集合 A、B、C 都成立的是？","options":[{"id":"A","text":"$A ∪ (B ∩ C) = (A ∪ B) ∩ C$"},{"id":"B","text":"$A − (B − C) = (A − B) − C$"},{"id":"C","text":"$(A − B) ∪ B = A$"},{"id":"D","text":"$A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)$"}],"answer":"D","explanation":"Intersection distributes over union. The others fail: the correct distributive law is $A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C); A − (B − C) = (A − B) ∪ (A ∩ C) ≠ (A − B) − C$; and $(A − B) ∪ B = A ∪ B$, not A.","explanationZh":"交集对并集可分配。其余均错：正确形式是 $A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)；A − (B − C) = (A − B) ∪ (A ∩ C) ≠ (A − B) − C；(A − B) ∪ B = A ∪ B$ 而非 A。","tags":["hard","identities","de-morgan"],"status":"published"},{"id":"math-set-relations-025","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"How many sets X satisfy ${1} ⊆ X ⊆ {1, 2, 3, 4}$?","stemZh":"满足 ${1} ⊆ X ⊆ {1, 2, 3, 4}$ 的集合 X 有多少个？","options":[{"id":"A","text":"8"},{"id":"B","text":"4"},{"id":"C","text":"16"},{"id":"D","text":"7"}],"answer":"A","explanation":"X must contain 1, and each of 2, 3, 4 may be in or out independently, giving $2^{3} = 8$ possibilities.","explanationZh":"X 必须含有 1，而 2、3、4 各可选或不选，故共有 $2^{3} = 8$ 个。","tags":["hard","subsets","counting"],"status":"published"},{"id":"math-set-relations-026","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"Let $A = {x | x^{2} − 4 ≤ 0}$ and $B = {x | x^{2} − 2x − 3 > 0}$. Find $A ∩$ B.","stemZh":"已知 $A = {x | x^{2} − 4 ≤ 0}，B = {x | x^{2} − 2x − 3 > 0}，求 A ∩$ B。","options":[{"id":"A","text":"$[−2, −1]$"},{"id":"B","text":"$[−2, −1)$"},{"id":"C","text":"$(−2, −1)$"},{"id":"D","text":"$∅$"}],"answer":"B","explanation":"$A = [−2, 2]$ and $B = (−∞, −1) ∪ (3, +∞)$. Their intersection is $[−2, −1): −2$ is included because it satisfies both $x^{2} ≤ 4$ and $x < −1$, while −1 is excluded because $x^{2} − 2x − 3 = 0$ there and the inequality is strict.","explanationZh":"$A = [−2, 2]，B = (−∞, −1) ∪ (3, +∞)，交集为 [−2, −1)：−2$ 同时满足 $x^{2} ≤ 4 与 x < −1 故取闭；−1 处 x^{2} − 2x − 3 = 0，因不等式是严格的故取开。$","tags":["hard","intersection","intervals"],"status":"published"},{"id":"math-set-relations-027","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":5,"stem":"Let $U = {1, 2, 3, 4, 5, 6}$. Given $A ∩$ ∁ᵤB $= {1, 3}$, ∁ᵤA $∩ B = {5}$ and $A ∩ B = {2}$, find $∁ᵤ(A ∪ B)$.","stemZh":"全集 $U = {1, 2, 3, 4, 5, 6}，已知 A ∩$ ∁ᵤB $= {1, 3}，∁ᵤA ∩ B = {5}，A ∩ B = {2}，求 ∁ᵤ(A ∪ B)。$","options":[{"id":"A","text":"{4}"},{"id":"B","text":"{6}"},{"id":"C","text":"{4, 6}"},{"id":"D","text":"{1, 4, 6}"}],"answer":"C","explanation":"The three regions give $A ∪ B = {1, 3} ∪ {2} ∪ {5} = {1, 2, 3, 5}$, so the complement in $U = {1, 2, 3, 4, 5, 6}$ is {4, 6}.","explanationZh":"三部分相加得 $A ∪ B = {1, 3} ∪ {2} ∪ {5} = {1, 2, 3, 5}，故在 U$ 中的补集为 {4, 6}。","tags":["hard","venn","complement"],"status":"published"},{"id":"math-set-relations-028","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"Let $A = {x | x^{2} − 3x − 4 < 0}$ and $B = {x | x^{2} − 2x − 8 ≤ 0}$. Which statement is true?","stemZh":"已知 $A = {x | x^{2} − 3x − 4 < 0}，B = {x | x^{2} − 2x − 8 ≤ 0}，下列关系正确的是？$","options":[{"id":"A","text":"$B ⊆ A$"},{"id":"B","text":"$A ∩ B = ∅$"},{"id":"C","text":"$A ∪ B = ℝ$"},{"id":"D","text":"$A ⊆ B$"}],"answer":"D","explanation":"$A = (−1, 4) (from (x − 4)(x + 1) < 0)$ and $B = [−2, 4] (from (x − 4)(x + 2) ≤ 0)$. Since $(−1, 4)$ lies inside $[−2, 4]$, we get $A ⊂ B$, hence $A ⊆$ B.","explanationZh":"$由 (x − 4)(x + 1) < 0 得 A = (−1, 4)；由 (x − 4)(x + 2) ≤ 0 得 B = [−2, 4]。(−1, 4)$ 包含于 $[−2, 4]，故 A ⊂$ B，即 $A ⊆$ B。","tags":["hard","subset","intervals"],"status":"published"},{"id":"math-set-relations-029","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":4,"stem":"Let $A = {x | x = 2k + 1, k ∈ ℤ}$ and $B = {x | x = 4k + 1, k ∈ ℤ}$. Which statement is true?","stemZh":"已知 $A = {x | x = 2k + 1, k ∈ ℤ}，B = {x | x = 4k + 1, k ∈ ℤ}，下列关系正确的是？$","options":[{"id":"A","text":"$A ⊂ B$"},{"id":"B","text":"$B ⊂ A$"},{"id":"C","text":"$A = B$"},{"id":"D","text":"$A ∩ B = ∅$"}],"answer":"B","explanation":"A is the set of odd integers. Every number $4k + 1 = 2(2k) + 1$ is odd, so $B ⊆$ A. But $3 ∈ A$ is not of the form 4k + 1, so the inclusion is $proper: B ⊂$ A.","explanationZh":"$A 是全体奇数。4k + 1 = 2(2k) + 1$ 恒为奇数，故 $B ⊆$ A；但 $3 ∈ A 且 3$ 不能写成 $4k + 1，故为真包含：B ⊂$ A。","tags":["hard","subset","number-sets"],"status":"published"},{"id":"math-set-relations-030","subject":"math","moduleId":"sets-inequalities","unitId":"sets","topicId":"set-relations","type":"single-choice","difficulty":5,"stem":"The solution set $A = {x | x^{2} +\\text{ px }+ q = 0}$ consists of exactly one $element, A = {1}$. Find p and q.","stemZh":"$若 A = {x | x^{2} +$ px $+ q = 0}$ 恰有一个元素且 $A = {1}，求$ p、q。","options":[{"id":"A","text":"$p = −2, q = 1$"},{"id":"B","text":"$p = 2, q = 1$"},{"id":"C","text":"$p = −2, q = −1$"},{"id":"D","text":"$p = 1, q = −2$"}],"answer":"A","explanation":"A single-element solution set means the quadratic has a double root at $x = 1$, so $x^{2} +$ px $+ q = (x − 1)^{2} = x^{2} − 2x + 1$. Hence $p = −2$ and $q = 1$.","explanationZh":"解集只有一个元素说明方程有二重根 $x = 1，即 x^{2} +$ px $+ q = (x − 1)^{2} = x^{2} − 2x + 1，故 p = −2，q = 1。$","tags":["hard","parameter","quadratic"],"status":"published"},{"id":"math-quadratic-inequality-001","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":1,"stem":"Solve $x^{2} − x − 6 ≥ 0$.","stemZh":"解不等式 $x^{2} − x − 6 ≥ 0。$","options":[{"id":"A","text":"$(−∞, −2) ∪ (3, +∞)$"},{"id":"B","text":"$[−2, 3]$"},{"id":"C","text":"$(−2, 3)$"},{"id":"D","text":"$(−∞, −2] ∪ [3, +∞)$"}],"answer":"D","explanation":"$x^{2} − x − 6 = (x − 3)(x + 2) ≥ 0$. The parabola opens upward, so the expression is non-negative outside the $roots: x ≤ −2$ or $x ≥ 3$. The roots themselves are included because the inequality is not strict.","explanationZh":"$x^{2} − x − 6 = (x − 3)(x + 2) ≥ 0，开口向上，两根之外非负，即 x ≤ −2 或 x ≥ 3；由于不等式带等号，两根要取到。$","tags":["easy","quadratic","solution-set"],"status":"published"},{"id":"math-quadratic-inequality-002","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":1,"stem":"Solve $x^{2} − 4 < 0$.","stemZh":"解不等式 $x^{2} − 4 < 0。$","options":[{"id":"A","text":"$x < 2$"},{"id":"B","text":"$−2 < x < 2$"},{"id":"C","text":"$x < −2$ or $x > 2$"},{"id":"D","text":"$x ≠ 2$"}],"answer":"B","explanation":"$(x − 2)(x + 2) < 0$ means the two factors have opposite signs, which happens strictly between the $roots: −2 < x < 2$.","explanationZh":"$(x − 2)(x + 2) < 0 表示两因式异号，解集在两根之间：−2 < x < 2。$","tags":["easy","quadratic","factorisation"],"status":"published"},{"id":"math-quadratic-inequality-003","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":1,"stem":"Solve $x^{2} − 5x + 6 ≤ 0$.","stemZh":"解不等式 $x^{2} − 5x + 6 ≤ 0。$","options":[{"id":"A","text":"$[2, 3]$"},{"id":"B","text":"$(2, 3)$"},{"id":"C","text":"$(−∞, 2] ∪ [3, +∞)$"},{"id":"D","text":"$(−∞, 2) ∪ (3, +∞)$"}],"answer":"A","explanation":"$x^{2} − 5x + 6 = (x − 2)(x − 3) ≤ 0$, so x lies between the roots $inclusive: 2 ≤ x ≤ 3$.","explanationZh":"$x^{2} − 5x + 6 = (x − 2)(x − 3) ≤ 0，解集为两根之间（含端点）：2 ≤ x ≤ 3。$","tags":["easy","quadratic","solution-set"],"status":"published"},{"id":"math-quadratic-inequality-004","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":2,"stem":"Solve $(x − 1)(x + 3) > 0$.","stemZh":"解不等式 $(x − 1)(x + 3) > 0。$","options":[{"id":"A","text":"$(−3, 1)$"},{"id":"B","text":"$[−3, 1]$"},{"id":"C","text":"$(−∞, −3) ∪ (1, +∞)$"},{"id":"D","text":"$(−∞, 1) ∪ (3, +∞)$"}],"answer":"C","explanation":"The roots are −3 and 1 and the product is positive outside $them: x < −3$ or $x > 1$. Both endpoints are excluded because the inequality is strict.","explanationZh":"两根为 $−3 与 1，乘积在两根之外为正：x < −3 或 x > 1；不等式严格，端点不取。$","tags":["easy","quadratic","sign-chart"],"status":"published"},{"id":"math-quadratic-inequality-005","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":2,"stem":"Solve $x^{2} + 4x + 4 ≤ 0$.","stemZh":"解不等式 $x^{2} + 4x + 4 ≤ 0。