Complement of a Set and Its Properties

First combine the sets: \(A\cup B=\{1,2,4,6,8,9,10,12,14,16\}\). The complement contains every element of U not appearing in this union. Checking the integers 1 through 16 leaves \(\{3,5,7,11,13,15\}\). Therefore option A is correct. Notice that repeated elements such as 4 and 16 are written only once in a union.

The set \(B\) consists of all odd numbers from 1 to 20. The odd prime numbers in this range are 3, 5, 7, 11, 13, 17, and 19. Removing these from B leaves the odd, non-prime numbers \(\{1,9,15\}\). Note that 1 is not prime, because a prime number must have exactly two distinct positive divisors.

The condition \(x^2<9\) means \(-3<x<3\). Because x is restricted to integers, \(A=\{-2,-1,0,1,2\}\). The universal set contains all integers from -5 through 5, so removing A leaves \(A^c=\{-5,-4,-3,3,4,5\}\). The endpoints -3 and 3 are included in the complement because their squares equal 9, not a value less than 9.

The complement is taken inside the universal interval \([-3,7]\). Set A contains every real number strictly between -1 and 4, but it excludes the endpoints -1 and 4. Therefore those endpoints belong to the complement. The portions outside A, while remaining inside U, are \([-3,-1]\) and \([4,7]\), so \(A^c=[-3,-1]\cup[4,7]\).

The complement must contain points of U that are not in A. Since A=[0,2] includes both 0 and 2, those endpoints must be excluded from the complement. The left part is therefore (-4,0), while the right part is (2,6]. The endpoint -4 is excluded because it is excluded from U, and 6 is included because U includes 6. Hence option A is correct.

Taking the complement reverses the direction of set inclusion. Since A ⊆ B, every element outside B is also outside A, so Bᶜ ⊆ Aᶜ. Similarly, B ⊆ C gives Cᶜ ⊆ Bᶜ. Combining these results gives Cᶜ ⊆ Bᶜ ⊆ Aᶜ. Therefore, option A is correct. Equality is not necessary, and the intersection of two complements cannot generally be the whole universal set.

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.

The elements of A are the positive multiples of 6 not exceeding 60: 6, 12, 18, …, 60. Their number is 60 ÷ 6 = 10. Since U has 60 elements, the complement Aᶜ contains all elements of U that are not divisible by 6. Hence n(Aᶜ) = n(U) − n(A) = 60 − 10 = 50. Therefore, option A is correct; option B counts A itself.

By definition, the complement of a set A contains all elements of U that are not in A. If A is the empty set, it contains no elements, so every element of U lies outside A. Therefore, ∅ᶜ = U. The complement of U is instead ∅, and a singleton or infinite set generally has a complement that is neither necessarily empty nor the whole universal set. Hence option A is the unique correct answer.

The integers from −6 through 6 form the universal set U, so n(U) = 13 because there are 6 negative integers, zero, and 6 positive integers. Solving x² = 16 gives x = −4 or x = 4, both of which belong to U. Thus A = {−4,4} and n(A) = 2. Therefore, n(Aᶜ) = n(U) − n(A) = 13 − 2 = 11. 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.

The complement has one-third as many elements as the universal set, so n(Aᶜ) = (1/3) × 96 = 32. A set and its complement partition the universal set into two non-overlapping parts. Therefore, n(A) + n(Aᶜ) = n(U), and n(A) = 96 − 32 = 64. Thus, option A is correct; 32 is the size of Aᶜ, not A.

The universal set U contains all integers from 1 through 36, so n(U) = 36. The perfect squares in this range are 1, 4, 9, 16, 25, and 36; hence n(A) = 6. The complement contains every element of U that is not a perfect square. Therefore, n(Aᶜ) = n(U) − n(A) = 36 − 6 = 30, so 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}.