Operations on Sets (Union, Intersection, Difference)

Evaluate the parenthesized union first. A ∪ B = {1, 2, 3}, since the repeated element 2 is written only once. Next, intersect this result with C = {3, 4}. The only element common to {1, 2, 3} and {3, 4} is 3. Thus, (A ∪ B) ∩ C = {3}. The answer is not {2}, because 2 is not an element of C.

First calculate A ∩ B, the set of elements common to A and B. The common elements are 3 and 5, so A ∩ B = {3, 5}. Now take the union of this result with C = {3, 4, 5}. Including every distinct element gives {3, 5} ∪ {3, 4, 5} = {3, 4, 5}. Therefore, option B is correct. Elements such as 2, 7, 1, and 9 are not included because they are not in the intermediate intersection or in C.

The difference A \ B contains elements present in A but absent from B, so A \ B = {1, 2}. Similarly, B \ A contains elements present in B but absent from A, giving B \ A = {5, 6}. Their union is {1, 2} ∪ {5, 6} = {1, 2, 5, 6}. Thus the expression collects the elements belonging to exactly one of the two sets and excludes the common elements 3 and 4.

Taking N as the positive natural numbers, A = {1, 2, 3, 4, 5, 6, 7} because x < 8. The conditions 4 < x < 10 give B = {5, 6, 7, 8, 9}. Set difference A \ B keeps only the elements of A that do not occur in B. Removing 5, 6, and 7 from A leaves {1, 2, 3, 4}. Hence, option A is correct. The strict inequalities exclude 4 from B and 10 from consideration.

The union A ∪ B contains every element of A together with every element of B. If this union is exactly A, adding B has introduced no new element. Therefore every element of B must already belong to A, which means B ⊆ A. The other statements do not necessarily follow from the given equality.

The intersection A ∩ B contains exactly the elements common to A and B. If the intersection equals all of A, then every element of A must also be present in B. This is precisely the definition of A being a subset of B, so A ⊆ B. The equality does not require B to be empty or equal to A.

A \ B consists of elements that are in A but not in B. Removing 2, 4, and 6 from A leaves {1, 3, 5}. For B \ A, we look for elements in B that are absent from A. Since B = {2, 4, 6} is completely contained in A, no such elements exist, so B \ A = ∅. Hence the ordered pair is ({1, 3, 5}, ∅), making option A correct.

Because x is an integer and x² ≤ 4, we have |x| ≤ 2, so -2 ≤ x ≤ 2. Therefore A = {-2, -1, 0, 1, 2}. The intersection keeps only elements common to A and B. Comparing B = {-1, 0, 1, 3} with A, the common elements are -1, 0, and 1; 3 is not in A. Hence A ∩ B = {-1, 0, 1}, so option B is correct.

Every element of A ∩ B belongs to both A and B. Since the union A ∪ B contains every element that belongs to at least one of the two sets, every element of the intersection must also be in the union. Therefore A ∩ B is always a subset of A ∪ B, regardless of whether the sets are equal, disjoint, finite, or empty.

First calculate the union: A ∪ B = {m, n, o, p, q}. The difference (A ∪ B) \ A removes every element that belongs to A, namely m, n, and o. The remaining elements are p and q. Equivalently, since A is already included in the union, (A ∪ B) \ A = B \ A. Therefore the answer is {p, q}, which is option A; the other choices retain elements that should be removed.

The union contains all distinct elements: A ∪ B = {1, 2, 3, 4, 5, 6, 7}. Subtracting B removes 4, 5, 6, and 7 from this union, leaving {1, 2, 3}. This is also the identity (A ∪ B) \ B = A \ B, because all elements of B are removed. Therefore option A is correct; option C gives B \ A instead, while D is the un-subtracted union.

The union is divided into three mutually disjoint regions: elements only in A, elements in both A and B, and elements only in B. Their cardinalities are 9, 6, and 11 respectively. Since these regions together contain every element of A ∪ B, add them: n(A ∪ B) = 9 + 6 + 11 = 26. Therefore option C is correct. Adding only two regions would omit one part of the union.

