Complement of a Set and Its Properties

Correct answer: A. {1, 2, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22}

Explanation: 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.

Correct answer: A. 24

Explanation: 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.

Correct answer: A. 96

Explanation: 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.

Correct answer: A. (-∞, 2] ∪ [4, ∞)

Explanation: 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.