$","options":[{"id":"A","text":"$ℝ$"},{"id":"B","text":"$[−2, 2]$"},{"id":"C","text":"$∅$"},{"id":"D","text":"{−2}"}],"answer":"D","explanation":"$x^{2} + 4x + 4 = (x + 2)^{2}$, which is never negative. It equals 0 only at $x = −2$, so the solution set is the single-element set {−2}.","explanationZh":"$x^{2} + 4x + 4 = (x + 2)^{2} ≥ 0$ 恒成立，仅在 $x = −2$ 时等于 $0，故解集为单元素集 {−2}。$","tags":["easy","quadratic","perfect-square"],"status":"published"},{"id":"math-quadratic-inequality-006","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":1,"stem":"Solve $x^{2} + 5x + 6 < 0$.","stemZh":"解不等式 $x^{2} + 5x + 6 < 0。$","options":[{"id":"A","text":"$(−3, −2)$"},{"id":"B","text":"$(2, 3)$"},{"id":"C","text":"$(−∞, −3) ∪ (−2, +∞)$"},{"id":"D","text":"$[−3, −2]$"}],"answer":"A","explanation":"$x^{2} + 5x + 6 = (x + 2)(x + 3) < 0$ has roots −3 and −2, and the solution lies strictly between $them: −3 < x < −2$.","explanationZh":"$x^{2} + 5x + 6 = (x + 2)(x + 3) < 0，两根为 −3 与 −2，解集在两根之间：−3 < x < −2。$","tags":["easy","quadratic","factorisation"],"status":"published"},{"id":"math-quadratic-inequality-007","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":2,"stem":"Solve $x^{2} + 2x + 1 > 0$.","stemZh":"解不等式 $x^{2} + 2x + 1 > 0。$","options":[{"id":"A","text":"$ℝ$"},{"id":"B","text":"$(−∞, −1)$"},{"id":"C","text":"$(−∞, −1) ∪ (−1, +∞)$"},{"id":"D","text":"$∅$"}],"answer":"C","explanation":"$x^{2} + 2x + 1 = (x + 1)^{2}$ is positive for every x except $x = −1$, where it is 0. So the solution set is $x ≠ −1$, i.e. $(−∞, −1) ∪ (−1, +∞)$.","explanationZh":"$x^{2} + 2x + 1 = (x + 1)^{2} 除 x = −1$ 时等于 0 外恒为正，故解集为 $x ≠ −1，即 (−∞, −1) ∪ (−1, +∞)。$","tags":["easy","quadratic","perfect-square"],"status":"published"},{"id":"math-quadratic-inequality-008","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":1,"stem":"Solve $x(x − 1) < 0$.","stemZh":"解不等式 $x(x − 1) < 0。$","options":[{"id":"A","text":"$x < 0$"},{"id":"B","text":"$0 < x < 1$"},{"id":"C","text":"$x > 1$"},{"id":"D","text":"$x < 0$ or $x > 1$"}],"answer":"B","explanation":"The roots are 0 and 1 and the product is negative only between $them: 0 < x < 1$.","explanationZh":"两根为 $0 与 1，乘积仅在两根之间为负：0 < x < 1。$","tags":["easy","quadratic","sign-chart"],"status":"published"},{"id":"math-quadratic-inequality-009","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":2,"stem":"Solve $x^{2} + 3x ≥ 0$.","stemZh":"解不等式 $x^{2} + 3x ≥ 0。$","options":[{"id":"A","text":"$[−3, 0]$"},{"id":"B","text":"$(−∞, −3) ∪ (0, +∞)$"},{"id":"C","text":"$[0, +∞)$"},{"id":"D","text":"$(−∞, −3] ∪ [0, +∞)$"}],"answer":"D","explanation":"$x(x + 3) ≥ 0$ is non-negative outside the roots −3 and 0, and both roots satisfy the equality, so $x ≤ −3$ or $x ≥ 0$.","explanationZh":"$x(x + 3) ≥ 0$ 在两根 −3 与 0 之外非负，且两交点处等号成立，故 $x ≤ −3 或 x ≥ 0。$","tags":["easy","quadratic","solution-set"],"status":"published"},{"id":"math-quadratic-inequality-010","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":2,"stem":"Solve $x^{2} + x + 1 > 0$.","stemZh":"解不等式 $x^{2} + x + 1 > 0。$","options":[{"id":"A","text":"$ℝ$"},{"id":"B","text":"$∅$"},{"id":"C","text":"$(−∞, −1) ∪ (1, +∞)$"},{"id":"D","text":"$(0, +∞)$"}],"answer":"A","explanation":"The discriminant is $Δ = 1 − 4 = −3 < 0$ and the leading coefficient is positive, so the quadratic is positive for every real x.","explanationZh":"判别式 $Δ = 1 − 4 = −3 < 0，且二次项系数为正，故对一切实数 x$ 该式恒大于 0。","tags":["easy","quadratic","discriminant"],"status":"published"},{"id":"math-quadratic-inequality-011","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $2x^{2} − 5x − 3 ≤ 0$.","stemZh":"解不等式 $2x^{2} − 5x − 3 ≤ 0。$","options":[{"id":"A","text":"$(−∞, −1/2] ∪ [3, +∞)$"},{"id":"B","text":"$[−1/2, 3]$"},{"id":"C","text":"$[−3, 1/2]$"},{"id":"D","text":"$(−1/2, 3)$"}],"answer":"B","explanation":"$2x^{2} − 5x − 3 = (2x + 1)(x − 3)$, with roots −1/2 and 3. Since the leading coefficient is positive, the expression is $≤ 0$ between the $roots: [−1/2, 3]$.","explanationZh":"$2x^{2} − 5x − 3 = (2x + 1)(x − 3)，两根为 −1/2 与 3；二次项系数为正，故两根之间 ≤ 0，解集为 [−1/2, 3]。$","tags":["medium","quadratic","factorisation"],"status":"published"},{"id":"math-quadratic-inequality-012","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $−x^{2} + 2x + 3 ≥ 0$.","stemZh":"解不等式 $−x^{2} + 2x + 3 ≥ 0。$","options":[{"id":"A","text":"$[−1, 3]$"},{"id":"B","text":"$(−∞, −1] ∪ [3, +∞)$"},{"id":"C","text":"$(−1, 3)$"},{"id":"D","text":"$[−3, 1]$"}],"answer":"A","explanation":"Multiplying by −1 reverses the $sign: x^{2} − 2x − 3 ≤ 0$, i.e. $(x − 3)(x + 1) ≤ 0$, giving $−1 ≤ x ≤ 3$.","explanationZh":"两边同乘 $−1 不等号反向：x^{2} − 2x − 3 ≤ 0，即 (x − 3)(x + 1) ≤ 0，得 −1 ≤ x ≤ 3。$","tags":["medium","quadratic","sign-flip"],"status":"published"},{"id":"math-quadratic-inequality-013","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $x^{2} − 4x + 3 < 0$.","stemZh":"解不等式 $x^{2} − 4x + 3 < 0。$","options":[{"id":"A","text":"$(−∞, 1) ∪ (3, +∞)$"},{"id":"B","text":"$[1, 3]$"},{"id":"C","text":"$(−3, −1)$"},{"id":"D","text":"$(1, 3)$"}],"answer":"D","explanation":"$(x − 1)(x − 3) < 0$ gives $1 < x < 3$; both endpoints are excluded.","explanationZh":"$(x − 1)(x − 3) < 0，解得 1 < x < 3，端点不取。$","tags":["medium","quadratic","sign-chart"],"status":"published"},{"id":"math-quadratic-inequality-014","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(2 − x)(x + 1) ≥ 0$.","stemZh":"解不等式 $(2 − x)(x + 1) ≥ 0。$","options":[{"id":"A","text":"$(−∞, −1] ∪ [2, +∞)$"},{"id":"B","text":"$(−∞, −1) ∪ (2, +∞)$"},{"id":"C","text":"$[−1, 2]$"},{"id":"D","text":"$(−1, 2)$"}],"answer":"C","explanation":"Rewrite as $−(x − 2)(x + 1) ≥ 0$, i.e. $(x − 2)(x + 1) ≤ 0$, which holds between the $roots: −1 ≤ x ≤ 2$. The sign of the leading coefficient reverses the usual 'outside the roots' answer.","explanationZh":"化为 $−(x − 2)(x + 1) ≥ 0，即 (x − 2)(x + 1) ≤ 0，解在两根之间：−1 ≤ x ≤ 2。注意首项系数为负时结论不是“两根之外”。$","tags":["medium","quadratic","sign-flip"],"status":"published"},{"id":"math-quadratic-inequality-015","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $x^{2} − 6x + 9 > 0$.","stemZh":"解不等式 $x^{2} − 6x + 9 > 0。$","options":[{"id":"A","text":"$x ≠ 3$"},{"id":"B","text":"$(−∞, 3) ∪ (3, +∞)$"},{"id":"C","text":"$ℝ$"},{"id":"D","text":"$∅$"}],"answer":"B","explanation":"$x^{2} − 6x + 9 = (x − 3)^{2} > 0$ holds for every x except $x = 3$, so the solution set is $(−∞, 3) ∪ (3, +∞)$.","explanationZh":"$x^{2} − 6x + 9 = (x − 3)^{2} > 0$ 对除 $x = 3$ 外的一切实数成立，故解集为 $(−∞, 3) ∪ (3, +∞)。$","tags":["medium","quadratic","perfect-square"],"status":"published"},{"id":"math-quadratic-inequality-016","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $3x^{2} − 12 ≤ 0$.","stemZh":"解不等式 $3x^{2} − 12 ≤ 0。$","options":[{"id":"A","text":"$(−∞, −2] ∪ [2, +∞)$"},{"id":"B","text":"$(−2, 2)$"},{"id":"C","text":"$(−∞, −4] ∪ [4, +∞)$"},{"id":"D","text":"$[−2, 2]$"}],"answer":"D","explanation":"$3x^{2} ≤ 12$ gives $x^{2} ≤ 4$, so $|x| ≤ 2$, i.e. $−2 ≤ x ≤ 2$.","explanationZh":"$3x^{2} ≤ 12 即 x^{2} ≤ 4，故 |x| ≤ 2，解得 −2 ≤ x ≤ 2。$","tags":["medium","quadratic","absolute-value"],"status":"published"},{"id":"math-quadratic-inequality-017","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $x^{2} + 2x − 8 > 0$.","stemZh":"解不等式 $x^{2} + 2x − 8 > 0。$","options":[{"id":"A","text":"$(−∞, −4) ∪ (2, +∞)$"},{"id":"B","text":"$(−4, 2)$"},{"id":"C","text":"$(−∞, −2) ∪ (4, +∞)$"},{"id":"D","text":"$(2, +∞)$"}],"answer":"A","explanation":"$x^{2} + 2x − 8 = (x + 4)(x − 2) > 0$, which is positive outside the roots −4 and $2: x < −4$ or $x > 2$.","explanationZh":"$x^{2} + 2x − 8 = (x + 4)(x − 2) > 0，在两根 −4 与 2$ 之外为正：x $< −4 或 x > 2。$","tags":["medium","quadratic","factorisation"],"status":"published"},{"id":"math-quadratic-inequality-018","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $x^{2} − x − 12 ≤ 0$.","stemZh":"解不等式 $x^{2} − x − 12 ≤ 0。