Operations on Sets (Union, Intersection, Difference)

The intersection contains numbers that belong to both intervals. The common range begins just greater than 1 because B excludes 1, so the left endpoint is open. The common range ends at 4, and 4 is included in A and also lies inside B because 1 < 4 < 6. Hence A ∩ B = (1, 4], making option A correct.

A contains every real number less than or equal to 3, while B contains every real number greater than or equal to 1. The intervals overlap on [1, 3], so there is no gap between them. Numbers below 1 are covered by A, and numbers above 3 are covered by B. Therefore, together they cover every real number, so A ∪ B = ℝ. Option A is correct.

First evaluate the expression inside the parentheses. B ∪ C = {1, 2, 4, 6, 7}, because union collects every distinct element from both sets. Now intersect this result with A = {1, 2, 3, 4, 5}. The common elements are 1, 2, and 4; numbers 6 and 7 are not in A. Hence A ∩ (B ∪ C) = {1, 2, 4}, so option A is correct.

The condition A ∩ B = ∅ means that A and B are disjoint, so they have no common elements. For any two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives n(A ∪ B) = 21 + 17 − 0 = 38. Equivalently, because no element is counted twice, we can simply add the two cardinalities. Therefore, option A is correct.

The sets A and B can be divided into three non-overlapping regions: elements only in A, elements only in B, and elements in both sets. Their sizes are 12, 9, and 6 respectively. The union contains all three regions, so n(A ∪ B) = 12 + 9 + 6 = 27. Thus, option A is correct.

For any two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Here A ∩ B is the empty set, so it has zero elements: n(A ∩ B) = 0. Substituting the given values gives n(A ∪ B) = p + q − 0 = p + q. Because the sets are disjoint, no common element is counted twice.

A contains every real number less than 2, and B contains every real number greater than −1. If a real number is less than 2, it belongs to A. If it is not less than 2, then it is at least 2 and is certainly greater than −1, so it belongs to B. Thus every real number is covered, and A ∪ B = R.

Set difference is not commutative. A \ B contains elements of A that are not in B, whereas B \ A contains elements of B that are not in A; these sets can be different. For example, if A = {1, 2} and B = {2, 3}, then A \ B = {1}, while B \ A = {3}. The other three statements are standard identities or inclusion properties.

The elements of A are the positive multiples of 2 not exceeding 15: {2, 4, 6, 8, 10, 12, 14}, so n(A) = 7. The elements of B are {5, 10, 15}, so n(B) = 3. The common element is 10, hence n(A ∩ B) = 1. By the inclusion–exclusion formula, n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 7 + 3 − 1 = 9. The common element must not be counted twice.

The union A ∪ B contains every element of A together with every element of B. Removing A removes all elements contributed by A. Since A and B are disjoint, no element of B is removed while subtracting A. The elements left are exactly those of B, so (A ∪ B) \ A = B. Therefore, option A is correct.

For a real number to belong to A ∩ B, it must satisfy both conditions simultaneously: x ≤ 0 and x > 2. These conditions are incompatible, because every number greater than 2 is also greater than 0 and therefore cannot be at most 0. No real number satisfies both conditions, so the intersection is the empty set ∅.

The expression shows that union with A distributes over the intersection of B and C. In general, the distributive identity is A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). The operation outside the parentheses, union, appears with A in both right-hand parentheses, which is the characteristic pattern of distribution. Hence option A is correct.

By definition, A \ B consists only of elements that belong to A and do not belong to B. Therefore every element of A \ B is automatically an element of A, which means A \ B ⊆ A. Since C = A \ B, it follows directly that C ⊆ A. The additional condition C ∩ B = ∅ is consistent with the definition but is not needed for this conclusion.

The primes not exceeding 20 are \(\{2,3,5,7,11,13,17,19\}\). Set \(B\) contains the odd natural numbers, so all the odd primes are removed from \(A\). The only prime that is not odd is 2, because 2 is the unique even prime. Hence \(A\setminus B=\{2\}\), making option A correct.

Because \(A\subseteq C\), every element of \(A\) belongs to \(C\). Similarly, every element of \(B\) belongs to \(C\). Therefore every element of \(A\cup B\) is also in \(C\). Set difference removes elements that are in \(C\), so no element remains: \((A\cup B)\setminus C=\varnothing\). Thus option A is the only correct answer.

For any two finite sets, \(|A\cup B|=|A|+|B|-|A\cap B|\). Here \(A\cap B=\varnothing\), so the intersection has cardinality zero. Substitution gives \(|A\cup B|=|A|+|B|-0=|A|+|B|\). The sets are disjoint, so no element is counted twice. Hence option A is correct.

The statement A ∩ B ⊆ C means that every element belonging to both A and B also belongs to C. Therefore, there cannot be any element of A ∩ B that lies outside C. The difference set (A ∩ B) \ C contains exactly those elements of A ∩ B that are not in C, so it must be the empty set, ∅.

A set with cardinality zero is the empty set, so \(n(A)=0\) implies \(A=\varnothing\). The intersection of the empty set with any set is empty because there is no element that can belong to both sets. Therefore \(A\cap B=\varnothing\). The cardinality equation is also consistent with this: \(n(A\cup B)=n(B)=0+n(B)\).

Because A ⊆ B, every element of A is also an element of B. The condition B ∩ C = ∅ says that B and C have no common elements. Consequently, no element of A can belong to C either, because any such element would also belong to B and would contradict the disjointness of B and C. Hence A ∩ C = ∅.

The correct answer is B. Taking ℕ as the positive natural numbers, A = {1,2,3,4,5,6,7,8,9,10,11,12}. The even elements form B = {2,4,6,8,10,12}, so |B| = 6. The prime elements form C = {2,3,5,7,11}, so |C| = 5. Their common element is 2, hence |B ∩ C| = 1. In a union, a shared element is counted only once: |B ∪ C| = |B| + |C| − |B ∩ C| = 6 + 5 − 1 = 10. Thus B ∪ C = {2,3,4,5,6,7,8,10,11,12}. A gives 9, C gives 8, and D gives 7; each misses some element. Memory cue: union means either set, but never double-count the overlap.

First form the union A ∪ B by listing every element appearing in either set: A ∪ B = {1,2,3,4}. Set difference removes from this result every element that belongs to C. Since 2 and 3 are in C, remove them; 1 and 4 are not in C and remain. Thus (A ∪ B) \ C = {1,4}, so option A is correct.

The condition A \ B = ∅ means that A has no element outside B; equivalently, A ⊆ B. When A is a subset of B, their union is simply B, so A ∪ B = B. The problem also states A ∪ B = U. Combining these equalities gives B = U. The other statements may occur in special cases but are not forced by the conditions.

The three pairwise intersection conditions show that no element is shared by any two of the sets. Thus A, B, and C are pairwise disjoint. For disjoint finite sets, the cardinality of their union is the sum of their cardinalities, because no element is counted twice. Therefore |A ∪ B ∪ C| = 9 + 11 + 13 = 33, so option A is correct.

The difference \(U-A\) consists of elements that belong to \(U\) but do not belong to \(A\). Starting with \(U=\{2,4,6,8,10,12\}\), remove the elements \(4,8,12\) listed in \(A\). The elements left are \(2,6,10\). Thus, \(U-A=\{2,6,10\}\), making option A correct.