Complement of a Set and Its Properties

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.

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}.

De Morgan’s law states that the complement of an intersection is the union of the complements: (A ∩ B)′ = A′ ∪ B′. An element fails to belong to A ∩ B whenever it is absent from A or absent from B. Option B is incorrect because (A ∪ B)′ = A′ ∩ B′. Also, (A′)′ = A and A ∪ A′ = U, not the empty set.

Taking complements reverses the direction of inclusion. Since every element of A is in B and every element of B is in C, any element outside C is certainly outside B and outside A. Therefore C′ ⊆ B′ ⊆ A′. Equality is not guaranteed because the original inclusions may be proper.

An element in A ∩ B must be divisible by both 5 and 8, so it must be a multiple of lcm(5,8) = 40. Within U = {1,2,...,40}, the only such element is 40 itself. Therefore n(A ∩ B) = 1. The universal set has 40 elements, so the complement contains 40 − 1 = 39 elements. Hence n((A ∩ B)′) = 39 and option A is correct.

First form the union: A ∪ B = {a, b, c, e, f, i}. The complement is taken with respect to the universal set U, so we select the elements of U that do not occur in this union. The remaining elements are d and g; therefore, (A ∪ B)' = {d, g}. Option D is wrong because f belongs to A and hence to the union.

The perfect squares in U are A = {1, 4, 9, 16, 25}. Set B contains every even number from 2 through 30. To find A' ∩ B, retain the even numbers but remove the perfect squares 4 and 16, because they are not in A'. Thus the result is {2, 6, 8, 10, 12, 14, 18, 20, 22, 24, 26, 28, 30}.

A contains all non-prime numbers in U. Therefore, its complement A' contains exactly the prime numbers from 1 to 36. These primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, and 31. There are 11 such numbers. Note that 1 is neither prime nor composite, but it still belongs to A because it is not prime.

A consists of the numbers 1 through 12, so its complement within U is A' = {13, 14, 15, 16, 17, 18, 19, 20}. Set B contains the odd numbers. Intersecting A' with B keeps only the odd members greater than 12: {13, 15, 17, 19}. Option D gives the whole complement without applying the intersection with B.

Given A ∪ B′ = U, take complements on both sides. By De Morgan’s law, (A ∪ B′)′ = A′ ∩ B, while U′ = ∅. Thus A′ ∩ B = ∅, meaning that no element of B lies outside A. Therefore every element of B belongs to A, so B ⊆ A. The other statements are not necessarily true.

Factor the quadratic: x² + 2x − 8 = (x + 4)(x − 2). Since the parabola opens upward, the inequality (x + 4)(x − 2) < 0 holds strictly between the roots, so A = (−4, 2). Because the universal set is R, the complement contains all real numbers outside this open interval, including the endpoints −4 and 2. Hence A′ = (−∞, −4] ∪ [2, ∞).

The multiples of 4 in U = {1, ..., 64} are 4, 8, 12, ..., 64, so their number is 64/4 = 16. Among them, the members of A are powers of 2: 4 = 2², 8 = 2³, 16 = 2⁴, 32 = 2⁵, and 64 = 2⁶, giving five numbers. These five are excluded from A′. Hence the required count is 16 − 5 = 11, so option A is correct.

Relative to U, A contains every even number, so A′ is the odd-number set {1, 3, 5, 7, 9, 11, 13, 15, 17}. To obtain A′ ∩ B′, remove from A′ every member that occurs in B. The odd members of B are 1, 3, 5, and 13. Removing them leaves {7, 9, 11, 15, 17}. Therefore option A is correct; option B is simply A′, and option C is A ∩ B.

By De Morgan’s law, A′ ∪ B′ = (A ∩ B)′. A number belongs to A ∩ B when it is divisible by both 10 and 15, hence by their least common multiple, lcm(10, 15) = 30. The multiples of 30 from 1 to 100 are 30, 60, and 90, so n(A ∩ B) = 3. Therefore n(A′ ∪ B′) = 100 − 3 = 97.

The intersection must contain numbers common to both intervals. A extends from −7 inclusive to −1 exclusive, while B extends from −3 exclusive to 5 inclusive. Therefore A ∩ B = (−3, −1); neither endpoint is included because −3 is excluded from B and −1 is excluded from A. The complement of this open interval in R includes both endpoints, giving (−∞, −3] ∪ [−1, ∞).

A ∩ B′ consists of numbers that are odd and not prime. The odd numbers from 1 to 24 are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, and 23. Removing the odd primes 3, 5, 7, 11, 13, 17, 19, and 23 leaves {1, 9, 15, 21}. Note that 1 is not prime, so it remains.

Complements are taken with respect to the universal set \(\mathbb{R}\). Set \(A\) contains every real number except 0, so \(A'=\{0\}\). Similarly, \(B\) contains every real number except 1, so \(B'=\{1\}\). Taking the union gives \(A'\cup B'=\{0\}\cup\{1\}=\{0,1\}\). Thus option A is the only correct answer.

Within the given universal set, \(B\) consists of all odd numbers. Therefore, \(B'\) consists of all even numbers from 1 through 35. The multiples of 5 in \(A\) are \(5,10,15,20,25,30,35\). Selecting only the even members of this list gives \(10,20,30\). Hence \(B'\cap A=\{10,20,30\}\), so option A is correct.

First form the union: \(A\cup B=\{1,2,3,4,6,8,9,12,15,16\}\). The complement relative to \(U\) contains every element of \(U\) that does not occur in this union. Checking the integers from 1 to 16 leaves \(5,7,10,11,13,14\). Therefore, \((A\cup B)'=\{5,7,10,11,13,14\}\), which is option A.

The set \(A\) contains all multiples of 3 in \(U\): \(\{3,6,9,12,15,18,21,24,27\}\). The set \(B\) contains the multiples of 9: \(\{9,18,27\}\). Thus \(B'\) excludes exactly these three numbers. Removing them from A leaves \(\{3,6,12,15,21,24\}\). Therefore, option A is correct.

De Morgan’s law states that the complement of a union is the intersection of the complements: (A ∪ B)' = A' ∩ B'. An element belongs to the left side precisely when it belongs to neither A nor B, which means it belongs to both A' and B'. Option A is incorrect because (A ∩ B)' = A' ∪ B', not A' ∩ B'. Also, the double-complement law gives (A')' = A, not A'.

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.