$","options":[{"id":"A","text":"$(−∞, −3] ∪ [4, +∞)$"},{"id":"B","text":"$(−3, 4)$"},{"id":"C","text":"$[−3, 4]$"},{"id":"D","text":"$[−4, 3]$"}],"answer":"C","explanation":"$(x − 4)(x + 3) ≤ 0$ gives $−3 ≤ x ≤ 4$, with both roots included.","explanationZh":"$(x − 4)(x + 3) ≤ 0，解得 −3 ≤ x ≤ 4，两根均取到。$","tags":["medium","quadratic","solution-set"],"status":"published"},{"id":"math-quadratic-inequality-019","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $4x^{2} − 4x + 1 ≥ 0$.","stemZh":"解不等式 $4x^{2} − 4x + 1 ≥ 0。$","options":[{"id":"A","text":"$ℝ$"},{"id":"B","text":"$[1/2, +∞)$"},{"id":"C","text":"$(−∞, 1/2]$"},{"id":"D","text":"{1/2}"}],"answer":"A","explanation":"$4x^{2} − 4x + 1 = (2x − 1)^{2} ≥ 0$ for every real x, so the solution set is all of $ℝ$. The value $x = 1/2$ makes it zero, which still satisfies $≥ 0$.","explanationZh":"$4x^{2} − 4x + 1 = (2x − 1)^{2} ≥ 0$ 对一切实数成立，故解集为 $ℝ；x = 1/2$ 时等于 $0，也满足 ≥ 0。$","tags":["medium","quadratic","perfect-square"],"status":"published"},{"id":"math-quadratic-inequality-020","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":3,"stem":"Solve $−2x^{2} + 3x + 2 < 0$.","stemZh":"解不等式 $−2x^{2} + 3x + 2 < 0。$","options":[{"id":"A","text":"$(−1/2, 2)$"},{"id":"B","text":"$[−1/2, 2]$"},{"id":"C","text":"$(−∞, −2) ∪ (1/2, +∞)$"},{"id":"D","text":"$(−∞, −1/2) ∪ (2, +∞)$"}],"answer":"D","explanation":"Multiply by $−1: 2x^{2} − 3x − 2 > 0$, i.e. $(2x + 1)(x − 2) > 0$, which holds outside the $roots: x < −1/2$ or $x > 2$.","explanationZh":"同乘 $−1 得 2x^{2} − 3x − 2 > 0，即 (2x + 1)(x − 2) > 0，解在两根之外：x < −1/2 或 x > 2。$","tags":["medium","quadratic","sign-flip"],"status":"published"},{"id":"math-quadratic-inequality-021","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":4,"stem":"For which values of a does $x^{2} − 2x + a > 0$ hold for every real x?","stemZh":"$当 a 取何值时，x^{2} − 2x + a > 0$ 对一切实数 x 恒成立？","options":[{"id":"A","text":"$a ≥ 1$"},{"id":"B","text":"$a > 0$"},{"id":"C","text":"$a > 1$"},{"id":"D","text":"$a < 1$"}],"answer":"C","explanation":"With leading coefficient $1 > 0$ the quadratic is always positive exactly when $Δ < 0: (−2)^{2} − 4a < 0$, so $4 − 4a < 0$ and $a > 1$. At $a = 1$ the expression is $(x − 1)^{2}$, which vanishes at $x = 1$, so the endpoint is excluded.","explanationZh":"二次项系数 $1 > 0，恒正的条件是 Δ < 0：(−2)^{2} − 4a < 0，即 4 − 4a < 0，得 a > 1。a = 1$ 时原式为 $(x − 1)^{2}，在 x = 1$ 处等于 0，故端点不取。","tags":["hard","quadratic","parameter"],"status":"published"},{"id":"math-quadratic-inequality-022","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":4,"stem":"If $x^{2} +$ bx $+ 1 > 0$ for every real x, find the range of b.","stemZh":"$若 x^{2} +$ bx $+ 1 > 0$ 对一切实数 x 恒成立，求 b 的取值范围。","options":[{"id":"A","text":"$b ≤ 2$"},{"id":"B","text":"$−2 < b < 2$"},{"id":"C","text":"$−2 ≤ b ≤ 2$"},{"id":"D","text":"$b > 0$"}],"answer":"B","explanation":"Always positive requires $Δ < 0: b^{2} − 4 < 0$, i.e. $b^{2} < 4$, giving $−2 < b < 2$. Equality $b = \\pm2$ would give a double root where the expression is 0.","explanationZh":"恒成立需 $Δ < 0：b^{2} − 4 < 0，即 b^{2} < 4，得 −2 < b < 2；b = \\pm2$ 时有二重根，原式可取到 0，不满足严格大于 0。","tags":["hard","quadratic","parameter"],"status":"published"},{"id":"math-quadratic-inequality-023","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":4,"stem":"How many integers x satisfy $x^{2} − 3x − 4 < 0$?","stemZh":"满足 $x^{2} − 3x − 4 < 0$ 的整数 x 有多少个？","options":[{"id":"A","text":"5"},{"id":"B","text":"6"},{"id":"C","text":"3"},{"id":"D","text":"4"}],"answer":"D","explanation":"$(x − 4)(x + 1) < 0$ gives $−1 < x < 4$. The integers strictly between −1 and 4 are 0, 1, 2, 3 — four of them.","explanationZh":"$(x − 4)(x + 1) < 0 得 −1 < x < 4，其间整数为 0、1、2、3，共 4$ 个。","tags":["hard","quadratic","integer-solutions"],"status":"published"},{"id":"math-quadratic-inequality-024","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":4,"stem":"Let $a < 0$ and suppose $ax^{2} +$ bx $+ c = 0$ has two roots $r_{1} < r_{2}$. The solution set of $ax^{2} +$ bx $+ c > 0 is$:","stemZh":"$设 a < 0，且 ax^{2} +$ bx $+ c = 0$ 有两根 $r_{1} < r_{2}，则 ax^{2} +$ bx $+ c > 0$ 的解集为？","options":[{"id":"A","text":"$(r_{1}, r_{2})$"},{"id":"B","text":"$(−∞, r_{1}) ∪ (r_{2}, +∞)$"},{"id":"C","text":"$(−∞, r_{1})$"},{"id":"D","text":"$∅$"}],"answer":"A","explanation":"With $a < 0$ the parabola opens downward, so the expression is positive between the two roots and negative outside them — the opposite of the $a > 0$ case.","explanationZh":"$a < 0$ 时抛物线开口向下，故两根之间为正、两根之外为负，与 $a > 0$ 的情形正好相反。","tags":["hard","quadratic","sign-chart"],"status":"published"},{"id":"math-quadratic-inequality-025","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":4,"stem":"If the solution set of $(x − 1)(x − a) < 0$ is $(1, 3)$, find a.","stemZh":"$若 (x − 1)(x − a) < 0$ 的解集为 $(1, 3)，求$ a。","options":[{"id":"A","text":"−3"},{"id":"B","text":"1"},{"id":"C","text":"3"},{"id":"D","text":"−1"}],"answer":"C","explanation":"The solution of $(x − 1)(x − a) < 0$ is the open interval between 1 and a. Matching $(1, 3)$ gives $a = 3. (If a = 1\\text{ the set would be }empty.)$","explanationZh":"$(x − 1)(x − a) < 0$ 的解集是 1 与 a 之间的开区间，与 $(1, 3)$ 对比得 $a = 3（若 a = 1$ 则解集为空）。","tags":["hard","quadratic","parameter"],"status":"published"},{"id":"math-quadratic-inequality-026","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":5,"stem":"Find the set of x satisfying both $x^{2} − 2x − 3 > 0$ and $x^{2} − 9 < 0$.","stemZh":"求同时满足 $x^{2} − 2x − 3 > 0 与 x^{2} − 9 < 0 的 x$ 的集合。","options":[{"id":"A","text":"$(−3, 3)$"},{"id":"B","text":"$(−3, −1)$"},{"id":"C","text":"$(−3, −1]$"},{"id":"D","text":"$(−1, 3)$"}],"answer":"B","explanation":"$x^{2} − 2x − 3 > 0$ gives $x < −1$ or $x > 3; x^{2} − 9 < 0$ gives $−3 < x < 3. Intersecting: (−3, −1)$. Both endpoints are open since −3 and −1 are excluded by the strict inequalities.","explanationZh":"$由 x^{2} − 2x − 3 > 0 得 x < −1 或 x > 3；由 x^{2} − 9 < 0 得 −3 < x < 3。取交集为 (−3, −1)，两端点因不等式严格均为开。$","tags":["hard","quadratic","system"],"status":"published"},{"id":"math-quadratic-inequality-027","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":5,"stem":"For which values of m does $x^{2} −$ mx $+ 4 ≥ 0$ hold for every real x?","stemZh":"$当 m 取何值时，x^{2} −$ mx $+ 4 ≥ 0$ 对一切实数 x 恒成立？","options":[{"id":"A","text":"$−4 < m < 4$"},{"id":"B","text":"$m ≤ 4$"},{"id":"C","text":"$m ≥ −4$"},{"id":"D","text":"$−4 ≤ m ≤ 4$"}],"answer":"D","explanation":"Because the inequality is non-strict, we only need the quadratic to never go below $zero: Δ ≤ 0$. Here $Δ = m^{2} − 16 ≤ 0$, so $−4 ≤ m ≤ 4. (Strict\\text{ positivity would require strict }inequalities.)$","explanationZh":"由于不等式不严格，只需 $Δ ≤ 0：Δ = m^{2} − 16 ≤ 0，得 −4 ≤ m ≤ 4（若为严格大于 0，则端点不取）。$","tags":["hard","quadratic","parameter"],"status":"published"},{"id":"math-quadratic-inequality-028","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":5,"stem":"The solution set of $x^{2} +$ ax $+ 4 < 0$ is $(1, 4)$. Find a.","stemZh":"$若 x^{2} +$ ax $+ 4 < 0$ 的解集为 $(1, 4)，求$ a。","options":[{"id":"A","text":"−5"},{"id":"B","text":"−4"},{"id":"C","text":"5"},{"id":"D","text":"4"}],"answer":"A","explanation":"The endpoints 1 and 4 are the roots, and their product is 4, matching the constant term. Their sum is $5 = −a$, so $a = −5. Check: x^{2} − 5x + 4 = (x − 1)(x − 4) < 0$ gives $1 < x < 4$.","explanationZh":"端点 1 与 4 即两根，其积为 4 与常数项一致；其和为 $5 =$ −a，故 $a = −5。验证：x^{2} − 5x + 4 = (x − 1)(x − 4) < 0 得 1 < x < 4。