Complement of a Set and Its Properties

First find the union of A and B: A ∪ B = {1, 2, 3, 4, 5}. The complement is always taken with respect to the universal set U, so (A ∪ B)′ = U \ (A ∪ B) = {6, 7, 8}. This set contains 3 elements, giving option A. Equivalently, because U has 8 elements and the union has 5 elements, its complement has 8 − 5 = 3 elements.

The intersection consists of elements common to both sets. Thus, A ∩ B = {3, 4}, which has 2 elements. The complement is taken in U, which has 10 elements. Therefore, n((A ∩ B)′) = n(U) − n(A ∩ B) = 10 − 2 = 8. Hence option A is correct; the value 2 is the size of the intersection itself, not its complement.

The complement A′ contains all elements of the universal set U that are not elements of A. Subtracting A = {1,3,5} from U = {1,2,3,4,5,6} gives A′ = {2,4,6}. Among the choices, only 2 belongs to this complement. The numbers 1, 3, and 5 are already in A, so they cannot be in A′.

The complement A′ contains every element of the universal set U that is not an element of A. Since A = {n, p}, its complement is A′ = {m, o, q}. Therefore, n is not in A′, and option A is correct. The element p would also be absent from A′, but it is not among the answer choices. Careful reading of “will not be” is essential here.

The condition x ≤ 4 describes the elements 1, 2, 3, and 4, so A = {1, 2, 3, 4}. The complement contains the remaining elements of U: {5, 6, 7, 8, 9, 10}. These are exactly the elements satisfying x > 4. The equality case x = 4 must be excluded from the complement because 4 already belongs to A; therefore option A is the only correct form.

The positive divisors of 12 that belong to U are 1, 2, 3, 4, 6, and 12. Hence A = {1, 2, 3, 4, 6, 12}. The complement A′ consists of all elements of U that are not divisors of 12. Removing these divisors from U leaves A′ = {5, 7, 8, 9, 10, 11}. Therefore, option A is correct. Option B lists A itself, not its complement.

The universal set U contains all seven days of the week. W contains the two weekend days, Saturday and Sunday. The complement W′ therefore contains every day in U that is not in W: Monday, Tuesday, Wednesday, Thursday, and Friday. Thus, W′ has five elements, and option A is correct.

First calculate the union: A ∪ B = {1, 2, 3, 4, 6, 7}. The complement is taken relative to U, so we remove every member of this union from U. The elements left are 5, 8, and 9. Thus, (A ∪ B)′ = {5, 8, 9}, which is option A. Option B is the union itself, while option C incorrectly includes 6 and 7, which are already in the union.

First find the intersection, because the complement is applied to the entire expression. The elements common to A and B are 2 and 4, so A ∩ B = {2,4}. The complement is taken relative to U; therefore remove 2 and 4 from U. The remaining elements are {1,3,5,6,7,8,9,10}. Option B is only the intersection, C is B′, and D is A ∪ B.

A set and its complement partition the universal set into two disjoint parts. Consequently, their cardinalities satisfy n(A) + n(A′) = n(U). Substituting the given values gives n(A) + 19 = 64, so n(A) = 64 − 19 = 45. Option C gives the size of the complement, while option B incorrectly adds the two numbers. Option D would be possible only if A′ were empty.

The complement A′ contains all elements of U that are not members of A. If A′ equals the whole universal set U, then every element of U lies outside A. Therefore A contains no elements and must be the empty set, A = ∅. If A were U, its complement would be empty, not U. The statement A = A′ is not generally implied, and A ≠ ∅ directly contradicts the condition.

The universal set contains the natural numbers from 1 through 14. Set A consists of the odd numbers 1, 3, 5, 7, 9, 11, and 13. Its complement within U therefore consists of the even numbers 2, 4, 6, 8, 10, 12, and 14. There are seven such numbers, so n(A′) = 7. The answer depends on the stated universal set; numbers outside U are not considered.

The complement Aᶜ contains every element of the universal set U that is not present in A. Starting with U = {1,2,3,4,5}, remove the elements 1 and 3 because they belong to A. The remaining elements are 2, 4, and 5, so Aᶜ = {2,4,5}. Option B is A itself, option C is only a partial set, and option D would be correct only if A equalled U.

The complement of A is formed by selecting elements of U that do not belong to A. The universal set has four elements: a, b, c, and d. Since b and d are already in A, the elements left in U are a and c. Hence Aᶜ = {a,c}. Option B repeats A, while options C and D contain one element from A or omit a required complementary element.

By definition, the complement of A contains elements that are in the universal set U but are not in A. Set difference expresses exactly this idea, so Aᶜ = U \ A. Since A is normally a subset of U, A \ U is empty rather than the complement. Also, A ∩ U equals A, and A ∪ U equals U. Thus only option A has the required meaning.

An element belongs to the complement Aᶜ precisely when it is in the universal set U but not in A. The conditions x ∈ U and x ∉ A state exactly this membership rule, so x ∈ Aᶜ. It cannot be concluded that x belongs to A, because the second condition says the opposite. The empty set has no elements, and A ∩ Aᶜ is always empty.

The complement of A contains elements of U that are not in A. Here A is exactly equal to U, so every element of the universal set is already included in A. No element remains outside A, and therefore Aᶜ is the empty set ∅. This is the standard identity Uᶜ = ∅. Options C and D are only partial subsets of U, not the complete complement.

The complement of A consists of all elements of U that are not in A. Since A is the empty set, it contains none of the elements 1, 2, or 3. Thus every element of U belongs to Aᶜ, giving Aᶜ = {1,2,3} = U. This is the standard identity ∅ᶜ = U. The other options omit some elements or incorrectly claim that the complement is empty.

For a set A contained in the universal set U, the complement Aᶜ contains every element of U that is not in A. Therefore, the number of elements in the complement is found by subtracting the number of elements of A from the number of elements of U: n(Aᶜ) = n(U) − n(A) = 20 − 7 = 13. Hence, option A is correct. Option C gives n(A), while option D gives n(U), not the complement.

The complement Aᶜ consists of the elements of the universal set U that are not in A. Therefore, A contains all elements of U that are not listed in Aᶜ. Removing 5 and 6 from U gives A = U − Aᶜ = {1, 2, 3, 4, 5, 6} − {5, 6} = {1, 2, 3, 4}. Thus option A is correct. Option B is the given complement, option C incorrectly includes elements of Aᶜ, and option D would be possible only if Aᶜ were equal to U.