Complement of a Set and Its Properties

Complementation reverses the direction of inclusion. In general, if X ⊆ Y, then Yᶜ ⊆ Xᶜ; conversely, if Yᶜ ⊆ Xᶜ, then X ⊆ Y. Applying this converse rule to Bᶜ ⊆ Aᶜ gives A ⊆ B. The statement B ⊆ A reverses the result incorrectly, while A = Bᶜ and A ∩ B = ∅ are stronger claims that do not necessarily follow from the given information. Hence option A is correct.

First use the inclusion–exclusion formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 42 + 38 − 20 = 60. The complement of A ∪ B contains the elements of U that are not in either A or B. Therefore, n((A ∪ B)ᶜ) = n(U) − n(A ∪ B) = 75 − 60 = 15. Thus option A is correct. The value 60 represents the union, not its complement.

First find the cardinality of Aᶜ ∪ Bᶜ using the inclusion–exclusion formula: n(Aᶜ ∪ Bᶜ) = n(Aᶜ) + n(Bᶜ) − n(Aᶜ ∩ Bᶜ) = 35 + 50 − 18 = 67. By De Morgan’s law, Aᶜ ∪ Bᶜ = (A ∩ B)ᶜ. Therefore, n((A ∩ B)ᶜ) = 67. Since the universal set has 90 elements, n(A ∩ B) = 90 − 67 = 23. Hence, option A is correct.

All complements are taken with respect to U. Thus Aᶜ = {6,7,8,9,10}, while Bᶜ = {1,2,3,9,10}. The difference Aᶜ − Bᶜ consists of elements present in Aᶜ but absent from Bᶜ. Removing 9 and 10 from Aᶜ leaves {6,7,8}. Equivalently, Aᶜ − Bᶜ = Aᶜ ∩ B. Therefore, option A is correct.

The difference B − A contains elements that belong to B but not to A. From B = {1,2,3,4,5}, removing 2, 3, and 5 gives B − A = {1,4}. Its complement relative to U contains every element of U except 1 and 4. Consequently, (B − A)ᶜ = {2,3,5,6,7,8,9}. Option B is the difference itself, not its complement, so option A is correct.

Among the integers from 1 to 40, there are 40/4 = 10 multiples of 4 and 40/5 = 8 multiples of 5. Multiples of both 4 and 5 are multiples of 20, and there are 40/20 = 2 of them. By inclusion–exclusion, n(A) = 10 + 8 − 2 = 16. Therefore, n(Aᶜ) = n(U) − n(A) = 40 − 16 = 24, so option A is correct.

Let T be the set of people who like tea and C be the set of people who like coffee. By inclusion–exclusion, n(T ∪ C) = n(T) + n(C) − n(T ∩ C) = 85 + 70 − 40 = 115. Therefore, the number who like neither is the complement of T ∪ C in the survey: 150 − 115 = 35. Hence, option A is correct.

Within U, the multiples of 4 are A = {4, 8, 12, 16, 20}. Its complement contains the numbers from 1 to 20 that are not divisible by 4. Intersecting this complement with the even numbers keeps only even numbers that are not multiples of 4: 2, 6, 10, 14, and 18. Therefore, option A is correct.

The complement of A with respect to U consists of elements in U that are not in A. Since A contains 1 through 6, we get Aᶜ = {7, 8, 9, 10}. The condition Aᶜ ⊆ B requires B to contain all four of these elements, while B may contain additional elements from U. The smallest such B contains no extras, so B = Aᶜ = {7, 8, 9, 10}.

The condition P ∩ Q = ∅ says that P and Q have no common elements, while P ∪ Q = U says that together they contain every element of the universal set. Thus, every element not in P must be in Q, and every element of Q is outside P. Therefore, Q is exactly the complement of P in U: Pᶜ = Q. Hence, option A is correct.

The complement A^c contains all elements of U that are not in A. Thus, A^c={1,2,4,5,7,8,10,11}, which has 8 elements. The elements 6 and 12 belong to A, so neither is already in A^c. Adding both of them to A^c gives {1,2,4,5,6,7,8,10,11,12}, containing 10 elements. Therefore, option A is correct.

A number divisible by both 2 and 5 must be divisible by their least common multiple, 10. Between 1 and 18, the only multiple of 10 is 10, so A={10}. The complement contains every element of U except 10. Hence A^c={1,2,3,4,5,6,7,8,9,11,12,13,14,15,16,17,18}, which is option A.