$","tags":["hard","quadratic","parameter"],"status":"published"},{"id":"math-quadratic-inequality-029","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":5,"stem":"If x satisfies $x^{2} − 3x + 2 < 0$, what can be said about the value of $x^{2} − 5x + 6$?","stemZh":"若 x 满足 $x^{2} − 3x + 2 < 0，则 x^{2} − 5x + 6$ 的值？","options":[{"id":"A","text":"It is negative"},{"id":"B","text":"It is zero"},{"id":"C","text":"It is positive"},{"id":"D","text":"Its sign cannot be determined"}],"answer":"C","explanation":"$x^{2} − 3x + 2 < 0$ gives $1 < x < 2$. Then $x^{2} − 5x + 6 = (x − 2)(x − 3)$, where both factors are negative on $(1, 2)$, so the product is positive.","explanationZh":"$由 x^{2} − 3x + 2 < 0 得 1 < x < 2；此时 x^{2} − 5x + 6 = (x − 2)(x − 3)$ 的两个因式均为负，故乘积为正。","tags":["hard","quadratic","reasoning"],"status":"published"},{"id":"math-quadratic-inequality-030","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"quadratic-inequality","type":"single-choice","difficulty":5,"stem":"The solution set of $ax^{2} +$ bx $+ c < 0$ is $(−∞, 1) ∪ (3, +∞)$. What is the sign of $ax^{2} +$ bx + c at $x = 0$?","stemZh":"$若 ax^{2} +$ bx $+ c < 0$ 的解集为 $(−∞, 1) ∪ (3, +∞)，则该式在 x = 0$ 处的符号为？","options":[{"id":"A","text":"Positive"},{"id":"B","text":"Negative"},{"id":"C","text":"Zero"},{"id":"D","text":"It depends on the value of a"}],"answer":"B","explanation":"Being negative outside the roots 1 and 3 means $a < 0$ and $ax^{2} +$ bx $+ c = a(x − 1)(x − 3)$. At $x = 0$ this is $a(−1)(−3) = 3a < 0$, so the value is negative.","explanationZh":"在两根 1、3 之外为负，说明 $a < 0 且 ax^{2} +$ bx $+ c = a(x − 1)(x − 3)；x = 0$ 时其值为 $a(−1)(−3) = 3a < 0，故为负。$","tags":["hard","quadratic","parameter"],"status":"published"},{"id":"math-rational-inequality-001","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":1,"stem":"Solve $(x − 1)/(x + 2) > 0$.","stemZh":"解不等式 $(x − 1)/(x + 2) > 0。$","options":[{"id":"A","text":"$(−2, 1)$"},{"id":"B","text":"$(−∞, 1) ∪ (2, +∞)$"},{"id":"C","text":"$(−∞, −2] ∪ [1, +∞)$"},{"id":"D","text":"$(−∞, −2) ∪ (1, +∞)$"}],"answer":"D","explanation":"The critical points are $−2 (denominator zero$, always excluded) and $1 (numerator zero). A$ quotient is positive when numerator and denominator share a sign, which happens outside the two $points: x < −2$ or $x > 1$.","explanationZh":"临界点为 $−2（分母为零，必排除）与 1（分子为零）。分式为正即分子分母同号，解在两点之外：x < −2 或 x > 1。$","tags":["easy","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-002","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":1,"stem":"Solve $1/x > 0$.","stemZh":"解不等式 $1/x > 0。$","options":[{"id":"A","text":"$(−∞, 0)$"},{"id":"B","text":"$(0, +∞)$"},{"id":"C","text":"$(−∞, 0) ∪ (0, +∞)$"},{"id":"D","text":"$ℝ$"}],"answer":"B","explanation":"The numerator 1 is positive, so the quotient is positive exactly when $x > 0$. The value $x = 0$ is not allowed.","explanationZh":"分子 1 为正，故分式为正当且仅当 $x > 0；x = 0$ 无意义。","tags":["easy","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-003","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":2,"stem":"Solve $(x + 1)/(x − 3) ≤ 0$.","stemZh":"解不等式 $(x + 1)/(x − 3) ≤ 0。$","options":[{"id":"A","text":"$[−1, 3)$"},{"id":"B","text":"$[−1, 3]$"},{"id":"C","text":"$(−1, 3]$"},{"id":"D","text":"$(−∞, −1] ∪ [3, +∞)$"}],"answer":"A","explanation":"The quotient is non-positive between the critical points −1 and 3. The endpoint −1 is included (numerator zero) but 3 is not, since the denominator vanishes there.","explanationZh":"分式在临界点 −1 与 3 之间非正；−1 处分子为零可取，3 处分母为零必须排除，故为 $[−1, 3)。$","tags":["easy","rational","endpoints"],"status":"published"},{"id":"math-rational-inequality-004","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":1,"stem":"Solve $(x − 2)/(x + 1) < 0$.","stemZh":"解不等式 $(x − 2)/(x + 1) < 0。$","options":[{"id":"A","text":"$(−∞, −1) ∪ (2, +∞)$"},{"id":"B","text":"$(−1, 2]$"},{"id":"C","text":"$(−1, 2)$"},{"id":"D","text":"$[−1, 2)$"}],"answer":"C","explanation":"Numerator and denominator must have opposite signs, which happens strictly between −1 and 2; both endpoints are excluded (the denominator zero at −1, and the inequality is strict at $2)$.","explanationZh":"分子分母异号时位于 −1 与 2 之间；−1 处分母为零、2 处不等式严格，两端点都不取，故为 $(−1, 2)。$","tags":["easy","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-005","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":2,"stem":"Solve $2/(x − 1) ≥ 0$.","stemZh":"解不等式 $2/(x − 1) ≥ 0。$","options":[{"id":"A","text":"$[1, +∞)$"},{"id":"B","text":"$(−∞, 1)$"},{"id":"C","text":"$(−∞, 1]$"},{"id":"D","text":"$(1, +∞)$"}],"answer":"D","explanation":"The numerator 2 is always positive and never zero, so the quotient is positive exactly when $x − 1 > 0$, i.e. $x > 1$. The value $x = 1$ is excluded by the denominator.","explanationZh":"分子 2 恒正且不为零，故分式非负当且仅当 $x − 1 > 0，即 x > 1；x = 1$ 时分母为零须排除。","tags":["easy","rational","endpoints"],"status":"published"},{"id":"math-rational-inequality-006","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":2,"stem":"Solve $(x − 3)/(x − 1) ≥ 0$.","stemZh":"解不等式 $(x − 3)/(x − 1) ≥ 0。$","options":[{"id":"A","text":"$(−∞, 1) ∪ [3, +∞)$"},{"id":"B","text":"$(−∞, 1] ∪ [3, +∞)$"},{"id":"C","text":"$(1, 3]$"},{"id":"D","text":"$[1, 3)$"}],"answer":"A","explanation":"Same signs outside the critical points 1 and 3. The point $x = 3$ is included (numerator zero) but $x = 1$ must be excluded because the denominator is zero there.","explanationZh":"同号区间在临界点 1、3 之外；x $= 3$ 处分子为零可取，x $= 1$ 处分母为零须排除，故为 $(−∞, 1) ∪ [3, +∞)。$","tags":["easy","rational","endpoints"],"status":"published"},{"id":"math-rational-inequality-007","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":1,"stem":"Solve $(x + 2)/(x − 2) ≤ 0$.","stemZh":"解不等式 $(x + 2)/(x − 2) ≤ 0。$","options":[{"id":"A","text":"$(−2, 2)$"},{"id":"B","text":"$[−2, 2)$"},{"id":"C","text":"$[−2, 2]$"},{"id":"D","text":"$(−∞, −2] ∪ [2, +∞)$"}],"answer":"B","explanation":"Between −2 and 2 the quotient is non-positive; −2 is allowed (numerator zero) while 2 is forbidden (denominator zero).","explanationZh":"在 −2 与 2 之间分式非正；−2 处分子为零可取，2 处分母为零须排除，故为 $[−2, 2)。$","tags":["easy","rational","endpoints"],"status":"published"},{"id":"math-rational-inequality-008","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":1,"stem":"Solve $1/(x + 3) < 0$.","stemZh":"解不等式 $1/(x + 3) < 0。$","options":[{"id":"A","text":"$(−3, +∞)$"},{"id":"B","text":"$(−∞, 3)$"},{"id":"C","text":"$(−∞, −3)$"},{"id":"D","text":"$(3, +∞)$"}],"answer":"C","explanation":"The numerator is positive, so the quotient is negative exactly when the denominator is $negative: x + 3 < 0$, i.e. $x < −3$.","explanationZh":"分子为正，分式为负当且仅当分母为负：x $+ 3 < 0，即 x < −3。$","tags":["easy","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-009","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":2,"stem":"Solve $x/(x − 4) > 0$.","stemZh":"解不等式 $x/(x − 4) > 0。$","options":[{"id":"A","text":"$(0, 4)$"},{"id":"B","text":"$(−∞, 0] ∪ [4, +∞)$"},{"id":"C","text":"$(0, +∞)$"},{"id":"D","text":"$(−∞, 0) ∪ (4, +∞)$"}],"answer":"D","explanation":"The critical points are 0 and 4. The quotient is positive outside $them: x < 0$ or $x > 4$; neither endpoint is included.","explanationZh":"临界点为 $0 与 4，同号区间在两点之外：x < 0 或 x > 4，两端点均不取。$","tags":["easy","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-010","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":2,"stem":"Solve $(x − 1)/(2 − x) ≥ 0$.","stemZh":"解不等式 $(x − 1)/(2 − x) ≥ 0。$","options":[{"id":"A","text":"$(1, 2]$"},{"id":"B","text":"$[1, 2]$"},{"id":"C","text":"$[1, 2)$"},{"id":"D","text":"$(−∞, 1] ∪ [2, +∞)$"}],"answer":"A","explanation":"Critical points are 1 and 2. Testing the middle: at $x = 3/2$ the value is $(1/2)/(1/2) = 1 > 0$, so the quotient is non-negative on $(1, 2]$. The point $x = 1$ is included (numerator zero) and $x = 2$ excluded (denominator zero).","explanationZh":"临界点为 $1 与 2。取 x = 3/2 检验：(1/2)/(1/2) = 1 > 0，故解集为 (1, 2]；x = 1$ 处分子为零可取，x $= 2$ 处分母为零须排除。","tags":["easy","rational","endpoints"],"status":"published"},{"id":"math-rational-inequality-011","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(x − 1)/(x + 1) ≤ 1$.","stemZh":"解不等式 $(x − 1)/(x + 1) ≤ 1。$","options":[{"id":"A","text":"$(−∞, −1)$"},{"id":"B","text":"$(−1, +∞)$"},{"id":"C","text":"$[1, +∞)$"},{"id":"D","text":"$(−∞, −1) ∪ [1, +∞)$"}],"answer":"B","explanation":"Move everything to one $side: (x − 1)/(x + 1) − 1 ≤ 0 → −2/(x + 1) ≤ 0 → 2/(x + 1) ≥ 0$. Since $2 > 0$ this needs $x + 1 > 0$, i.e. $x > −1$. Never multiply across by a denominator whose sign is unknown.","explanationZh":"$移项通分：(x − 1)/(x + 1) − 1 ≤ 0 → −2/(x + 1) ≤ 0 → 2/(x + 1) ≥ 0。因 2 > 0，故需 x + 1 > 0，即 x > −1。