Complement of a Set and Its Properties

The multiples of 3 in \(U\) are \(3,6,9,12,15,18,21,24,27\), so \(|A|=9\) and \(|A'|=27-9=18\). The multiples of 9 are \(9,18,27\), so \(|B|=3\). Because every multiple of 9 is also a multiple of 3, \(B\subseteq A\), which means \(A'\cap B=\varnothing\). Therefore, \(|A'\cup B|=|A'|+|B|=18+3=21\). Thus, option D is correct.

The allowed values of \(n\) are 0, 1, 2, 3, 4, and 5. Hence, \(A=\{2^0,2^1,2^2,2^3,2^4,2^5\}=\{1,2,4,8,16,32\}\), which contains six distinct elements. Since the complement is taken relative to \(U\), every element of U not in A belongs to \(A'\). Therefore, \(|A'|=|U|-|A|=32-6=26\). Thus, option B is correct. The value 32 is included in A because the condition permits \(n=5\).

The intervals A = (1,4] and B = [4,7) meet at 4, and 4 is included in both sets. Therefore, their union is (1,7): 1 is excluded and 7 is excluded. Since the universal set is ℝ, the complement contains every real number outside this open interval. Hence (A ∪ B)' = (-∞,1] ∪ [7,∞). The endpoints 1 and 7 belong to the complement because they do not belong to A ∪ B.

The complement (A ∪ B)' consists of elements of U that are not in A and not in B. Thus, its elements are numbers from 1 to 30 that are divisible by neither 2 nor 3. The number 1 is not divisible by 2 or 3, so it belongs to the complement. Because 1 is the smallest element of U, it is automatically the smallest element of (A ∪ B)'. Therefore, option A is correct.

Relative to U, A contains every odd number from 1 through 13, so A' = {2,4,6,8,10,12,14}. The set B contains 2,3,5,7,11,13, so B' = {1,4,6,8,9,10,12,14}. The elements common to A' and B' are therefore {4,6,8,10,12,14}. This also follows directly from De Morgan's law: A' ∩ B' = (A ∪ B)'.

The set A contains the open interval (0,2), so its complement contains x ≤ 0 and x ≥ 2 around that interval. It also contains [5,9), so its complement contains x < 5 and x ≥ 9 in the corresponding region; combining the portions gives [2,5) between the two intervals. Therefore, over ℝ, A' = (-∞,0] ∪ [2,5) ∪ [9,∞). Endpoint symbols are determined by whether the endpoint was included in A.

Apply De Morgan's law to the left side: A' ∪ B' = (A ∩ B)'. The condition says that (A ∩ B)' is the entire universal set U. The only subset whose complement is U is the empty set, because no element can belong to A ∩ B. Hence A ∩ B = ∅. This means A and B are disjoint, although it does not imply that either set itself is empty.

Let C be the set of students who play cricket and F the set who play football. The number playing at least one game is found by inclusion–exclusion: |C ∪ F| = |C| + |F| − |C ∩ F| = 46 + 38 − 20 = 64. Students playing neither game form the complement of C ∪ F in the class. Thus, 80 − 64 = 16 students play neither game, so option B is correct.

Since A ⊆ B, every element of A is also an element of B. Therefore, if an element is not in B, it cannot be in A. The elements outside B are consequently contained among the elements outside A, giving B′ ⊆ A′. Complementation reverses the direction of a subset relation; this is a fundamental property of complements.

By De Morgan’s law, A′ ∩ B′ = (A ∪ B)′. Among the numbers 1 to 36, 18 are divisible by 2, 12 are divisible by 3, and 6 are divisible by both because they are multiples of 6. Hence |A ∪ B| = 18 + 12 − 6 = 24. Its complement in U therefore has 36 − 24 = 12 elements, so option C is correct.

The first interval includes −3 but excludes 4, while the second includes 10 but excludes 7. Therefore, points not in A are all real numbers less than −3, the interval from 4 through 7 including both endpoints, and numbers greater than 10. Thus A′ = (−∞, −3) ∪ [4, 7] ∪ (10, ∞), which is option A.

A number belongs to both A and B precisely when it is divisible by the least common multiple of 6 and 14. Since lcm(6, 14) = 42, the common elements in U are 42 and 84 only. Thus |A ∩ B| = 2, and the complement contains the remaining 84 − 2 = 82 elements. Therefore, option C is correct.

There are floor(96/6) = 16 multiples of 6 and floor(96/8) = 12 multiples of 8. Numbers counted in both sets are multiples of lcm(6,8) = 24, giving floor(96/24) = 4. Hence |A ∪ B| = 16 + 12 − 4 = 24. The complement therefore has 96 − 24 = 72 elements, so option C is correct.

The two intervals overlap, so their union extends continuously from −6 to 5. The left endpoint −6 is excluded because it is excluded from A, and the right endpoint 5 is excluded because it is excluded from B. Thus A ∪ B = (−6, 5). Taking the complement in ℝ gives (−∞, −6] ∪ [5, ∞), including both boundary points in the complement.

The complement operation reverses inclusion. Starting with A′ ⊆ B′ and taking complements of both sides gives (B′)′ ⊆ (A′)′, which simplifies to B ⊆ A. No equality or disjointness is forced by the given condition, so the other choices need not hold. Therefore, option A is the only conclusion that is always valid.

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 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.