All complements are taken with respect to U. Since C=A^c, taking the complement again gives C^c=(A^c)^c=A, by the double-complement law. Therefore, C^c∩{1,2,3,4}=A∩{1,2,3,4}. The elements common to A={2,4,6,8,10} and the second set are 2 and 4, so the answer is {2,4}, option A.

A partition of U requires two essential conditions: its parts must be pairwise disjoint, and their union must be the whole universal set. A and A^c are disjoint because no element can belong to both, so A∩A^c=∅. Together they contain every element of U, so A∪A^c=U. Therefore, option A proves that they form a partition.

A set and its complement are disjoint, so A ∩ Aᶜ = ∅. If A = Aᶜ, replacing Aᶜ by A would give A ∩ A = ∅, which means A = ∅. But then Aᶜ = U, and the equality A = Aᶜ would require ∅ = U. This contradicts U ≠ ∅. Equivalently, no nonempty universal set can be equal to its own complement. Therefore, option A is correct.

A complement is defined relative to a particular universal set: A^c=U−A. If U is enlarged while A stays unchanged, any newly included elements that are not members of A will belong to the new complement. Thus the complement may gain new elements. It is not necessarily empty, equal to A, or unchanged. Therefore, option A is correct.

The complement depends on the universal set. Relative to U1, A^c=U1−A={1,3,4}. Relative to U2, the complement is A^c=U2−A={1,3,4,6,7}. Comparing these two complements, the elements that appear only in the second one are 6 and 7. Thus the extra elements are {6,7}, making option A correct.

All complements are taken with respect to U. First, Aᶜ = U − A = {5, 6, 7, 8, 9, 10, 11, 12}. Intersecting this with B = {3, 4, 5, 6, 7} gives Aᶜ ∩ B = {5, 6, 7}. Now take the complement of this result in U: (Aᶜ ∩ B)ᶜ = U − {5, 6, 7} = {1, 2, 3, 4, 8, 9, 10, 11, 12}. Thus, option A is correct.

The complement of A with respect to U is Aᶜ = {1,3,5,7,9}. Taking the union with B gives Aᶜ ∪ B = {1,2,3,4,5,7,9}. The elements of U not present in this union are {6,8,10}, so (Aᶜ ∪ B)ᶜ = {6,8,10}. This also follows from De Morgan’s law: (Aᶜ ∪ B)ᶜ = A ∩ Bᶜ.

Among the numbers 1 to 24, six numbers are divisible by 4: 4, 8, 12, 16, 20, and 24. Four are divisible by 6: 6, 12, 18, and 24. The common elements are 12 and 24, so |A ∪ B| = 6 + 4 − 2 = 8. By De Morgan’s law, Aᶜ ∩ Bᶜ = (A ∪ B)ᶜ, and its size is 24 − 8 = 16.

De Morgan’s first law states that the complement of a union is the intersection of the complements: (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ. An element is outside A ∪ B precisely when it is outside A and also outside B. Option B incorrectly retains the union sign, option C uses the wrong operation, and option D is false because (Aᶜ)ᶜ = A.

The complement A^c contains the elements of U that are not in A, while B^c contains the elements of U that are not in B. By De Morgan’s law, A^c ∩ B^c = (A ∪ B)^c. Here, A ∪ B = {1,2,4,6,7,8,10}. Removing these elements from U={1,2,3,...,11} leaves {3,5,9,11}. Therefore, option A is correct.

A and Aᶜ are disjoint and together contain every element of U, so n(A) + n(Aᶜ) = n(U) = 84. Let n(Aᶜ) = x. The given relation gives n(A) = 2x. Hence 2x + x = 84, so 3x = 84 and x = 28. Therefore, n(Aᶜ) is 28. The finite-set complement formula is essential here.

Let n(A)=x. For a finite universal set, A and its complement A^c are disjoint and together contain every element of U. Therefore, n(A)+n(A^c)=n(U). Substituting n(A^c)=3n(A), n(U)=72, and n(A)=x gives x+3x=72, so 4x=72 and x=18. Thus, option A is the only correct answer.

The complement of A relative to U contains every element of U that is absent from A. Therefore Aᶜ = U \ A = {1,4,6,7,8,9}. Option B lists exactly these six elements, and options C and D express the same set using difference notation and set-builder notation. Option A omits 9, so it is not the complement.