Complement of a Set and Its Properties

The complement A′ is defined with respect to the universal set U. It contains every element of U that is not present in A. Removing a, e, and h from U leaves b, c, d, and g. The element f must not be included because f is not an element of the given universal set U. Therefore, A′ = {b, c, d, g}.

The members of A whose last digit is 0 or 5 are 5, 10, 15, 20, 25, and 30, so n(A) = 6. The universal set contains 30 elements. Since A′ contains all elements of U not in A, the complement formula gives n(A′) = n(U) − n(A) = 30 − 6 = 24. Hence option A is correct.

The complement U′ consists of all elements in the universal framework that are not in U. Because U already contains every element under consideration, its complement is the empty set: U′ = ∅. Therefore, A ∩ U′ = A ∩ ∅ = ∅, since the intersection of any set with the empty set is empty. Thus option A is correct.

When A ⊆ B, every element of A is also an element of B. Therefore, any element that is outside B must certainly be outside A. This means B′ ⊆ A′. Complementation reverses the direction of subset inclusion. The other statements are not always true: for example, A′ ∩ B′ equals (A ∪ B)′, not necessarily the empty set.

Within U, set A contains 1 through 8, while set B contains 14 through 20. Their union therefore contains the two outer ranges: {1, ..., 8, 14, ..., 20}. The elements of U that are missing from this union are exactly 9, 10, 11, 12, and 13. Hence (A ∪ B)′ = {9, 10, 11, 12, 13}.

The universal set U contains all integers from −10 through 10, including both endpoints, so n(U) = 10 − (−10) + 1 = 21. The condition |x| ≤ 3 means −3 ≤ x ≤ 3, giving A = {−3, −2, −1, 0, 1, 2, 3}, which has 7 elements. Since A' contains the elements of U that are not in A, n(A') = n(U) − n(A) = 21 − 7 = 14. Therefore, option A is correct.

Every multiple of 12 is automatically a multiple of 6, because 12 = 2 × 6. Thus every multiple of 12 belonging to U is already an element of A. The multiples of 12 in U are 12, 24, and 36, and none lies outside A. Consequently, no multiple of 12 belongs to A′, so the required number is 0.

Complements are taken relative to ℝ. Since A includes all real numbers up to and including 0, A′ = (0, ∞). Since B includes all real numbers from 2 onward, including 2, B′ = (−∞, 2). Their intersection is therefore the real interval (0, 2). Both endpoints are excluded: 0 is in A and 2 is in B.

First factor the quadratic: x² − 11x + 30 = (x − 5)(x − 6). Thus the roots are x = 5 and x = 6, and both belong to U, so A = {5, 6}. The complement A' in U contains every element of U except 5 and 6. Intersecting A' with {5, 6, 7, 8} removes 5 and 6 and leaves {7, 8}. Therefore, option A is correct.

Assuming ℕ begins at 1, U has 48 elements. There are 12 multiples of 4 and 8 multiples of 6 up to 48. Their common elements are multiples of lcm(4,6) = 12, namely 4 numbers: 12, 24, 36, and 48. Hence n(A ∪ B) = 12 + 8 − 4 = 16. Therefore n((A ∪ B)′) = 48 − 16 = 32.

Factor the quadratic: x² − 2x − 15 = (x − 5)(x + 3). The inequality (x − 5)(x + 3) ≤ 0 holds for −3 ≤ x ≤ 5. Therefore, A = {−3, −2, −1, 0, 1, 2, 3, 4, 5}, which has 9 integers. The universal set U contains all integers from −8 through 8, so n(U) = 17. Hence n(A′) = n(U) − n(A) = 17 − 9 = 8.

The complement contains every real number that is not in A. Because (−∞, −2) excludes −2, the point −2 belongs to A′. The interval [3, 7] includes both endpoints 3 and 7, so neither endpoint belongs to the complement. Numbers between −2 and 3, together with numbers greater than 7, are excluded from A and therefore form A′ = [−2, 3) ∪ (7, ∞).

The expression A′ ∩ B represents the elements that belong to B but do not belong to A. Since A is a subset of B, removing all 31 elements of A from the 54 elements of B leaves B − A. Therefore, n(A′ ∩ B) = n(B) − n(A) = 54 − 31 = 23. The value n(U) is not needed for this calculation.

A contains all even numbers in U, so A′ contains all odd numbers from 1 to 20. The prime numbers in U are 2, 3, 5, 7, 11, 13, 17, and 19. Intersecting this prime set with A′ removes 2 because it is even. The remaining odd primes are {3, 5, 7, 11, 13, 17, 19}, which is therefore A′ ∩ B.

There are floor(60/4) = 15 multiples of 4 and floor(60/9) = 6 multiples of 9. Numbers in both sets must be multiples of lcm(4, 9) = 36; only 36 occurs up to 60. Thus n(A ∪ B) = 15 + 6 − 1 = 20 by inclusion–exclusion. The complement contains the remaining 60 − 20 = 40 elements.

The condition x² ≤ 16 is equivalent to |x| ≤ 4, or −4 ≤ x ≤ 4. Since x must be an integer, A = {−4, −3, −2, −1, 0, 1, 2, 3, 4}, so n(A) = 9. The universal set contains the 21 integers from −10 to 10 inclusive. Therefore, n(A′) = n(U) − n(A) = 21 − 9 = 12.

The first part of A contains all real numbers up to and including −4, so the complement begins just after −4; therefore −4 is excluded. The second part contains numbers strictly between 2 and 6, so both 2 and 6 are excluded from A and included in its complement. Hence A′ = (−4, 2] ∪ [6, ∞).

Because A is a subset of B, the elements in B that are outside A are exactly B − A. The set identity A′ ∩ B = B − A therefore applies. Removing the 64 elements of A from the 97 elements of B gives n(A′ ∩ B) = 97 − 64 = 33. The size of U is extra information and does not affect this difference.

The complement A′ consists of all odd numbers in U. The primes from 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29. Since 2 is even, it is not in A′. Every other prime in this list is odd, so the intersection is {3, 5, 7, 11, 13, 17, 19, 23, 29}.

A′ consists of the elements of U not in A: {b, c, e, f, h}. B′ consists of the elements of U not in B: {a, c, f, g}. Their intersection contains the elements common to both complements, namely c and f. Equivalently, by De Morgan’s law, A′ ∩ B′ = (A ∪ B)′; since A ∪ B = {a, b, d, e, g, h}, its complement in U is {c, f}.