Complement of a Set and Its Properties

There are 13 even numbers from 1 through 27: 2, 4, 6, ..., 26, since floor(27/2) = 13. The even numbers that are divisible by 3 are 6, 12, 18, and 24; these belong to A and therefore are excluded from A′. Consequently, the number of even elements in A′ is 13 − 4 = 9. Option A is correct. Option B counts all even numbers without removing those in A, while option C counts only the excluded even multiples of 3.

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.

By De Morgan’s law, A′ ∩ B′ = (A ∪ B)′. The set A contains the multiples of 6 up to 84, so n(A) = 84/6 = 14. The set B contains the multiples of 14, so n(B) = 84/14 = 6. Their common elements are multiples of lcm(6,14) = 42, giving n(A ∩ B) = 84/42 = 2. Therefore n(A ∪ B) = 14 + 6 − 2 = 18, and n(A′ ∩ B′) = 84 − 18 = 66.

Using De Morgan’s law, A′ ∩ B′ = (A ∪ B)′. There are 96/8 = 12 multiples of 8 and 96/12 = 8 multiples of 12 in U. Numbers counted in both sets are multiples of lcm(8,12) = 24; there are 96/24 = 4 of them. Thus n(A ∪ B) = 12 + 8 − 4 = 16. Since U has 96 elements, n(A′ ∩ B′) = 96 − 16 = 80.

Factor the quadratic inequality: x² − 9x + 18 = (x − 3)(x − 6). Since the parabola opens upward, the inequality is non-positive for 3 ≤ x ≤ 6. The integer elements of A are therefore 3, 4, 5, and 6, so n(A) = 4. The universal set contains all integers from −12 through 12, giving 25 elements. Hence n(A′) = 25 − 4 = 21, so option D is correct.

To find the complement in ℝ, reverse the inclusion of each interval and include the boundary points that were excluded from A. The point −5 is included in the complement, while 1 is excluded because it belongs to A. Similarly, 4 is excluded and 9 is included. Therefore A′ = [−5, 1) ∪ (4, 9].

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.

The set A contains the multiples of 3 up to 42, so n(A) = 42/3 = 14. The elements removed in A − B are those that are also multiples of 7, namely the common multiples of 3 and 7. These are multiples of lcm(3,7) = 21; there are 42/21 = 2 such elements, 21 and 42. Therefore n(A − B) = 14 − 2 = 12. Since U has 42 elements, n((A − B)′) = 42 − 12 = 30. Thus option A is correct.

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 inequality |x + 1| ≥ 4 means x + 1 ≤ -4 or x + 1 ≥ 4. Therefore A = (-∞, -5] ∪ [3, ∞). Since the universal set is R, its complement consists of the real numbers strictly between -5 and 3: A' = (-5, 3). The endpoints are excluded because they belong to A.

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}.

B and B' are disjoint because no element can simultaneously belong to a set and its complement. Consequently, A ∩ B and A ∩ B' are also disjoint. The question states that these two disjoint sets are equal. The only set equal to two disjoint copies of itself is the empty set, so A ∩ B = A ∩ B' = ∅, which forces A = ∅.

Use inclusion–exclusion. There are 18 multiples of 4, 12 of 6, and 8 of 9. Pairwise intersections have sizes 6, 2, and 4 because the relevant LCMs are 12, 36, and 18. The triple intersection has size 2, from multiples of 36. Thus n(A ∪ B ∪ C) = 18 + 12 + 8 - 6 - 2 - 4 + 2 = 28. Therefore the complement has 72 - 28 = 44 elements.

Using De Morgan's law, A' ∪ B' = (A ∩ B)'. The intersection of A = (-2, 6] and B = [0, 9) is [0, 6]. Taking its complement in ℝ excludes every number from 0 through 6, including both endpoints. Therefore the result is (-∞, 0) ∪ (6, ∞). The open endpoints are essential.

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.

The integers from -5 through 15 inclusive number 15 - (-5) + 1 = 21, so n(U) = 21. The members congruent to 2 modulo 5 in this interval are -3, 2, 7, and 12, giving n(A) = 4. Since A' is the complement within U, n(A') = n(U) - n(A) = 21 - 4 = 17. Therefore option A is correct.

First use inclusion–exclusion for the complements: n(A′ ∩ B′) = n(A′) + n(B′) − n(A′ ∪ B′) = 120 + 150 − 210 = 60. De Morgan’s law gives (A ∪ B)′ = A′ ∩ B′, so the complement of A ∪ B has 60 elements. Therefore n(A ∪ B) = n(U) − n((A ∪ B)′) = 300 − 60 = 240. Option A is correct; 210 is the given union of complements, not the requested union.

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.