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].

The union includes all elements from both sets, but it does not include numbers in the gap between 2 and 3. Since the intervals are disjoint, they cannot be combined into the single interval (1, 4), which would incorrectly include the gap. Therefore the union must remain written as (1, 2) ∪ (3, 4).

An intersection contains only the numbers common to both sets. Set A contains all numbers up to and including 4, while set B contains all numbers strictly greater than 1. A common number must therefore be greater than 1 and less than or equal to 4. The lower endpoint is open because 1 is excluded from B, and the upper endpoint is closed because 4 is included in A. Thus A ∩ B = (1, 4], so option A is correct.

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.

The relation A ⊆ B means that every element of A is already an element of B. Therefore, taking the union of A and B adds no new element to B, so A ∪ B = B. Also, A ∩ B = A and A \ B = ∅. Option B would require B ⊆ A, while option C would also require B ⊆ A. Option D is not generally possible because A \ B is empty, not equal to B.

The equality A ∪ B = B says that adding all elements of A to B does not enlarge B. To prove the result element by element, take any x ∈ A. Then x ∈ A ∪ B, and because the union equals B, x must belong to B. Thus every element of A lies in B, so A ⊆ B. The other options require additional conditions and are not forced by the given equality.

The difference A − B consists of elements that belong to A but do not belong to B. If this difference is empty, there is no element of A outside B. Consequently, every element of A must also belong to B, which is exactly the statement A ⊆ B. Equality is not implied because B may contain additional elements. The reverse subset relation and disjointness are also not guaranteed.

Because subset inclusion is transitive, \(A\subseteq B\) and \(B\subseteq C\) imply \(A\subseteq C\). When one set is a subset of another, their intersection is the smaller set: if \(X\subseteq Y\), then \(X\cap Y=X\). Therefore, \(A\cap C=A\). The other choices are not generally correct: \(B\) and \(C\) may contain additional elements, while the empty set would occur only if \(A\) itself were empty.

The difference \(A\setminus B\) consists of elements that belong to \(A\) but do not belong to \(B\). Since \(A\subseteq B\), every element of \(A\) is already in \(B\). Consequently, there is no element of \(A\) outside \(B\), so the difference contains no elements and equals \(\emptyset\). For example, if \(A=\{1,2\}\) and \(B=\{1,2,3\}\), then removing all elements of \(B\) from \(A\) leaves nothing.

The symmetric difference is defined by A △ B = (A − B) ∪ (B − A). Since A ⊆ B, no element of A lies outside B, so A − B = ∅. Therefore A △ B = B − A. Given that A △ B = {7}, it follows immediately that B − A = {7}. Hence option A is correct; the empty-set option would contradict the given nonempty symmetric difference.

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.