Complement of a Set and Its Properties

The numbers in U that are divisible by 4 are exactly 4, 8, 12, 16, 20, and 24. These six numbers are all listed in A', the complement of A. Since A and A' are disjoint and together contain U, none of these divisible-by-4 elements can belong to A. Therefore the required number of elements is 0, so option A is correct.

Both complements are taken with respect to the same universal set U. Taking the complement of both sides of A' = B' gives (A')' = (B')'. By the double-complement law, (A')' = A and (B')' = B, so A = B. The other statements do not necessarily follow: equal sets need not be disjoint, their union need not be U, and A' = B is a different condition.

The set B contains the multiples of 9 from 1 to 54: B = {9, 18, 27, 36, 45, 54}. The complement A′ contains all elements of U that are not divisible by 6. Among the elements of B, 18, 36, and 54 are divisible by 6, so they are excluded from A′. The remaining elements are 9, 27, and 45. Therefore, A′ ∩ B = {9, 27, 45}, and its cardinality is n(A′ ∩ B) = 3.

Solving x² = 4 over the real numbers gives x = −2 or x = 2. Thus A = {−2, 2}. The complement is always determined relative to the stated universal set. Since U = ℝ, remove these two elements from all real numbers: A' = ℝ \ {−2, 2}. Option B is A itself, while options C and D incorrectly ignore the two excluded real numbers.

By De Morgan’s law, A' ∩ B' = (A ∪ B)'. Taking the complement once more gives (A' ∩ B')' = ((A ∪ B)')' = A ∪ B, using the double-complement law. The question already gives n(A ∪ B) = 50, so n((A' ∩ B')') = 50. The values of n(U), n(A), and n(B) are not needed after this identity is applied.

The perfect squares in U are 1, 4, 9, and 16. The even perfect squares, which belong to both A and B, are therefore A ∩ B = {4, 16}. The complement is taken within U, so remove 4 and 16 from {1, 2, ..., 22}. The remaining 20 elements are exactly the set listed in option A. Option B is the intersection itself, not its complement.

The set A contains all multiples of 3 from 1 to 81, so n(A) = 81 ÷ 3 = 27. The set B contains the multiples of 27: 27, 54, and 81, so n(B) = 81 ÷ 27 = 3. Since every multiple of 27 is also a multiple of 3, B is a subset of A. Therefore B' ∩ A consists of the elements of A not in B, and its cardinality is 27 − 3 = 24.

The double-complement law states that (A′)′ = A for every subset A of a universal set U. Here, A consists of all real numbers less than or equal to -1 together with all real numbers greater than or equal to 6. Its complement in ℝ is (-1, 6), because -1 and 6 already belong to A. Taking the complement again restores the original set, so the answer is option A.

By De Morgan’s law, A′ ∩ B′ = (A ∪ B)′. In U, the multiples of 10 are 10, 20, ..., 90, so there are 9; the multiples of 18 are 18, 36, 54, 72, 90, so there are 5. Their only common element is 90 because lcm(10,18) = 90. Thus n(A ∪ B) = 9 + 5 - 1 = 13, and n(A′ ∩ B′) = 90 - 13 = 77. Therefore, option A is correct.

Count the union by inclusion–exclusion. There are 120/8 = 15 multiples of 8, 120/15 = 8 multiples of 15, and 120/20 = 6 multiples of 20. The pairwise intersections contain 2 multiples of lcm(8,15)=120, 1 multiple of lcm(8,20)=40, and 2 multiples of lcm(15,20)=60. The triple intersection contains only 120. Hence n(A ∪ B ∪ C) = 15 + 8 + 6 - 1 - 3 - 2 + 1 = 24. Therefore, its complement has 120 - 24 = 96 elements.

The inequality (x - 2)(x + 5) ≤ 0 holds between the roots, including both roots, so A = [-5, 2]. Also, |x - 1| < 3 gives -3 < x - 1 < 3, hence -2 < x < 4 and B = (-2, 4). Therefore A′ = (-∞, -5) ∪ (2, ∞), and A′ ∩ B = (2, 4). Taking the complement in ℝ gives (A′ ∩ B)′ = (-∞, 2] ∪ [4, ∞), which is option A.