Operations on Sets (Union, Intersection, Difference)

The intersection contains only the real numbers that belong to both intervals. The common range begins just greater than 2 because B excludes 2, so the left endpoint is open. The common range ends at 4, and 4 belongs to both A and B, so the right endpoint is closed. Therefore A ∩ B = (2, 4].

The intervals meet at the number 3, but 3 is not included in A because A has an open right endpoint. Although 3 is included in B, it must belong to both sets to be in the intersection. No other number is common to the intervals, so A ∩ B is the empty set, ∅.

The union contains every number that belongs to A or B or to both. The intervals overlap from 0 to 2, so there is no gap between them. The smallest included endpoint is −3 and the largest included endpoint is 5; both are closed because both original intervals include their endpoints. Therefore A ∪ B = [−3, 5].

A consists of all real numbers less than or equal to zero, while B consists of all real numbers greater than or equal to zero. A number in the intersection must satisfy both x ≤ 0 and x ≥ 0 simultaneously. The only real number satisfying both inequalities is x = 0. Since both inequalities include equality, zero belongs to both sets, so A ∩ B = {0}.

The difference A \ B consists of elements that belong to A but do not belong to B. Set A contains all numbers greater than −4 up to and including 1. Set B contains every number from −2, including −2, through values less than 3. Removing B from A removes [−2,1], leaving numbers greater than −4 and less than −2. Since −2 belongs to B, it is excluded. Hence A \ B = (−4,−2), so A is correct.

List the positive multiples of 2 less than 12: 2, 4, 6, 8, and 10. The positive multiples of 3 less than 12 are 3, 6, and 9. Because the condition uses “or,” take the union of these lists and write the repeated element 6 only once. Thus A = {2, 3, 4, 6, 8, 9, 10}; 12 is excluded because it is not less than 12.

The difference \(A-B\) consists of all elements that belong to \(A\) but do not belong to \(B\). From \(A=\{1,2,3,4\}\), remove 2 and 3 because both are in \(B\). The remaining elements are 1 and 4, so \(A-B=\{1,4\}\). Every element of this difference already belongs to \(A\), so it is a subset of \(A\). Thus option A is correct.

Let M be the mathematics-club set and S be the science-club set. By the inclusion–exclusion principle, n(M ∪ S) = n(M) + n(S) − n(M ∩ S) = 18 + 15 − 6 = 27. Thus, 27 students belong to at least one club. The students in neither club are outside this union, so their number is 40 − 27 = 13. Therefore, option B is correct.

De Morgan’s law states that the complement of a union equals the intersection of the complements: (A ∪ B)′ = A′ ∩ B′. An element lies outside A ∪ B only when it belongs to neither A nor B. That means it must be in A′ and also in B′, which is precisely membership in A′ ∩ B′. Therefore, option B is correct; A′ ∪ B′ would represent the complement of A ∩ B instead.

Let \(M\) be the music group and \(D\) the drama group. By inclusion–exclusion, \(n(M\cup D)=n(M)+n(D)-n(M\cap D)=32+28-10=50\). Thus 50 students participate in at least one activity. The number participating in neither is \(75-50=25\), so option B is correct. The overlap is subtracted once because it was counted twice.

Use De Morgan’s law: A'∪B'=(A∩B)'. The common elements of A and B are {2,4}, so |A∩B|=2. Since U has 10 elements, the complement of this intersection has 10−2=8 elements. Directly, A'={6,7,8,9,10} and B'={1,3,5,7,9}; their union is {1,3,5,6,7,8,9,10}, which also contains eight elements. Thus option C is correct.

The intersection A ∩ B consists of numbers that are multiples of both 2 and 3, hence multiples of lcm(2,3) = 6. Within U = {1, 2, ..., 24}, these are 6, 12, 18, and 24, so the intersection has 4 elements. The complement is taken in U, which has 24 elements. Therefore, |(A ∩ B)'| = 24 − 4 = 20, making option C correct.

