Operations on Sets (Union, Intersection, Difference)

Set A contains all integers divisible by 2, and set B contains all integers divisible by 3. An element of their intersection must satisfy both divisibility conditions at the same time. Since 2 and 3 are coprime, every integer divisible by both is divisible by their least common multiple, 6. Thus A ∩ B is the set of all integer multiples of 6, including 0 and negative multiples.

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.

Consider any element x of A. If x is not in B, then x belongs to A \ B. If x is in B, then x belongs to A ∩ B. These two cases cover every element of A, and both resulting sets contain only elements of A. Therefore their union is exactly A. This is the partition identity A = (A \ B) ∪ (A ∩ B).

The union A ∪ B is divided into three mutually disjoint regions: elements belonging only to A, represented by A \ B; elements belonging only to B, represented by B \ A; and elements common to both, represented by A ∩ B. Therefore no element is counted twice when these three numbers are added: n(A ∪ B) = 5 + 7 + 4 = 16.

First calculate the number of elements in A ∪ B. Because the 11 elements in A ∩ B are included in both A and B, they would be counted twice in n(A) + n(B), so subtract them once: n(A ∪ B) = 32 + 27 − 11 = 48. The complement (A ∪ B)′ contains all elements of U that are not in the union. Therefore, n((A ∪ B)′) = n(U) − n(A ∪ B) = 60 − 48 = 12. Thus, option A is correct.

For A \ B to contain exactly 2 elements, we must choose exactly 2 elements of A to be excluded from B. Once those two elements are chosen, the other two elements must belong to B, so each choice determines exactly one valid subset B. The number of choices is the combination 4 choose 2, equal to 4!/(2!2!) = 6. Hence six subsets satisfy the condition.

Factor the quadratic: x² − 4x + 3 = (x − 1)(x − 3). Since the parabola opens upward, the inequality (x − 1)(x − 3) ≤ 0 holds for 1 ≤ x ≤ 3. Thus B = [1, 3]. This interval is already contained in A = [0, 4], so their intersection is [1, 3]. Therefore, option A is correct.

Let the common set be S, where S = A ∪ B = A ∩ C. Because A ⊆ A ∪ B, we have A ⊆ S. Also, A ∩ C ⊆ A, so S ⊆ A. Therefore S = A. From A ∪ B = A, every element of B must already belong to A, giving B ⊆ A. From A ∩ C = A, every element of A must belong to C, giving A ⊆ C. Hence option A is the only definite conclusion; the other options need not hold.

The sets A ∩ B and A \ B represent two non-overlapping parts of A: the elements inside B and the elements outside B, respectively. Their intersection is always empty. If these two disjoint sets are equal, the common set must itself be empty. Their union is A, so A must also be empty. Hence option A is correct.

The difference A \ B contains elements that belong to A but not B, whereas B \ A contains elements that belong to B but not A. These two differences are always disjoint. They therefore cannot be equal to the same non-empty set {2, 5}, because that would make 2 and 5 belong to both differences simultaneously. The stated condition is impossible.

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.

A number belongs to A ∩ B precisely when it is divisible by both 4 and 6. Such numbers are multiples of lcm(4, 6) = 12. The positive multiples of 12 not exceeding 50 are 12, 24, 36, and 48. There are four numbers, so n(A ∩ B) = 4. The definition of N as positive natural numbers removes any ambiguity about including zero.

The condition |x| ≤ 3, with x an integer, gives A = {−3, −2, −1, 0, 1, 2, 3}. For B, x² − 1 = 0 means x² = 1, so x = −1 or x = 1; hence B = {−1, 1}. The difference A \ B contains the elements of A that are not in B. Removing −1 and 1 from A leaves {−3, −2, 0, 2, 3}. 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 equality A ∩ B = A ∩ C says that the elements common to A and B are exactly the same as the elements common to A and C. The symmetric difference B △ C contains elements belonging to exactly one of B or C. Since no such differing element can lie in A, their intersection with A is empty. However, B and C may still differ outside A, so B = C and the union statement are not necessarily true.

Separate the elements of A ∪ B into three disjoint regions: A \ B, A ∩ B, and B \ A. Subtracting A \ B removes only the elements that belong exclusively to A. The remaining two regions, A ∩ B and B \ A, together contain every element of B and no element outside B. Therefore, (A ∪ B) \ (A \ B) = B.

The union A ∪ B consists of three mutually disjoint regions: the elements only in A, the elements only in B, and the elements common to both sets. Therefore, n(A ∪ B) = n(A \ B) + n(B \ A) + n(A ∩ B). Substituting the values gives 54 = 17 + 21 + n(A ∩ B), so n(A ∩ B) = 54 − 38 = 16. Hence, option A is correct.

To find A \ B, retain points of A that do not belong to B. From [−4, 2), remove (−1, 2), but −1 remains because B is open at −1; this leaves [−4, −1]. From (5, 9], remove (5, 6], including 6 because B contains 6; this leaves (6, 9]. Hence A \ B = [−4, −1] ∪ (6, 9].