Complement of a Set and Its Properties

The only common element of A and B is 5, so A ∩ B = {5}. The complement is taken relative to U, which contains all integers from 1 through 15. Removing 5 from U gives (A ∩ B)′ = {1,2,3,4,6,7,8,9,10,11,12,13,14,15}. Thus option A is correct; option C represents A ∪ B.

Because the universal set is the real numbers, the complement of A contains all real numbers that are not in (-∞,0) or (4,∞). The endpoints 0 and 4 are excluded from A because both defining intervals are open there. Therefore A′ = [0,4]. Since [0,4] is wholly contained in B = [-2,6], their intersection is [0,4].

The universal set U contains the natural numbers from 1 through 32, so n(U) = 32. Since k belongs to ℕ₀, the relevant powers are 2⁰, 2¹, 2², 2³, 2⁴ and 2⁵, giving A = {1, 2, 4, 8, 16, 32}. Thus n(A) = 6. The complement A′ contains all elements of U that are not in A, so n(A′) = n(U) − n(A) = 32 − 6 = 26.

A and A′ are complementary subsets of U, so A = U \ A′. Remove 2, 3, 5, 7, 11, 13, 17, and 19 from the numbers 1 through 20. The remaining elements are 1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, and 20. Notice that 1 remains because it is neither prime nor listed in A′.

The difference A \ B consists of elements that belong to A but do not belong to B. If this difference is empty, there is no element of A outside B. Therefore every element of A must also be an element of B, which is precisely the definition of A ⊆ B. The condition does not imply that B ⊆ A or that the two sets are complements.

By De Morgan’s law, A′ ∩ B′ = (A ∪ B)′. In U = {1, 2, ..., 54}, the number of multiples of 9 is ⌊54/9⌋ = 6, and the number of multiples of 6 is ⌊54/6⌋ = 9. Numbers counted in both groups are multiples of lcm(9, 6) = 18; there are ⌊54/18⌋ = 3. Hence n(A ∪ B) = 6 + 9 − 3 = 12, and n(A′ ∩ B′) = 54 − 12 = 42.

The set A contains every real number less than or equal to −1 and every real number greater than 3. Its complement therefore consists of real numbers greater than −1 and less than or equal to 3. The point −1 is excluded because it belongs to A, while 3 is included because the condition x > 3 excludes it from A. Hence A′ = (-1,3].

If A ⊆ B, every element of A is also in B. Consider an element of B′: it is not in B. Since every element of A would have to be in B, that element cannot be in A either; therefore it belongs to A′. Thus every element of B′ is in A′, giving B′ ⊆ A′. Taking complements reverses the direction of inclusion.

The odd numbers from 1 through 22 are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, and 21, making 11 odd numbers. The odd primes in this range are 3, 5, 7, 11, 13, 17, and 19. Removing these seven primes leaves 1, 9, 15, and 21 in A′, so the required number is 4. Note that 1 is not prime.

The universal set U has 14 elements. Set A contains the seven even numbers from 2 through 14, so its complement A′ contains the seven odd numbers {1,3,5,7,9,11,13}. The Cartesian product A′×A consists of ordered pairs whose first element is chosen from A′ and second from A. Therefore, n(A′×A)=n(A′)×n(A)=7×7=49. Hence, option A is correct.

First solve the inequality (x − 1)(x + 2) ≥ 0. Its critical values are −2 and 1, so the product is non-negative when x ≤ −2 or x ≥ 1. Restricting to U gives A = {−4, −3, −2, 1, 2, 3, 4, 5, 6}. Therefore, the elements of U not in A are −1 and 0. Hence A′ = {−1, 0}.

De Morgan’s law states that (A∩B)′=A′∪B′. The inclusion–exclusion formula gives n(A′∪B′)=n(A′)+n(B′)−n(A′∩B′). Substituting the given values, we obtain 76+89−31=134. The intersection must be subtracted once because it is counted in both sets. Therefore, option A is correct.

With respect to U, A′ = {4, 5, 6, 7, 8, 9, 10}. Intersecting this set with B = {3, 4, 5} gives A′ ∩ B = {4, 5}. Now take the complement of {4, 5} in U. The remaining elements are {1, 2, 3, 6, 7, 8, 9, 10}. Thus option A is correct; option B is only the intermediate intersection, not its complement.

Complements are taken in the real-number universal set. Because A includes 1 and all larger numbers, A′=(−∞,1). Because B contains numbers strictly less than 4 but not 4 itself, B′=[4,∞). Their union is therefore (−∞,1)∪[4,∞). Endpoint brackets are essential: 1 is excluded from A′, while 4 is included in B′.

The numbers from 1 to 81 divisible by 9 are 9, 18, 27, 36, 45, 54, 63, 72, and 81, so there are 9 such numbers. Among them, 9 = 3², 36 = 6², and 81 = 9² are perfect squares and therefore belong to A. Removing these three from the nine multiples leaves 9 − 3 = 6 numbers in A′.

The multiples of 3 in U are {3,6,9,12,15,18}. The number 3 is prime, so it belongs to A and cannot belong to A′. Each of 6, 9, 12, 15, and 18 is composite, so all five belong to the complement of the prime-number set. Consequently, the required set is {6,9,12,15,18}, which is option A.

The universal set U contains every integer from 1 through 30, so n(U) = 30. The integers congruent to 1 modulo 4 in this interval are 1, 5, 9, 13, 17, 21, 25, and 29; hence n(A) = 8. Since A′ contains all elements of U not in A, n(A′) = n(U) − n(A) = 30 − 8 = 22. Therefore option A is correct.

A set and its complement together contain every element of the universal set, so A′∪A=U. Taking the complement of both sides with respect to U gives (A′∪A)′=U′. The complement of the universal set is the empty set because no element of U lies outside U. Thus, the expression equals ∅, and option A is correct.

Within U, the multiples of 4 are A = {4, 8, 12, 16, 20, 24}. The multiples of 8 form B = {8, 16, 24}, so B′ contains every element of U except those three. Therefore B′ ∩ A consists of the multiples of 4 that are not multiples of 8. These are 4, 12, and 20, giving option A.

The universal set is the set of all real numbers. Set A contains every real number except −2 and 5, because those two values are explicitly excluded by the conditions. Therefore, the complement A′, which contains elements of U that are not in A, consists exactly of the two excluded real numbers: A′ = {−2, 5}.