切勿直接乘以符号未知的分母。$","tags":["medium","rational","rearrangement"],"status":"published"},{"id":"math-rational-inequality-012","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(2x − 1)/(x + 3) > 0$.","stemZh":"解不等式 $(2x − 1)/(x + 3) > 0。$","options":[{"id":"A","text":"$(−3, 1/2)$"},{"id":"B","text":"$(−∞, 1/2) ∪ (3, +∞)$"},{"id":"C","text":"$(−∞, −3) ∪ (1/2, +∞)$"},{"id":"D","text":"$(1/2, +∞)$"}],"answer":"C","explanation":"The numerator vanishes at $x = 1/2$ and the denominator at $x = −3$. Same signs occur outside these $points: x < −3$ or $x > 1/2$.","explanationZh":"分子零点 $x = 1/2，分母零点 x = −3；同号区间在两点之外：x < −3 或 x > 1/2。$","tags":["medium","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-013","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(x + 1)/(x − 2) ≥ 2$.","stemZh":"解不等式 $(x + 1)/(x − 2) ≥ 2。$","options":[{"id":"A","text":"$(−∞, 2) ∪ [5, +∞)$"},{"id":"B","text":"$[2, 5]$"},{"id":"C","text":"$(2, 5)$"},{"id":"D","text":"$(2, 5]$"}],"answer":"D","explanation":"$(x + 1)/(x − 2) − 2 ≥ 0 → (5 − x)/(x − 2) ≥ 0 → (x − 5)/(x − 2) ≤ 0$, which holds on $(2, 5]: x = 2$ is excluded by the $denominator, x = 5$ is included since the numerator vanishes.","explanationZh":"$移项：(x + 1)/(x − 2) − 2 ≥ 0 → (5 − x)/(x − 2) ≥ 0 → (x − 5)/(x − 2) ≤ 0，解集为 (2, 5]；x = 2$ 分母为零须排除，x $= 5$ 分子为零可取。","tags":["medium","rational","rearrangement"],"status":"published"},{"id":"math-rational-inequality-014","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(x − 1)/(x^{2} + 1) > 0$.","stemZh":"解不等式 $(x − 1)/(x^{2} + 1) > 0。$","options":[{"id":"A","text":"$(1, +∞)$"},{"id":"B","text":"$(−∞, 1)$"},{"id":"C","text":"$(−1, 1)$"},{"id":"D","text":"$(−∞, −1) ∪ (1, +∞)$"}],"answer":"A","explanation":"$x^{2} + 1 > 0$ for every real x, so the sign of the fraction is the sign of x − 1: the solution is $x > 1$. Here the denominator can safely be ignored because it is always positive.","explanationZh":"$x^{2} + 1$ 对一切实数恒正，故分式的符号由 x − 1 决定，解为 $x > 1；此处分母恒正，可以放心“忽略”。$","tags":["medium","rational","always-positive-denominator"],"status":"published"},{"id":"math-rational-inequality-015","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(x + 2)/(x − 1) < 2$.","stemZh":"解不等式 $(x + 2)/(x − 1) < 2。$","options":[{"id":"A","text":"$(−∞, 1) ∪ (4, +∞)$"},{"id":"B","text":"$(1, 4)$"},{"id":"C","text":"$(−∞, 1) ∪ [4, +∞)$"},{"id":"D","text":"$(1, 4]$"}],"answer":"A","explanation":"$(x + 2)/(x − 1) − 2 < 0 → (4 − x)/(x − 1) < 0 → (x − 4)/(x − 1) > 0$, so $x < 1$ or $x > 4$, with both endpoints open.","explanationZh":"移项得 $(x + 2)/(x − 1) − 2 < 0 → (4 − x)/(x − 1) < 0 → (x − 4)/(x − 1) > 0，故 x < 1 或 x > 4，两端点均开。$","tags":["medium","rational","rearrangement"],"status":"published"},{"id":"math-rational-inequality-016","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(3x − 1)/(2 − x) ≤ 0$.","stemZh":"解不等式 $(3x − 1)/(2 − x) ≤ 0。$","options":[{"id":"A","text":"$[1/3, 2]$"},{"id":"B","text":"$(−∞, 1/3] ∪ [2, +∞)$"},{"id":"C","text":"$(−∞, 1/3] ∪ (2, +∞)$"},{"id":"D","text":"$(1/3, 2)$"}],"answer":"C","explanation":"Rewrite as $(3x − 1)/(x − 2) ≥ 0 (multiplying\\text{ by }−1)$. Critical points 1/3 and 2; same signs outside them, so $x ≤ 1/3$ or $x > 2$, with $x = 2$ excluded by the denominator.","explanationZh":"同乘 −1 化为 $(3x − 1)/(x − 2) ≥ 0，临界点为 1/3 与 2；同号在两点之外，故 x ≤ 1/3 或 x > 2，其中 x = 2$ 分母为零须排除。","tags":["medium","rational","sign-flip"],"status":"published"},{"id":"math-rational-inequality-017","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(x^{2} − 9)/(x + 1) ≤ 0$.","stemZh":"解不等式 $(x^{2} − 9)/(x + 1) ≤ 0。$","options":[{"id":"A","text":"$[−3, −1) ∪ (3, +∞)$"},{"id":"B","text":"$(−∞, −3] ∪ (−1, 3]$"},{"id":"C","text":"$(−∞, −3] ∪ [−1, 3]$"},{"id":"D","text":"$[−3, −1] ∪ [3, +∞)$"}],"answer":"B","explanation":"Critical points are $−3, −1 (denominator)$ and 3. A sign chart gives: negative on $(−∞, −3)$, zero at −3, positive on $(−3, −1)$, undefined at −1, negative on $(−1, 3)$, zero at 3, positive on $(3, +∞)$. Hence $(−∞, −3] ∪ (−1, 3]$.","explanationZh":"临界点为 $−3、−1（分母）与 3。符号表：(−∞, −3) 为负，−3 处为零，(−3, −1) 为正，−1 处无意义，(−1, 3) 为负，3 处为零，(3, +∞)$ 为正，故解集为 $(−∞, −3] ∪ (−1, 3]。$","tags":["medium","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-018","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $3/(x − 2) ≤ 1$.","stemZh":"解不等式 $3/(x − 2) ≤ 1。$","options":[{"id":"A","text":"$(2, 5]$"},{"id":"B","text":"$(−∞, 2] ∪ [5, +∞)$"},{"id":"C","text":"$[2, 5]$"},{"id":"D","text":"$(−∞, 2) ∪ [5, +∞)$"}],"answer":"D","explanation":"$3/(x − 2) − 1 ≤ 0 → (5 − x)/(x − 2) ≤ 0 → (x − 5)/(x − 2) ≥ 0$, so $x < 2$ or $x ≥ 5$. Note $x = 2$ is excluded and $x = 5$ is included.","explanationZh":"$移项：3/(x − 2) − 1 ≤ 0 → (5 − x)/(x − 2) ≤ 0 → (x − 5)/(x − 2) ≥ 0，故 x < 2 或 x ≥ 5；其中 x = 2$ 排除，x $= 5$ 可取。","tags":["medium","rational","rearrangement"],"status":"published"},{"id":"math-rational-inequality-019","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $1/(x − 1) < 1/(x + 1)$.","stemZh":"解不等式 $1/(x − 1) < 1/(x + 1)。$","options":[{"id":"A","text":"$(1, +∞)$"},{"id":"B","text":"$(−1, 1)$"},{"id":"C","text":"$(−∞, −1)$"},{"id":"D","text":"$(−∞, −1) ∪ (1, +∞)$"}],"answer":"B","explanation":"Bring to one $side: 1/(x − 1) − 1/(x + 1) < 0 → 2/((x − 1)(x + 1)) < 0 → (x − 1)(x + 1) < 0$, giving $−1 < x < 1$.","explanationZh":"$移项通分：1/(x − 1) − 1/(x + 1) < 0 → 2/((x − 1)(x + 1)) < 0 → (x − 1)(x + 1) < 0，得 −1 < x < 1。$","tags":["medium","rational","rearrangement"],"status":"published"},{"id":"math-rational-inequality-020","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":3,"stem":"Solve $(x − 1)/(x + 2) ≥ 1$.","stemZh":"解不等式 $(x − 1)/(x + 2) ≥ 1。$","options":[{"id":"A","text":"$(−2, +∞)$"},{"id":"B","text":"$(−∞, −2]$"},{"id":"C","text":"$∅$"},{"id":"D","text":"$(−∞, −2)$"}],"answer":"D","explanation":"$(x − 1)/(x + 2) − 1 ≥ 0 → −3/(x + 2) ≥ 0 → x + 2 < 0$, so $x < −2$. The endpoint −2 is excluded because the denominator is zero there.","explanationZh":"$移项：(x − 1)/(x + 2) − 1 ≥ 0 → −3/(x + 2) ≥ 0 → x + 2 < 0，故 x < −2；−2$ 处分母为零须排除。","tags":["medium","rational","rearrangement"],"status":"published"},{"id":"math-rational-inequality-021","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":4,"stem":"If the solution set of $(x − a)/(x − 2) < 0$ is $(a, 2)$, what condition must a satisfy?","stemZh":"$若 (x − a)/(x − 2) < 0$ 的解集为 $(a, 2)，则 a$ 应满足？","options":[{"id":"A","text":"$a > 2$"},{"id":"B","text":"$a < 2$"},{"id":"C","text":"$a ≤ 2$"},{"id":"D","text":"$a = 2$"}],"answer":"B","explanation":"The solution always lies between the two critical points a and 2, so we need $a < 2$ for the interval to be $(a, 2)$. If $a > 2$ the interval would be $(2, a)$, and if $a = 2$ the fraction equals 1, giving no solution.","explanationZh":"解集总在两临界点之间，要写成 $(a, 2)$ 必须 $a < 2；若 a > 2$ 则为 $(2, a)，若 a = 2$ 则分式恒为 $1（x ≠ 2），无解。$","tags":["hard","rational","parameter"],"status":"published"},{"id":"math-rational-inequality-022","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":4,"stem":"Solve $(x^{2} − 4)/(x^{2} − 1) ≤ 0$.","stemZh":"解不等式 $(x^{2} − 4)/(x^{2} − 1) ≤ 0。$","options":[{"id":"A","text":"$[−2, −1] ∪ [1, 2]$"},{"id":"B","text":"$(−2, −1) ∪ (1, 2)$"},{"id":"C","text":"$[−2, −1) ∪ (1, 2]$"},{"id":"D","text":"$(−∞, −2] ∪ [2, +∞)$"}],"answer":"C","explanation":"Critical points: −2, −1, 1, 2, where $\\pm1$ come from the denominator and are always excluded. The quotient is $≤ 0$ on $[−2, −1)$ and on $(1, 2]$; the zeros −2 and 2 are included, −1 and 1 are not.","explanationZh":"临界点为 $−2、−1、1、2，其中 \\pm1$ 来自分母必须排除；分式 $≤ 0$ 的区间为 $[−2, −1) 与 (1, 2]（零点 −2、2 可取，−1、1$ 不可取）。","tags":["hard","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-023","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":4,"stem":"Solve $(x − 1)/(x^{2} − x − 6) ≥ 0$.","stemZh":"解不等式 $(x − 1)/(x^{2} − x − 6) ≥ 0。