Complement of a Set and Its Properties

First calculate the intersection: A ∩ B = {2, 5}, because 2 and 5 are the only elements common to A and B. The complement of a set is taken with respect to the universal set U, so remove 2 and 5 from U. This gives (A ∩ B)′ = U − {2, 5} = {1, 3, 4, 6, 7, 8, 9, 10}. Option A is the intersection itself, option C is A ∪ B, and option D omits several elements that belong to the complement.

The governing concept is that a complement is always taken with respect to the stated universal set. First combine all distinct elements: A ∪ B = {1, 2, 3, 5, 7, 9, 11}. Now remove these elements from U = {1, 2, ..., 12}. The elements left are 4, 6, 8, 10, and 12, so (A ∪ B)′ = {4, 6, 8, 10, 12}. Option B lists the union itself, while option C is only the intersection.

The complement of A is defined relative to the universal set U. It contains every element of U that is not an element of A. Starting with U = {1, 2, 3, 4, 5} and removing 1, 3, and 5 leaves {2, 4}. Therefore A′ = U \ A = {2, 4}, making option A correct. The universal set must always be identified before finding a complement.

The complement B′ consists of all elements in the universal set U that are not in B. From U = {a, b, c, d}, remove b and d, the elements of B. The remaining elements are a and c. Thus B′ = U \ B = {a, c}, so option A is the only correct answer. The complement depends on the specified universal set.

To find A′, take the elements of the universal set U that are absent from A. The universal set contains 2, 4, 6, 8, and 10; removing 4 and 8 leaves 2, 6, and 10. Therefore A′ = U \ A = {2, 6, 10}. Option B repeats A itself, and option D omits 2, so option A is correct.

Under the usual school convention in this question, N denotes the positive natural numbers. Thus U = {1, 2, 3, 4, 5, 6}. The complement A′ contains the members of U that are not in A = {1, 2, 3}. Removing those three elements leaves {4, 5, 6}; hence option A is correct. Zero is not included under the stated convention.

First list every integer in the universal set: U = {−2, −1, 0, 1, 2}. The complement A′ contains exactly those elements of U that are not present in A. Since A = {−1, 0, 1}, removing these three elements from U leaves −2 and 2. Therefore, A′ = {−2, 2}. The complement is always defined relative to the stated universal set.

The complement A′ of a set A contains all elements of the universal set U that do not belong to A. Here A is the empty set, so it contains no elements at all. Consequently, none of the elements p, q, or r is excluded from the complement. Every element of U therefore belongs to A′, giving A′ = U = {p, q, r}. This illustrates the standard identity ∅′ = U.

The complement of \(A\) is defined as \(A'=U\setminus A\), meaning that it contains all elements of the universal set that are not in \(A\). Taking the complement again gives \((A')'=U\setminus(U\setminus A)=A\). Thus, the double-complement law states that complementing a set twice returns the original set. Therefore, option A is correct.

The set \(A'\) contains the elements of the universal set that are not in \(A\). Since \(U=\{1,2,3,4,5,6,7\}\) and \(A'=\{2,4,6\}\), the elements remaining in \(U\) are \(1,3,5,7\). These remaining elements form \(A\). Equivalently, \(A=U\setminus A'\). Hence \(A=\{1,3,5,7\}\), so option A is correct.

The complement \(V'\) consists of the elements of the universal set \(U\) that do not belong to \(V\). Here, \(U=\{a,e,i,o,u\}\) and the excluded elements are \(e\) and \(o\). Removing them from \(U\) leaves \(\{a,i,u\}\). Therefore, \(V=U\setminus V'=\{a,i,u\}\), and option A is the correct answer.

The set-builder condition says that \(A\) contains the even elements of \(U\), namely \(\{2,4,6,8\}\). The complement \(A'\) contains every element of \(U\) that is not even. Within this finite universal set, those elements are the odd numbers \(1,3,5,7\). Hence \(A'=\{1,3,5,7\}\), making option A correct.

The condition \(x<5\) selects the elements \(1,2,3,4\) from the universal set, so \(A=\{1,2,3,4\}\). The complement contains all elements of \(U\) that are not in this set. Therefore, \(A'=\{5,6,7,8,9\}\). Notice that 5 is included in the complement because the condition is strictly less than 5, not less than or equal to 5.

The condition \(x\ge 7\) includes 7 and every larger element of the universal set. Thus, \(A=\{7,8,9,10\}\). The complement contains the elements of \(U\) that are not in \(A\), namely \(\{1,2,3,4,5,6\}\). Since the inequality includes equality, 7 belongs to \(A\), not to \(A'\). Therefore, option A is correct.

The universal set has 12 elements, so \(n(U)=12\). The set \(A=\{3,6,9,12\}\) has 4 elements, so \(n(A)=4\). For a finite universal set, a set and its complement partition the universal set; therefore, \(n(A')=n(U)-n(A)=12-4=8\). Hence option A is correct.

For a finite universal set, the set \(A\) and its complement \(A'\) are disjoint and together contain all elements of \(U\). Therefore, their cardinalities satisfy \(n(A)+n(A')=n(U)\). Substituting the given values gives \(18+n(A')=30\), so \(n(A')=30-18=12\). Thus, option A is correct.

A set and its complement are disjoint, and their union is the universal set. Consequently, their cardinalities satisfy \(n(A)+n(A')=n(U)\). Using the given values, \(n(A)+22=50\), so \(n(A)=50-22=28\). The number 22 counts the elements of the complement, while 50 counts all elements of the universal set. Therefore, option A is correct.

The set \(A\) and its complement \(A'\) are disjoint, and every element of the universal set belongs to exactly one of them. Hence, \(A\cup A'=U\) and \(n(U)=n(A)+n(A')\). Substituting the given values gives \(n(U)=15+35=50\). Therefore, the universal set contains 50 elements, so option A is correct.

First form the union by listing every element that belongs to A or B: A ∪ B = {1, 2, 3, 4, 5}. The complement is taken with respect to U, so we select the elements of U that are absent from this union. Only 6 is missing from A ∪ B. Therefore, (A ∪ B)′ = {6}, making option A correct. Option B is the union itself, not its complement.

The intersection contains elements common to both sets. Comparing A and B gives A ∩ B = {b, d}. The complement is formed by removing b and d from the universal set U = {a, b, c, d, e}. The remaining elements are a, c, and e. Hence (A ∩ B)′ = {a, c, e}, so option A is correct. Option B represents the intersection itself.