First form the union: A ∪ B = {a, c, e} ∪ {b, e} = {a, b, c, e}. The complement contains the elements of U that are absent from this union. Since d is the only element of U not in A ∪ B, (A ∪ B)' = {d}. Therefore option A is correct. The complement must be taken relative to U.

Complements are taken relative to U. Thus A' = U − A = {2,4,6,8}, and B' = U − B = {1,4,6,7}. Their intersection contains the elements appearing in both lists, namely 4 and 6. Therefore A' ∩ B' = {4,6}, so option A is correct. Option B is A ∩ B, while the larger options confuse union with intersection.

In U, there are 15 multiples of 2 and 6 multiples of 5. Their overlap consists of multiples of lcm(2,5)=10, namely 10, 20, and 30, so there are 3 common elements. By inclusion-exclusion, n(A∪B)=15+6−3=18. Therefore the complement has n((A∪B)')=30−18=12 elements. Hence option A is correct.

Use the inclusion–exclusion formula for three finite sets: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C). Substitution gives 10 + 8 + 6 − 3 − 2 − 1 + 0 = 18. Thus, the union contains 18 elements. Pairwise overlaps are subtracted to avoid counting common elements twice.

Use the three-set inclusion–exclusion formula: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C). Substitution gives 8 + 7 + 6 − 2 − 1 − 3 + 1 = 16. The triple intersection is added once because it was over-subtracted.

For three finite sets, the inclusion–exclusion principle is \(n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)\). Substituting the given values gives \(12+14+10-5-3-4+2=26\). The pairwise intersections are subtracted because their elements were counted twice, and the triple intersection is added once because it was then subtracted too many times. Therefore, option B, 26, is correct.

Five students are in neither A nor B, so the union contains 60 − 5 = 55 students. Apply the inclusion–exclusion formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Thus 55 = 25 + 30 − n(A ∩ B), giving n(A ∩ B) = 0. Therefore, the two sets are disjoint in this situation.

The governing concept is symmetric difference. The set A △ B contains elements belonging to exactly one of A or B, so A △ B = (A − B) ∪ (B − A). These two difference sets cannot overlap; hence their cardinalities are added. Thus n(A △ B) = 29 + 17 = 46. Options C and D represent only one part, and option A subtracts the parts incorrectly. Therefore, option B is correct.

The complement Aᶜ contains all elements that are not in A. Therefore, B ∩ Aᶜ contains those elements that belong to B and do not belong to A. This is exactly the definition of the set difference B − A, also written as B \ A, so option B is correct. In a Venn diagram, shade set B and remove its overlapping part with A. Option A represents elements in A but not B, while option C represents the common elements of A and B.

The set A can be divided into two non-overlapping regions: A − B, containing elements that are in A but not in B, and A intersection B, containing elements common to both sets. Their union is A, so their cardinalities add. Thus n(A) = n(A − B) + n(A intersection B) = 32 + 18 = 50. Subtracting the values would be incorrect because the two given regions are separate parts of A, not quantities to be removed from one another. Therefore, option B is correct.

The union of two sets consists of three disjoint Venn-diagram regions: A − B, A intersection B, and B − A. Thus n(A union B) = n(A − B) + n(A intersection B) + n(B − A). Substitution gives 92 = 35 + 24 + n(B − A). Solving, n(B − A) = 92 − 35 − 24 = 33. The value 59 is only 35 + 24 and does not represent the missing B-only region. Hence option A is correct.

Use the cardinality formula n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 76 = 31 + n(B) − 0. Therefore, n(B) = 76 − 31 = 45. The zero intersection means that A and B are disjoint, so no common elements are counted twice. Option A is n(A), option C is the union, and option D incorrectly adds 31 and 76.

Apply n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Rearranging gives n(A ∩ B) = n(A) + n(B) − n(A ∪ B). Substitution yields n(A ∩ B) = 68 + 53 − 91 = 121 − 91 = 30. The intersection is the part counted in both sets, so it is obtained by removing the union count from the sum of the two individual counts.