$","options":[{"id":"A","text":"$(−2, 1] ∪ (3, +∞)$"},{"id":"B","text":"$[−2, 1] ∪ [3, +∞)$"},{"id":"C","text":"$(−∞, −2) ∪ (1, 3)$"},{"id":"D","text":"$(−2, 3)$"}],"answer":"A","explanation":"The denominator factors as $(x − 3)(x + 2)$, so the critical points are −2 and $3 (excluded)$ and $1 (included$, numerator zero). A sign chart gives non-negative on $(−2, 1]$ and on $(3, +∞)$.","explanationZh":"分母分解为 $(x − 3)(x + 2)，临界点为 −2、3（须排除）与 1（可取）。符号表显示非负区间为 (−2, 1] 与 (3, +∞)。$","tags":["hard","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-024","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":4,"stem":"How many integers x satisfy $(x − 1)/(x + 3) ≤ 0$?","stemZh":"满足 $(x − 1)/(x + 3) ≤ 0$ 的整数 x 有多少个？","options":[{"id":"A","text":"3"},{"id":"B","text":"4"},{"id":"C","text":"5"},{"id":"D","text":"2"}],"answer":"B","explanation":"The solution set is $(−3, 1]: −3$ is excluded (denominator zero) and 1 is included. The integers are −2, −1, 0, 1, which is 4 of them.","explanationZh":"解集为 $(−3, 1]：−3 处分母为零须排除，1$ 处分子为零可取。整数为 −2、−1、0、1，共 4 个。","tags":["hard","rational","integer-solutions"],"status":"published"},{"id":"math-rational-inequality-025","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":5,"stem":"Solve $(x + 1)/(x − 2) > (x + 1)/(x − 3)$.","stemZh":"解不等式 $(x + 1)/(x − 2) > (x + 1)/(x − 3)。$","options":[{"id":"A","text":"$(−1, 2) ∪ (3, +∞)$"},{"id":"B","text":"$(−∞, −1) ∪ (2, 3]$"},{"id":"C","text":"$(−∞, −1) ∪ (2, 3)$"},{"id":"D","text":"$(−1, 3)$"}],"answer":"C","explanation":"$Subtract: (x + 1)[1/(x − 2) − 1/(x − 3)] > 0 → (x + 1)·(−1)/((x − 2)(x − 3)) > 0 → (x + 1)/((x − 2)(x − 3)) < 0$. With critical points −1, 2, 3 the expression is negative on $(−∞, −1)$ and on $(2, 3)$.","explanationZh":"$作差：(x + 1)[1/(x − 2) − 1/(x − 3)] > 0 → −(x + 1)/((x − 2)(x − 3)) > 0 → (x + 1)/((x − 2)(x − 3)) < 0。临界点 −1、2、3，负值区间为 (−∞, −1) 与 (2, 3)。$","tags":["hard","rational","comparison"],"status":"published"},{"id":"math-rational-inequality-026","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":4,"stem":"Solve $(x^{2} − 3x + 2)/(x − 3) ≤ 0$.","stemZh":"解不等式 $(x^{2} − 3x + 2)/(x − 3) ≤ 0。$","options":[{"id":"A","text":"$(−∞, 1] ∪ [2, 3)$"},{"id":"B","text":"$[1, 2] ∪ [3, +∞)$"},{"id":"C","text":"$(−∞, 1) ∪ (2, 3]$"},{"id":"D","text":"$(1, 2)$"}],"answer":"A","explanation":"The numerator factors as $(x − 1)(x − 2)$. Critical points $1, 2 (included)$ and $3 (excluded)$. The quotient is $≤ 0$ on $(−∞, 1]$ and on $[2, 3)$.","explanationZh":"分子分解为 $(x − 1)(x − 2)，临界点为 1、2（可取）与 3（须排除）。分式 ≤ 0$ 的区间为 $(−∞, 1] 与 [2, 3)。$","tags":["hard","rational","sign-chart"],"status":"published"},{"id":"math-rational-inequality-027","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":5,"stem":"If the solution set of $(x − a)/(x + 1) ≥ 0$ is $(−∞, −1) ∪ [a, +∞)$, what condition must a satisfy?","stemZh":"$若 (x − a)/(x + 1) ≥ 0$ 的解集为 $(−∞, −1) ∪ [a, +∞)，则 a$ 应满足？","options":[{"id":"A","text":"$a < −1$"},{"id":"B","text":"$a > −1$"},{"id":"C","text":"$a ≥ −1$"},{"id":"D","text":"$a = −1$"}],"answer":"B","explanation":"The solution lies outside the two critical points a and −1. For it to read $(−∞, −1) ∪ [a, +∞)$ the smaller point must be −1, so $a > −1$. If $a < −1$ the set would be $(−∞, a] ∪ (−1, +∞)$.","explanationZh":"解集在两临界点 a 与 −1 之外。要写成 $(−∞, −1) ∪ [a, +∞)，较小者必须是 −1，故 a > −1；若 a < −1，则解集为 (−∞, a] ∪ (−1, +∞)。$","tags":["hard","rational","parameter"],"status":"published"},{"id":"math-rational-inequality-028","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":5,"stem":"Solve $(x − 2)/(x^{2} − 4) ≥ 0$.","stemZh":"解不等式 $(x − 2)/(x^{2} − 4) ≥ 0。$","options":[{"id":"A","text":"$(−2, 2) ∪ (2, +∞)$"},{"id":"B","text":"$(−2, +∞)$"},{"id":"C","text":"$(−2, 2)$"},{"id":"D","text":"$(2, +∞)$"}],"answer":"A","explanation":"The domain excludes both $x = −2$ and $x = 2$. On that domain $(x − 2)/(x^{2} − 4)$ simplifies to $1/(x + 2)$, which is $≥ 0$ when $x + 2 > 0$. So $x > −2$ with $x ≠ 2$, i.e. $(−2, 2) ∪ (2, +∞)$. Cancelling the factor $(x − 2)$ without recording $x ≠ 2$ is the classic error.","explanationZh":"定义域要求 $x ≠ −2 且 x ≠ 2。在定义域内原式化为 1/(x + 2)，其 ≥ 0 当 x + 2 > 0，即 x > −2 且 x ≠ 2，故为 (−2, 2) ∪ (2, +∞)。约去 (x − 2)$ 却忘记 $x ≠ 2$ 是典型错误。","tags":["hard","rational","domain"],"status":"published"},{"id":"math-rational-inequality-029","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":5,"stem":"Solve $(x + 2)/(x − 1) ≥ (x + 3)/(x − 2)$.","stemZh":"解不等式 $(x + 2)/(x − 1) ≥ (x + 3)/(x − 2)。$","options":[{"id":"A","text":"$[−1/2, 1) ∪ (2, +∞)$"},{"id":"B","text":"$(−∞, −1/2) ∪ (1, 2]$"},{"id":"C","text":"$(−∞, −1/2] ∪ (1, 2)$"},{"id":"D","text":"$(−1/2, 1) ∪ (2, +∞)$"}],"answer":"C","explanation":"Subtract and $combine: [(x + 2)(x − 2) − (x + 3)(x − 1)]/((x − 1)(x − 2)) ≥ 0 → (−2x − 1)/((x − 1)(x − 2)) ≥ 0 → (2x + 1)/((x − 1)(x − 2)) ≤ 0$. Critical points $−1/2 (included), 1$ and $2 (excluded)$ give $(−∞, −1/2] ∪ (1, 2)$.","explanationZh":"$作差通分：[(x + 2)(x − 2) − (x + 3)(x − 1)]/((x − 1)(x − 2)) ≥ 0 → (−2x − 1)/((x − 1)(x − 2)) ≥ 0 → (2x + 1)/((x − 1)(x − 2)) ≤ 0。临界点 −1/2（可取）、1 与 2（须排除），得 (−∞, −1/2] ∪ (1, 2)。$","tags":["hard","rational","comparison"],"status":"published"},{"id":"math-rational-inequality-030","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"rational-inequality","type":"single-choice","difficulty":5,"stem":"How many integers x satisfy both $(x − 1)/(x + 2) ≥ 0$ and $x^{2} ≤ 9$?","stemZh":"同时满足 $(x − 1)/(x + 2) ≥ 0 与 x^{2} ≤ 9$ 的整数 x 有多少个？","options":[{"id":"A","text":"3"},{"id":"B","text":"2"},{"id":"C","text":"5"},{"id":"D","text":"4"}],"answer":"D","explanation":"The first inequality gives $(−∞, −2) ∪ [1, +∞)$; the second gives $[−3, 3]$. Their intersection is $[−3, −2) ∪ [1, 3]$, whose integers are −3, 1, 2, 3 — four values.","explanationZh":"第一个不等式解集为 $(−∞, −2) ∪ [1, +∞)，第二个为 [−3, 3]；交集为 [−3, −2) ∪ [1, 3]，其中整数为 −3、1、2、3，共 4$ 个。","tags":["hard","rational","integer-solutions"],"status":"published"},{"id":"math-absolute-inequality-001","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":1,"stem":"Solve $|x| < 3$.","stemZh":"解不等式 $|x| < 3。$","options":[{"id":"A","text":"$(−∞, −3) ∪ (3, +∞)$"},{"id":"B","text":"$[−3, 3]$"},{"id":"C","text":"$(−3, 3)$"},{"id":"D","text":"$(−∞, 3)$"}],"answer":"C","explanation":"$|x| < 3$ means the distance from x to 0 is less than 3, i.e. $−3 < x < 3$.","explanationZh":"$|x| < 3$ 表示 x 到 0 的距离小于 $3，即 −3 < x < 3。$","tags":["easy","absolute-value","basic"],"status":"published"},{"id":"math-absolute-inequality-002","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":1,"stem":"Solve $|x| ≥ 2$.","stemZh":"解不等式 $|x| ≥ 2。$","options":[{"id":"A","text":"$(−∞, −2] ∪ [2, +∞)$"},{"id":"B","text":"$(−2, 2)$"},{"id":"C","text":"$[−2, 2]$"},{"id":"D","text":"$(−∞, −2) ∪ (2, +∞)$"}],"answer":"A","explanation":"$|x| ≥ 2$ means $x ≤ −2$ or $x ≥ 2$. The endpoints are included because the inequality is not strict.","explanationZh":"$|x| ≥ 2 即 x ≤ −2 或 x ≥ 2；不等式带等号，端点取到。$","tags":["easy","absolute-value","basic"],"status":"published"},{"id":"math-absolute-inequality-003","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":2,"stem":"Solve $|x − 1| ≤ 2$.","stemZh":"解不等式 $|x − 1| ≤ 2。$","options":[{"id":"A","text":"$(−1, 3)$"},{"id":"B","text":"$(−∞, −1] ∪ [3, +∞)$"},{"id":"C","text":"$[−3, 1]$"},{"id":"D","text":"$[−1, 3]$"}],"answer":"D","explanation":"$−2 ≤ x − 1 ≤ 2$, so $−1 ≤ x ≤ 3. Equivalently, x$ is within distance 2 of the point 1.","explanationZh":"$由 −2 ≤ x − 1 ≤ 2 得 −1 ≤ x ≤ 3，即 x 与 1$ 的距离不超过 2。","tags":["easy","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-004","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":1,"stem":"Solve $|x| > −1$.","stemZh":"解不等式 $|x| > −1。$","options":[{"id":"A","text":"$∅$"},{"id":"B","text":"$ℝ$"},{"id":"C","text":"$(0, +∞)$"},{"id":"D","text":"$(−1, 1)$"}],"answer":"B","explanation":"$|x| ≥ 0$ for every real x, so $|x| > −1$ holds for all real numbers. An absolute value compared with a negative bound never gives the empty set.","explanationZh":"对任意实数 $|x| ≥ 0，故 |x| > −1$ 恒成立，解集为 $ℝ；绝对值与负数比较时解集不会是空集。