Set B is partitioned into two disjoint parts: the elements of B that are outside A, namely B \ A, and the elements common to both sets, namely A ∩ B. Thus n(B) = n(B \ A) + n(A ∩ B). Substituting the given values gives 35 = 24 + n(A ∩ B), so n(A ∩ B) = 35 − 24 = 11. Hence option A is correct; the union and n(A) values are unnecessary for this calculation.

The governing concepts are set difference and intersection. In B \ C, remove from B every element that also occurs in C. Since 3 and 5 occur in C, B \ C = {2, 7}. Now intersect this result with A, retaining only elements common to both sets. The element 2 belongs to A, but 7 does not. Therefore A ∩ (B \ C) = {2}, so option A is correct; option D incorrectly keeps 7.

Use the definitions of union and intersection systematically. First, A ∪ B = {a, b, c, d, e}, because all distinct elements from both sets are included. Next, B ∪ C = {b, d, e, f}. The common elements in these two unions are b, d, and e. Hence (A ∪ B) ∩ (B ∪ C) = {b, d, e}. Option A is correct; option D is the full union, not the intersection.

The operation inside parentheses is performed first. The elements common to A and B are 2, 4, and 8, so A ∩ B = {2, 4, 8}. Set difference then removes from this result every element found in C. Because 4 and 8 are in C, they are deleted, while 2 remains because it is not in C. Thus (A ∩ B) \ C = {2}, making option A correct.

A is the set of natural numbers from 1 through 12. First form the union B ∪ C, which contains every element appearing in either set: {2, 3, 4, 6, 8, 9, 10, 12}. Set difference A \ (B ∪ C) keeps elements of A that are absent from this union. Checking 1 through 12 leaves 1, 5, 7, and 11. Therefore option A is correct.

Since \(x\) is a natural number and \(x^2\le49\), the possible values are \(1,2,3,4,5,6,7\). Thus, \(A=\{1,2,3,4,5,6,7\}\). The difference \(A\setminus B\) contains elements that belong to A but do not belong to B. Removing 2, 4, and 6 from A leaves \(\{1,3,5,7\}\). The element 8 in B is irrelevant because it is not in A. Therefore, option A is correct.

The intersection A ∩ B ∩ C contains only those elements that occur in all three sets simultaneously. The element 3 occurs in A, B, and C, and the element 6 also occurs in A, B, and C. The elements 1, 2, 5, and 9 fail to occur in at least one of the sets. Therefore, A ∩ B ∩ C = {3, 6}, so option A is correct.

First calculate the union inside the parentheses: B ∪ C = {s, u, v, w}. Next, retain only the elements that are also present in A = {r, s, t, u}. The common elements are s and u, while v and w are not in A and r and t are not in B ∪ C. Hence A ∩ (B ∪ C) = {s, u}; option A is correct.

First form the union B ∪ C = {1, 2, 5, 6, 8, 9}. The difference A \ (B ∪ C) consists of elements that belong to A but do not belong to this union. From A, the elements 1, 2, 5, and 6 must be removed; 3 and 4 remain. Therefore A \ (B ∪ C) = {3, 4}, making option A correct.

For two finite sets, the inclusion–exclusion formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 26 + 19 − 8 = 37. The intersection is subtracted because the eight common elements were counted once in n(A) and once again in n(B). Thus the union contains 37 elements, so option A is correct.

Use the two-set formula n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the values gives 46 = 31 + 28 − n(A ∩ B), so n(A ∩ B) = 31 + 28 − 46 = 13. The common elements must be subtracted from the sum because they were counted twice. Therefore option A is correct.

The set B is divided into two disjoint parts: B \ A, which contains elements of B outside A, and A ∩ B, which contains elements common to A and B. Thus n(B) = n(B \ A) + n(A ∩ B). Consequently, 42 = 25 + n(A ∩ B), giving n(A ∩ B) = 17. Therefore option A is correct.

“At least one activity” means the union of the music and painting groups. Apply n(M ∪ P) = n(M) + n(P) − n(M ∩ P). Hence n(M ∪ P) = 36 + 29 − 14 = 51. The 14 students who learn both activities must be subtracted once because they were included in both original totals. Therefore option A is correct.