Complement of a Set and Its Properties

The complement \(A'\) contains every element of the universal set \(U\) that is not an element of \(A\). Starting with \(U=\{a,b,c,d,e,f\}\), remove \(a,c,f\), because those elements belong to \(A\). The remaining elements are \(b,d,e\). Hence \(A'=\{b,d,e\}\), making option A correct.

The complement \(A'\) contains all elements of the universal set \(U\) that are not present in \(A\). The universal set has 10 elements, while \(A\) has 3 elements: 1, 4, and 9. Therefore, \(n(A')=n(U)-n(A)=10-3=7\). In fact, \(A'=\{2,3,5,6,7,8,10\}\), which confirms that it contains seven elements. Hence, option C is correct.

De Morgan’s law states that the complement of an intersection is the union of the complements: \((A\cap B)'=A'\cup B'\). An element is outside \(A\cap B\) whenever it fails to belong to at least one of the two sets. Therefore it belongs to \(A'\cup B'\). Option B is correct; option A incorrectly uses intersection.

The complement is defined by \(A'=U\setminus A\). If \(A'=\varnothing\), then there is no element of \(U\) outside \(A\); hence every element of \(U\) belongs to \(A\), so \(U\subseteq A\). Since the question already gives \(A\subseteq U\), both inclusions imply \(A=U\). Therefore option B is correct.

A set and its complement are disjoint and together contain every element of the universal set. Therefore, their cardinalities satisfy \(n(A)+n(A')=n(U)\). Substituting the given values gives \(n(A)+19=80\), so \(n(A)=80-19=61\). Hence option C is correct. The answer cannot exceed 80 because \(A\subseteq U\).

The universal set U contains the integers 1 through 18, so it has 18 elements. The elements divisible by 4 are A = {4, 8, 12, 16}, giving |A| = 4. The complement A′ contains the elements of U that are not in A. Hence |A′| = |U| − |A| = 18 − 4 = 14, so option C is correct.

The complement of A relative to U is defined as A′ = U \ A, meaning all elements of U that are not in A. Since A is empty, none of the elements of U are removed. Therefore A′ = U = {2, 4, 6}. Option A is A itself, option C contains an element not in U, and option D is a power set rather than a complement.

The complement of a set is taken with respect to the universal set \(U\). Thus, \(A'=U\setminus A=\{1,3,5,7,8\}\), because these elements are in \(U\) but not in \(A\). Similarly, \(B'=U\setminus B=\{1,2,3,5,7\}\). The intersection contains only elements common to both complements: \(A'\cap B'=\{1,3,5,7\}\). Therefore, option A is correct. This also agrees with De Morgan’s law: \(A'\cap B'=(A\cup B)'\).

First find the intersection. The only element common to A = {p,q,r} and B = {r,s} is r, so A ∩ B = {r}. The complement is taken in U, not in an unrestricted universe. Removing r from U = {p,q,r,s,t} leaves {p,q,s,t}. Hence (A ∩ B)′ = {p,q,s,t}, so option B is correct.

Because A is a subset of B, every element of A is also an element of B. Now take any element x in B'. It is not in B. Since every element of A must lie in B, x cannot be in A either; therefore x belongs to A'. Hence B' ⊆ A'. This is the complement rule that reverses inclusion. Equality is not guaranteed unless A = B.

The complement A' contains the elements of the universal set U that are not in A. Therefore, A consists of all elements of U that are not in A'. Subtracting {4, 10} from {2, 4, 6, 8, 10, 12} leaves {2, 6, 8, 12}. Thus option A is correct. The result also satisfies A ∩ A' = ∅ and A ∪ A' = U.

First find the union of A and B by listing every element appearing in either set: A ∪ B = {1, 2, 3, 5, 6, 7}. The complement is taken relative to U, so remove these union elements from U. The elements left are {4, 8, 9, 10}. Hence option A is correct. Option B is the union itself, while option C is the intersection, not the complement.

The universal set U contains the natural numbers 1 through 30, so n(U) = 30. The positive divisors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30, giving n(A) = 8. The complement A' contains all elements of U that are not in A. Thus n(A') = n(U) − n(A) = 30 − 8 = 22. Hence option C is correct.

The governing definition is A′ = U − A: the complement contains precisely those elements of U that are absent from A. Removing 2, 4, 6, and 8 from U = {1, 2, 3, 4, 5, 6, 7, 8, 9} leaves the odd elements {1, 3, 5, 7, 9}. Hence option A is correct. Option B is A itself, option C is the whole universal set, and option D would apply only if A equalled U.

Use the complement-counting principle |A′| = |U| − |A|. The integers from −3 through 4 are U = {−3, −2, −1, 0, 1, 2, 3, 4}, so |U| = 8. The condition x² < 4 is equivalent to −2 < x < 2, giving A = {−1, 0, 1} and |A| = 3. Therefore |A′| = 8 − 3 = 5, so option C is correct.

First find the union. A contains a, c, and e, while B contains b, c, and d. Together they contain every element of U, so A ∪ B = {a,b,c,d,e} = U. The complement is defined relative to U. Therefore (A ∪ B)′ = U \ (A ∪ B) = U \ U = ∅. Option B is not correct because c is already included in both A and B, so it cannot remain in the complement.

Use the inclusion–exclusion formula: |A ∪ B| = |A| + |B| − |A ∩ B|. Therefore, |A ∪ B| = 12 + 9 − 5 = 16. The complement contains the elements of U that are not in A ∪ B, so |(A ∪ B)'| = |U| − |A ∪ B| = 20 − 16 = 4. Hence, option B is correct. Subtracting the intersection is essential because common elements would otherwise be counted twice.

The complement is always taken relative to U. Here A' = U − A = {2, 4, 6, 8}. Taking the complement of A' removes these even elements from U and returns the odd elements: (A')' = U − A' = {1, 3, 5, 7} = A. This is the double-complement law, (A')' = A. Therefore, option B is correct. Option A is only A', not the double complement.

The universal set consists of all integers from 0 through 10: U = {0,1,2,3,4,5,6,7,8,9,10}. The complement A' contains elements of U that are not in A. Since A contains all the even integers in this range, the remaining elements are the odd integers {1,3,5,7,9}. Thus A' = {1,3,5,7,9}, so option A is correct. Zero cannot appear in A' because it already belongs to A.

A set A and its complement A' divide the universal set U into two disjoint parts. Therefore, |A| + |A'| = |U|. Substituting the given values gives |A| + 6 = 15, so |A| = 15 − 6 = 9. Hence option B is correct. The answer cannot be 6 because that is the size of A', and it cannot be 15 because A is only one part of U. The complement relation also guarantees that the two cardinalities add to the universal-set cardinality.