$","tags":["easy","absolute-value","property"],"status":"published"},{"id":"math-absolute-inequality-005","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":2,"stem":"Solve $|x| ≤ 0$.","stemZh":"解不等式 $|x| ≤ 0。$","options":[{"id":"A","text":"{0}"},{"id":"B","text":"$∅$"},{"id":"C","text":"$ℝ$"},{"id":"D","text":"$(−∞, 0]$"}],"answer":"A","explanation":"$|x|$ is never negative, so $|x| ≤ 0$ forces $|x| = 0$, i.e. $x = 0$. The solution set is {0}, not the empty set.","explanationZh":"$|x|$ 不可能为负，故 $|x| ≤ 0$ 只能是 $|x| = 0，即 x = 0，解集为 {0}$ 而非空集。","tags":["easy","absolute-value","property"],"status":"published"},{"id":"math-absolute-inequality-006","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":1,"stem":"Solve $|2x| < 4$.","stemZh":"解不等式 $|2x| < 4。$","options":[{"id":"A","text":"$(−4, 4)$"},{"id":"B","text":"$(−∞, −2) ∪ (2, +∞)$"},{"id":"C","text":"$(−2, 2)$"},{"id":"D","text":"$[−2, 2]$"}],"answer":"C","explanation":"$|2x| = 2|x| < 4$ gives $|x| < 2$, so $−2 < x < 2$. Remember to divide the bound by the coefficient of x.","explanationZh":"$|2x| = 2|x| < 4 即 |x| < 2，故 −2 < x < 2；注意右端也要除以 x$ 的系数。","tags":["easy","absolute-value","basic"],"status":"published"},{"id":"math-absolute-inequality-007","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":2,"stem":"Solve $|x + 1| > 3$.","stemZh":"解不等式 $|x + 1| > 3。$","options":[{"id":"A","text":"$(−4, 2)$"},{"id":"B","text":"$(−∞, −4) ∪ (2, +∞)$"},{"id":"C","text":"$(−∞, −2) ∪ (4, +∞)$"},{"id":"D","text":"$[−4, 2]$"}],"answer":"B","explanation":"$|x + 1| > 3$ means $x + 1 > 3$ or $x + 1 < −3$, giving $x > 2$ or $x < −4$. Distances larger than 3 lie outside the interval $[−4, 2]$.","explanationZh":"$|x + 1| > 3 即 x + 1 > 3 或 x + 1 < −3，得 x > 2 或 x < −4；距离大于 3$ 的点在区间 $[−4, 2]$ 之外。","tags":["easy","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-008","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":1,"stem":"Solve $|x − 2| ≥ 1$.","stemZh":"解不等式 $|x − 2| ≥ 1。$","options":[{"id":"A","text":"$[1, 3]$"},{"id":"B","text":"$(1, 3)$"},{"id":"C","text":"$(−∞, 1) ∪ (3, +∞)$"},{"id":"D","text":"$(−∞, 1] ∪ [3, +∞)$"}],"answer":"D","explanation":"$x − 2 ≥ 1$ or $x − 2 ≤ −1$, i.e. $x ≥ 3$ or $x ≤ 1$. Endpoints included because of the non-strict sign.","explanationZh":"$x − 2 ≥ 1 或 x − 2 ≤ −1，即 x ≥ 3 或 x ≤ 1；因不等式不严格，端点取到。$","tags":["easy","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-009","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":2,"stem":"Solve $|x + 2| ≤ 3$.","stemZh":"解不等式 $|x + 2| ≤ 3。$","options":[{"id":"A","text":"$[−5, 1]$"},{"id":"B","text":"$(−5, 1)$"},{"id":"C","text":"$[−1, 5]$"},{"id":"D","text":"$(−∞, −5] ∪ [1, +∞)$"}],"answer":"A","explanation":"$−3 ≤ x + 2 ≤ 3$, so $−5 ≤ x ≤ 1$.","explanationZh":"$由 −3 ≤ x + 2 ≤ 3 得 −5 ≤ x ≤ 1。$","tags":["easy","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-010","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":2,"stem":"Solve $|x| > 0$.","stemZh":"解不等式 $|x| > 0。$","options":[{"id":"A","text":"$ℝ$"},{"id":"B","text":"$[0, +∞)$"},{"id":"C","text":"$(−∞, 0) ∪ (0, +∞)$"},{"id":"D","text":"$(0, +∞)$"}],"answer":"C","explanation":"$|x| > 0$ fails only at $x = 0$, so the solution set is all non-zero $reals: (−∞, 0) ∪ (0, +∞)$.","explanationZh":"$|x| > 0$ 仅在 $x = 0 处不成立，故解集为一切非零实数：(−∞, 0) ∪ (0, +∞)。$","tags":["easy","absolute-value","property"],"status":"published"},{"id":"math-absolute-inequality-011","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Solve $|2x − 1| ≤ 3$.","stemZh":"解不等式 $|2x − 1| ≤ 3。$","options":[{"id":"A","text":"$[−2, 1]$"},{"id":"B","text":"$[−1, 2]$"},{"id":"C","text":"$(−1, 2)$"},{"id":"D","text":"$(−∞, −1] ∪ [2, +∞)$"}],"answer":"B","explanation":"$−3 ≤ 2x − 1 ≤ 3 → −2 ≤ 2x ≤ 4 → −1 ≤ x ≤ 2$.","explanationZh":"$由 −3 ≤ 2x − 1 ≤ 3 得 −2 ≤ 2x ≤ 4，即 −1 ≤ x ≤ 2。$","tags":["medium","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-012","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Solve $|3 − x| < 2$.","stemZh":"解不等式 $|3 − x| < 2。$","options":[{"id":"A","text":"$(−5, −1)$"},{"id":"B","text":"$(−1, 5)$"},{"id":"C","text":"$[1, 5]$"},{"id":"D","text":"$(1, 5)$"}],"answer":"D","explanation":"$|3 − x| = |x − 3| < 2$ gives $−2 < x − 3 < 2$, so $1 < x < 5$.","explanationZh":"$|3 − x| = |x − 3| < 2，即 −2 < x − 3 < 2，得 1 < x < 5。$","tags":["medium","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-013","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Solve $|x − 1| ≥ 2$.","stemZh":"解不等式 $|x − 1| ≥ 2。$","options":[{"id":"A","text":"$[−1, 3]$"},{"id":"B","text":"$(−∞, −1) ∪ (3, +∞)$"},{"id":"C","text":"$(−1, 3)$"},{"id":"D","text":"$(−∞, −1] ∪ [3, +∞)$"}],"answer":"D","explanation":"$x − 1 ≥ 2$ or $x − 1 ≤ −2$, giving $x ≥ 3$ or $x ≤ −1$, with both endpoints included.","explanationZh":"$x − 1 ≥ 2 或 x − 1 ≤ −2，得 x ≥ 3 或 x ≤ −1，端点均取到。$","tags":["medium","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-014","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Solve $|2x + 4| > 6$.","stemZh":"解不等式 $|2x + 4| > 6。$","options":[{"id":"A","text":"$(−∞, −5) ∪ (1, +∞)$"},{"id":"B","text":"$(−5, 1)$"},{"id":"C","text":"$(−∞, 1) ∪ (5, +∞)$"},{"id":"D","text":"$[−5, 1]$"}],"answer":"A","explanation":"$2x + 4 > 6$ gives $x > 1; 2x + 4 < −6$ gives $x < −5$. So the solution is $(−∞, −5) ∪ (1, +∞)$.","explanationZh":"$2x + 4 > 6 得 x > 1；2x + 4 < −6 得 x < −5，故解集为 (−∞, −5) ∪ (1, +∞)。$","tags":["medium","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-015","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Solve $|x − 2| + 1 ≤ 4$.","stemZh":"解不等式 $|x − 2| + 1 ≤ 4。$","options":[{"id":"A","text":"$(−1, 5)$"},{"id":"B","text":"$[−5, 1]$"},{"id":"C","text":"$[−1, 5]$"},{"id":"D","text":"$(−∞, −1] ∪ [5, +∞)$"}],"answer":"C","explanation":"Isolate the absolute value $first: |x − 2| ≤ 3$, so $−3 ≤ x − 2 ≤ 3$ and $−1 ≤ x ≤ 5$.","explanationZh":"$先移项：|x − 2| ≤ 3，即 −3 ≤ x − 2 ≤ 3，得 −1 ≤ x ≤ 5。$","tags":["medium","absolute-value","rearrangement"],"status":"published"},{"id":"math-absolute-inequality-016","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Find the set of x satisfying both $|x| ≤ 3$ and $|x| > 1$.","stemZh":"求同时满足 $|x| ≤ 3 与 |x| > 1 的 x$ 的集合。","options":[{"id":"A","text":"$[−3, 3]$"},{"id":"B","text":"$[−3, −1) ∪ (1, 3]$"},{"id":"C","text":"$(−3, −1] ∪ [1, 3)$"},{"id":"D","text":"$(−∞, −3] ∪ [3, +∞)$"}],"answer":"B","explanation":"$|x| ≤ 3$ gives $[−3, 3]$ and $|x| > 1$ gives $(−∞, −1) ∪ (1, +∞)$. Their intersection is $[−3, −1) ∪ (1, 3]$.","explanationZh":"$|x| ≤ 3$ 的解集为 $[−3, 3]，|x| > 1$ 的解集为 $(−∞, −1) ∪ (1, +∞)，交集为 [−3, −1) ∪ (1, 3]。$","tags":["medium","absolute-value","system"],"status":"published"},{"id":"math-absolute-inequality-017","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Solve $|x − 1| ≤ 0$.","stemZh":"解不等式 $|x − 1| ≤ 0。$","options":[{"id":"A","text":"{1}"},{"id":"B","text":"$∅$"},{"id":"C","text":"$ℝ$"},{"id":"D","text":"$[0, 2]$"}],"answer":"A","explanation":"An absolute value is never negative, so $|x − 1| ≤ 0$ forces $|x − 1| = 0$, i.e. $x = 1$.","explanationZh":"绝对值非负，故 $|x − 1| ≤ 0$ 只能是 $|x − 1| = 0，即 x = 1。$","tags":["medium","absolute-value","property"],"status":"published"},{"id":"math-absolute-inequality-018","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"Solve $|1 − 2x| ≥ 3$.","stemZh":"解不等式 $|1 − 2x| ≥ 3。$","options":[{"id":"A","text":"$[−1, 2]$"},{"id":"B","text":"$(−1, 2)$"},{"id":"C","text":"$(−∞, −1] ∪ [2, +∞)$"},{"id":"D","text":"$(−∞, −2] ∪ [1, +∞)$"}],"answer":"C","explanation":"$|1 − 2x| = |2x − 1| ≥ 3$ gives $2x − 1 ≥ 3 (x ≥ 2)$ or $2x − 1 ≤ −3 (x ≤ −1)$, so $(−∞, −1] ∪ [2, +∞)$.","explanationZh":"$|1 − 2x| = |2x − 1| ≥ 3，即 2x − 1 ≥ 3（x ≥ 2）或 2x − 1 ≤ −3（x ≤ −1），故解集为 (−∞, −1] ∪ [2, +∞)。