Complement of a Set and Its Properties

Since \(|B'|=110\) in a universal set of 180 elements, \(|B|=|U|-|B'|=180-110=70\). Now apply inclusion-exclusion: \(|A\cup B|=|A|+|B|-|A\cap B|=85+70-28=127\). The complement of the union contains all elements of U that are in neither A nor B, so \(|(A\cup B)'|=|U|-|A\cup B|=180-127=53\). Therefore, option C is correct. The intersection must be subtracted once because its elements were counted in both |A| and |B|.

There are 25 odd integers from 1 through 49. The perfect squares in this universe are 1, 4, 9, 16, 25, 36, and 49; the odd ones among them are 1, 9, 25, and 49, four numbers. A′ ∩ B consists of odd numbers that are not perfect squares, so its cardinality is 25 − 4 = 21. Therefore, option B is correct.

Use De Morgan’s law together with the double-complement law. The complement of a union is the intersection of the complements, so (A ∪ B')' = A' ∩ (B')'. Since (B')' = B, the expression simplifies to A' ∩ B. Thus option A is correct. Option B uses a union instead of the required intersection, option C fails to complement A, and option D does not apply the outer complement.

Solve the absolute-value inequality: |x + 2| ≤ 6 means −6 ≤ x + 2 ≤ 6. Subtracting 2 gives −8 ≤ x ≤ 4, so A = [−8, 4]. Because the universal set is ℝ, its complement contains numbers strictly less than −8 or strictly greater than 4. Hence A′ = (−∞, −8) ∪ (4, ∞), option A.

The governing concept is De Morgan’s law: A′ ∪ B′ = (A ∩ B)′. Since A ∩ B = U, taking the complement relative to U gives (A ∩ B)′ = U′ = ∅. Equivalently, if the intersection of A and B is the entire universal set, then both A and B equal U, so each complement is empty and their union is also empty. Therefore option A is correct; U and A ∪ B are not empty in general, while option D is an intersection rather than the required union.

Use inclusion–exclusion. There are 30 multiples of 4, 20 of 6, and 12 of 10. Pairwise intersections contain 10 multiples of 12, 6 multiples of 20, and 4 multiples of 30; the triple intersection has 2 multiples of 60. Thus |A ∪ B ∪ C| = 30 + 20 + 12 − 10 − 6 − 4 + 2 = 44. Therefore the complement has 120 − 44 = 76 elements, so option C is correct.

Use De Morgan’s law, which states that (A ∪ B)′ = A′ ∩ B′. The given complements are A′ = {q, s, w} and B′ = {p, s, u}. Comparing the two lists, s is the only element appearing in both sets. Hence A′ ∩ B′ = {s}, and therefore (A ∪ B)′ = {s}. Option B is their union, not their intersection, while options C and D do not contain exactly the common elements.

The set A′ ∩ B′ contains elements that are neither multiples of 4 nor perfect squares, because A′ excludes multiples of 4 and B′ excludes perfect squares. The number 18 is not divisible by 4 and is not a perfect square. In contrast, 16 and 36 are multiples of 4 and perfect squares, while 25 is a perfect square. Hence option A is correct.

Set difference is defined by B − A = B ∩ A′. If B − A equals all of B, removing A has removed no element from B. Therefore no element can belong to both A and B, which means A ∩ B = ∅. This conclusion says that A and B are disjoint. The other statements do not necessarily follow, so option A is the only always-correct answer.

The set A contains all real numbers less than −4 together with all real numbers greater than or equal to 2. Its complement therefore contains the numbers that are not less than −4 and are also less than 2. Thus −4 is included, while 2 is excluded. Consequently, A′ = [−4, 2), so option A is correct.

Set inclusion reverses when complements are taken. From A ⊆ B, every element of A is in B, so any element outside B must also be outside A. Therefore B′ ⊆ A′. The condition B′ ⊆ C is additional information but is not needed for this conclusion. Hence option A is always true, while the other options need not hold.

The universal set is partitioned into four mutually disjoint regions: A ∩ B, A′ ∩ B, A′ ∩ B′, and A ∩ B′. Their cardinalities must add to |U|. Therefore |A ∩ B′| = 90 − 24 − 17 − 29 = 20. Hence option B is correct. This partition prevents any element from being counted twice or omitted.

By De Morgan’s law, A' ∪ B' = (A ∩ B)'. An element belongs to A ∩ B when it is divisible by both 3 and 7, so it must be divisible by lcm(3, 7) = 21. Between 1 and 42, the multiples of 21 are 21 and 42; hence |A ∩ B| = 2. Therefore, |A' ∪ B'| = |U| − |A ∩ B| = 42 − 2 = 40. Thus option B is correct.

The complement is taken in R. The interval (-∞, -1) excludes -1, so -1 belongs to the complement and the left endpoint is included. The interval [3, 8) includes 3 but excludes 8, so 3 is omitted from the complement while 8 is included. All real numbers from -1 through values less than 3, together with 8 and larger numbers, give A' = [-1, 3) ∪ [8, ∞). Hence option A is correct.

A set and its complement are disjoint, so A ∩ A' = ∅ for every set A. Since the universal set U is explicitly non-empty, ∅ cannot equal U. The other statements are standard complement identities: A ∪ A' = U, (A')' = A, and the complement of U relative to U is ∅. Therefore, statement C cannot always be true.

A' ∩ B consists of elements that are divisible by 7 but not divisible by 5. From 1 to 70, B contains 70/7 = 10 numbers. The numbers divisible by both 5 and 7 are multiples of lcm(5, 7) = 35, namely 35 and 70, so there are 2 such numbers. Therefore, |A' ∩ B| = 10 − 2 = 8, making option C correct.

The universal set U contains all integers from 1 through 24, so |U| = 24. The given complement A′ contains 8 elements. Since A and A′ are disjoint and together make U, |A| + |A′| = |U|. Therefore, |A| = 24 − 8 = 16. Hence, option C is correct. This uses the complement cardinality property for a finite universal set.

Factor the quadratic: x² − 4x − 12 = (x − 6)(x + 2). Because the parabola opens upward, the expression is negative between its roots, so A = (-2, 6). The complement in R contains the endpoints and the two outside intervals. Therefore, A' = (-∞, -2] ∪ [6, ∞), which is option A.

First calculate the difference B − A. The elements 6 and 12 occur in both sets, so they are removed from B, leaving B − A = {9, 15, 18}. The complement is taken with respect to U = {1, ..., 18}; therefore remove 9, 15, and 18 from U. The result is {1,2,3,4,5,6,7,8,10,11,12,13,14,16,17}, so option A is correct.

A consists of the numbers in U that are not divisible by 12. Therefore, its complement A′ consists exactly of the numbers in U that are divisible by 12. The positive multiples of 12 not exceeding 144 are 12, 24, 36, …, 144. Their number is 144 ÷ 12 = 12, because the kth multiple is 12k and 12k ≤ 144 gives k ≤ 12. Hence |A′| = 12, so option C is correct.