[{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"},{"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","access":"public"}]

Sets & Inequalities