$","tags":["medium","absolute-value","two-case"],"status":"published"},{"id":"math-absolute-inequality-019","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"How many integers x satisfy $|x| ≤ 2$?","stemZh":"满足 $|x| ≤ 2$ 的整数 x 有多少个？","options":[{"id":"A","text":"2"},{"id":"B","text":"3"},{"id":"C","text":"4"},{"id":"D","text":"5"}],"answer":"D","explanation":"$|x| ≤ 2$ gives $−2 ≤ x ≤ 2$, whose integers are −2, −1, 0, 1, 2 — five values.","explanationZh":"$|x| ≤ 2 即 −2 ≤ x ≤ 2，整数为 −2、−1、0、1、2，共 5$ 个。","tags":["medium","absolute-value","integer-solutions"],"status":"published"},{"id":"math-absolute-inequality-020","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":3,"stem":"For which real x does $|x| =$ −x hold?","stemZh":"等式 $|x| =$ −x 对哪些实数 x 成立？","options":[{"id":"A","text":"$[0, +∞)$"},{"id":"B","text":"$(−∞, 0]$"},{"id":"C","text":"{0}"},{"id":"D","text":"$∅$"}],"answer":"B","explanation":"Since $|x| ≥ 0$, the right side must satisfy −x $≥ 0$, i.e. $x ≤ 0$. For $x ≤ 0$ we indeed have $|x| = −x$, so the solution is $(−∞, 0]$.","explanationZh":"$由 |x| ≥ 0$ 知需 −x $≥ 0，即 x ≤ 0；当 x ≤ 0$ 时确有 $|x| =$ −x，故解集为 $(−∞, 0]。$","tags":["medium","absolute-value","property"],"status":"published"},{"id":"math-absolute-inequality-021","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":4,"stem":"Solve $|x − 1| + |x − 2| ≤ 1$.","stemZh":"解不等式 $|x − 1| + |x − 2| ≤ 1。$","options":[{"id":"A","text":"$(1, 2)$"},{"id":"B","text":"$(−∞, 1] ∪ [2, +∞)$"},{"id":"C","text":"$[1, 2]$"},{"id":"D","text":"$∅$"}],"answer":"C","explanation":"$|x − 1| + |x − 2|$ is the sum of the distances from x to 1 and to 2. Its minimum is the distance between the two points, 1, attained precisely on the segment $[1, 2]$. Outside it the sum exceeds 1.","explanationZh":"$|x − 1| + |x − 2|$ 表示 x 到 1 与 2 的距离之和，其最小值等于两点间距离 $1，且恰在 [1, 2]$ 上取得；区间外距离和大于 1。","tags":["hard","absolute-value","distance-reasoning"],"status":"published"},{"id":"math-absolute-inequality-022","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":5,"stem":"Solve $|x − 3| + |x + 1| ≥ 4$.","stemZh":"解不等式 $|x − 3| + |x + 1| ≥ 4。$","options":[{"id":"A","text":"$[−1, 3]$"},{"id":"B","text":"$(−∞, −1] ∪ [3, +∞)$"},{"id":"C","text":"$∅$"},{"id":"D","text":"$ℝ$"}],"answer":"D","explanation":"The sum of the distances from x to 3 and to −1 is at least the distance between the two points, which is 4, with equality on $[−1, 3]$. So the inequality holds for every real x.","explanationZh":"x 到 3 与 −1 的距离之和不小于两点间距离 $4（在 [−1, 3]$ 上取等号），故对一切实数成立，解集为 $ℝ。$","tags":["hard","absolute-value","distance-reasoning"],"status":"published"},{"id":"math-absolute-inequality-023","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":5,"stem":"Solve $|x − 1| < 2x$.","stemZh":"解不等式 $|x − 1| < 2x。$","options":[{"id":"A","text":"$(0, +∞)$"},{"id":"B","text":"$(1/3, +∞)$"},{"id":"C","text":"$(1/2, +∞)$"},{"id":"D","text":"$(−1, +∞)$"}],"answer":"B","explanation":"Since $|x − 1| ≥ 0$ we need $2x > 0$. Then $|x − 1| < 2x$ is equivalent to $−2x < x − 1 < 2x$, i.e. $3x > 1$ and $x > −1$. Together with $x > 0$ this gives $x > 1/3$.","explanationZh":"$因 |x − 1| ≥ 0，须 2x > 0；此时 |x − 1| < 2x$ 等价于 $−2x < x − 1 < 2x，即 3x > 1 且 x > −1。结合 x > 0 得 x > 1/3。$","tags":["hard","absolute-value","variable-bound"],"status":"published"},{"id":"math-absolute-inequality-024","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":4,"stem":"If $|x − 2| = 2 − x, then$:","stemZh":"$若 |x − 2| = 2 −$ x，则？","options":[{"id":"A","text":"$x ≤ 2$"},{"id":"B","text":"$x ≥ 2$"},{"id":"C","text":"$x < 2$"},{"id":"D","text":"$x = 2$"}],"answer":"A","explanation":"Since $2 − x = −(x − 2)$, the equation says $|x − 2| = −(x − 2)$, which holds exactly when $x − 2 ≤ 0$, i.e. $x ≤ 2 (the\\text{ case }x = 2\\text{ gives }0 = 0)$.","explanationZh":"$因 2 − x = −(x − 2)，等式即 |x − 2| = −(x − 2)，当且仅当 x − 2 ≤ 0，即 x ≤ 2（x = 2 时 0 = 0$ 也成立）。","tags":["hard","absolute-value","property"],"status":"published"},{"id":"math-absolute-inequality-025","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":4,"stem":"How many integers x satisfy $|x − 1| < 2$ and $|x| ≥ 1$?","stemZh":"同时满足 $|x − 1| < 2 与 |x| ≥ 1$ 的整数 x 有多少个？","options":[{"id":"A","text":"3"},{"id":"B","text":"2"},{"id":"C","text":"4"},{"id":"D","text":"1"}],"answer":"B","explanation":"$|x − 1| < 2$ gives $−1 < x < 3; |x| ≥ 1$ gives $x ≤ −1$ or $x ≥ 1$. The intersection is $[1, 3)$, whose integers are 1 and 2 — two values.","explanationZh":"$|x − 1| < 2 得 −1 < x < 3；|x| ≥ 1 得 x ≤ −1 或 x ≥ 1。交集为 [1, 3)，整数为 1 和 2，共 2$ 个。","tags":["hard","absolute-value","integer-solutions"],"status":"published"},{"id":"math-absolute-inequality-026","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":5,"stem":"Solve $|x − 1| ≤ 2x − 1$.","stemZh":"解不等式 $|x − 1| ≤ 2x − 1。$","options":[{"id":"A","text":"$[2/3, +∞)$"},{"id":"B","text":"$(2/3, +∞)$"},{"id":"C","text":"$[1/2, +∞)$"},{"id":"D","text":"$(−∞, 1/3] ∪ [2, +∞)$"}],"answer":"A","explanation":"First the right side must be $non-negative: 2x − 1 ≥ 0$, so $x ≥ 1/2$. Then $−(2x − 1) ≤ x − 1 ≤ 2x − 1$ gives $x ≥ 0$ and $3x ≥ 2$, i.e. $x ≥ 2/3. Combining: x ≥ 2/3$.","explanationZh":"$首先右端须非负：2x − 1 ≥ 0，即 x ≥ 1/2；再由 −(2x − 1) ≤ x − 1 ≤ 2x − 1 得 x ≥ 0 且 3x ≥ 2，即 x ≥ 2/3。综合得 x ≥ 2/3。$","tags":["hard","absolute-value","variable-bound"],"status":"published"},{"id":"math-absolute-inequality-027","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":5,"stem":"Solve $|x| + |x − 2| < 3$.","stemZh":"解不等式 $|x| + |x − 2| < 3。$","options":[{"id":"A","text":"$(−3/2, 3/2)$"},{"id":"B","text":"$(−1, 3)$"},{"id":"C","text":"$(0, 2)$"},{"id":"D","text":"$(−1/2, 5/2)$"}],"answer":"D","explanation":"On $[0, 2]$ the sum equals $2 < 3$. For $x > 2$ it is $x + (x − 2) = 2x − 2 < 3$, so $x < 5/2$. For $x < 0$ it is −x $+ (2 − x) = 2 − 2x < 3$, so $x > −1/2$. Hence $(−1/2, 5/2)$.","explanationZh":"$在 [0, 2]$ 上距离和为 $2 < 3；当 x > 2 时 x + (x − 2) = 2x − 2 < 3，得 x < 5/2；当 x < 0 时$ −x $+ (2 − x) = 2 − 2x < 3，得 x > −1/2。故解集为 (−1/2, 5/2)。$","tags":["hard","absolute-value","distance-reasoning"],"status":"published"},{"id":"math-absolute-inequality-028","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":4,"stem":"If the solution set of $|x − a| ≤ 2$ is $[−1, 3]$, find a.","stemZh":"$若 |x − a| ≤ 2$ 的解集为 $[−1, 3]，求$ a。","options":[{"id":"A","text":"−1"},{"id":"B","text":"2"},{"id":"C","text":"1"},{"id":"D","text":"3"}],"answer":"C","explanation":"$|x − a| ≤ 2$ has solution $[a − 2, a + 2]$. Matching $[−1, 3]$ gives $a − 2 = −1$ and $a + 2 = 3$, so $a = 1$.","explanationZh":"$|x − a| ≤ 2$ 的解集为 $[a − 2, a + 2]，与 [−1, 3]$ 对比得 $a − 2 = −1、a + 2 = 3，故 a = 1。$","tags":["hard","absolute-value","parameter"],"status":"published"},{"id":"math-absolute-inequality-029","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":5,"stem":"Solve $|2x − 1| ≥$ x.","stemZh":"解不等式 $|2x − 1| ≥$ x。","options":[{"id":"A","text":"$(−∞, 1/3) ∪ (1, +∞)$"},{"id":"B","text":"$(−∞, 1/3] ∪ [1, +∞)$"},{"id":"C","text":"$[1/3, 1]$"},{"id":"D","text":"$(0, 1/3] ∪ [1, +∞)$"}],"answer":"B","explanation":"For $x ≤ 0$ the inequality is automatic since $|2x − 1| ≥ 0 ≥$ x. For $x > 0$ we need $2x − 1 ≥ x$ or $2x − 1 ≤ −x$, i.e. $x ≥ 1$ or $x ≤ 1/3$. Altogether $(−∞, 1/3] ∪ [1, +∞)$.","explanationZh":"$当 x ≤ 0 时，|2x − 1| ≥ 0 ≥ x$ 自动成立；当 $x > 0$ 时需 $2x − 1 ≥ x 或 2x − 1 ≤$ −x，即 $x ≥ 1 或 x ≤ 1/3。合并得 (−∞, 1/3] ∪ [1, +∞)。$","tags":["hard","absolute-value","variable-bound"],"status":"published"},{"id":"math-absolute-inequality-030","subject":"math","moduleId":"sets-inequalities","unitId":"inequalities","topicId":"absolute-inequality","type":"single-choice","difficulty":5,"stem":"How many integers x satisfy $|x − 1| + |x + 1| ≤ 4$?","stemZh":"满足 $|x − 1| + |x + 1| ≤ 4$ 的整数 x 有多少个？","options":[{"id":"A","text":"5"},{"id":"B","text":"4"},{"id":"C","text":"6"},{"id":"D","text":"3"}],"answer":"A","explanation":"On $[−1, 1]$ the sum is $2 ≤ 4$. For $x > 1$ it is $2x ≤ 4$, so $x ≤ 2$; for $x < −1$ it is $−2x ≤ 4$, so $x ≥ −2$. Hence the solution set is $[−2, 2]$, containing the five integers −2, −1, 0, 1, 2.","explanationZh":"$在 [−1, 1]$ 上距离和为 $2 ≤ 4；x > 1$ 时为 $2x ≤ 4，得 x ≤ 2；x < −1$ 时为 $−2x ≤ 4，得 x ≥ −2。故解集为 [−2, 2]，含整数 −2、−1、0、1、2 共 5$ 个。","tags":["hard","absolute-value","integer-solutions"],"status":"published"}]

Sets & Inequalities