Operations on Sets (Union, Intersection, Difference)

Factor the quadratic: x² − 5x + 6 = (x − 2)(x − 3). Since the parabola opens upward, the inequality (x − 2)(x − 3) ≤ 0 holds for 2 ≤ x ≤ 3, so A = [2, 3]. Every number in [2, 3] is less than 4 and therefore belongs to B. Thus A is a subset of B, leaving no element in A − B; the answer is the empty set.

Assume \(A\cap C=\varnothing\) as well. Since \(A\cap B=\varnothing\), every element of \(B\) lies outside \(A\). Equality of the unions \(A\cup B\) and \(A\cup C\) then forces every element of \(B\) to occur in \(C\); otherwise it would appear only in the first union. Hence \(B\subseteq C\), so option A is sufficient.

An element belongs to \(A-(B\cup C)\) exactly when it is in A and is in neither B nor C. The same condition means that it belongs to both \(A-B\) and \(A-C\). Therefore, \(A-(B\cup C)=(A-B)\cap(A-C)\). This is a difference form of De Morgan’s law, so option C is always true.

The symmetric difference A △ B contains the elements that belong to exactly one of A or B, while A ∩ B contains the elements common to both. These two parts are disjoint and together make up A ∪ B. Therefore, n(A ∪ B) = n(A △ B) + n(A ∩ B) = 26 + 9 = 35. Hence option A is correct.

To prove the result, consider any element x. If x belongs to A, the equality of intersections tells us that x belongs to B exactly when it belongs to C. If x does not belong to A, the equality of unions tells us that x belongs to B exactly when it belongs to C. Thus every element has the same membership status in B and C, so B and C contain precisely the same elements. Therefore \(B=C\). The other options are not forced by the two given equalities.

Factor the quadratic: \(x^2-4x+3=(x-1)(x-3)\). Because the quadratic opens upward, it is positive outside its roots, so A is \((-infty,1)\cup(3,infty)\). Set B contains numbers greater than 1, represented by \((1,infty)\). Their common part is therefore only the interval \((3,infty)\). The endpoints 1 and 3 are excluded because the inequality is strict and the roots make the expression zero. Hence option A is correct.

The sets C and A ∪ B are disjoint by the condition that C contains no element of A ∪ B. Hence removing C from A ∪ B does not change A ∪ B, so (A ∪ B) − C = A ∪ B. The union A ∪ B consists of A − B, B − A, and A ∩ B, which are disjoint parts. Therefore its cardinality is 19 + 24 + 13 = 56. Option A is correct.

Since x² ≤ 16, the integer x must lie between −4 and 4, so B = {−4, −3, −2, −1, 0, 1, 2, 3, 4}. Therefore, removing B from A = {−7, …, 7} leaves A − B = {−7, −6, −5, 5, 6, 7}. Among these remaining elements, only −6 and 6 are divisible by 3. Hence (A − B) ∩ C = {−6, 6}, which is option A.

Every element of A belongs to the union \(A\cup B\). Since the union is given to equal \(A\cap B\), every element of A must also belong to B. Thus \(A\subseteq B\). Similarly, every element of B belongs to the union and therefore to A, giving \(B\subseteq A\). Mutual containment proves \(A=B\). The sets need not be empty, so options B and C are not always true, and D is also not necessary.

Use the difference identity \(A\setminus X=A\cap X'\). Thus \(A\setminus(B\cap C)=A\cap(B\cap C)'\). By De Morgan's law, \((B\cap C)'=B'\cup C'\), so the expression becomes \(A\cap(B'\cup C')\). Distributing intersection over union gives \((A\cap B')\cup(A\cap C')\), which is exactly \((A\setminus B)\cup(A\setminus C)\). Therefore option A is correct.

An element belongs to \(A\setminus(B\cup C)\) exactly when it is in A and is not in the union \(B\cup C\). Not being in a union means that it is in neither B nor C. Therefore the element is simultaneously in \(A\setminus B\) and in \(A\setminus C\), giving \((A\setminus B)\cap(A\setminus C)\). Equivalently, use \(A\cap(B\cup C)'=A\cap(B'\cap C')\).

An element belongs to A \ (B ∪ C) when it is in A but not in B ∪ C. Not being in B ∪ C means that it is neither in B nor in C. Thus the element must belong simultaneously to A \ B and A \ C. Consequently, A \ (B ∪ C) = (A \ B) ∩ (A \ C). This is the set-difference form of De Morgan’s law.

An element is in A \ (B ∩ C) if it belongs to A and does not belong to both B and C at the same time. Therefore, at least one of the conditions ‘not in B’ or ‘not in C’ must hold. The element is consequently in A \ B or in A \ C, giving A \ (B ∩ C) = (A \ B) ∪ (A \ C). The union is essential because either condition is sufficient.

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).

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 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.

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].

The condition A ∩ (B \ C) = ∅ says that no element can simultaneously belong to A and to B without belonging to C. Therefore, every element common to A and B must also be an element of C. Hence A ∩ B ⊆ C. The other options are stronger or unrelated statements and are not forced by the given condition.

Use the inclusion–exclusion formula for three finite sets: |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |B ∩ C| − |C ∩ A| + |A ∩ B ∩ C|. Substitution gives 12 + 18 + 20 − 5 − 7 − 4 + 2 = 36. The pairwise intersections are subtracted because they were counted twice, and the triple intersection is added once because it was then removed too many times.

The symmetric difference contains elements that belong to exactly one of A and B, so the common elements must be excluded from both sets. Its cardinality is |A △ B| = |A| + |B| − 2|A ∩ B|. Substituting the given values gives 21 + 19 − 2(8) = 40 − 16 = 24. Therefore, option A is correct.

The set (A ∪ B) \ (A ∩ B) consists of elements that belong to exactly one of A or B; it is the symmetric difference. If this set is empty, there are no elements occurring in only one set. Thus every element of A is in B and every element of B is in A, so A = B. The sets need not be empty.

The sets A \ B and B \ A represent the elements belonging exclusively to A and exclusively to B, respectively. They are always disjoint. If two disjoint sets are equal, their common value must be empty; hence A \ B = ∅ and B \ A = ∅. Therefore A has no element outside B and B has no element outside A, so A = B.

To prove \(B=C\), consider any element \(x\). If \(x\in A\), equality of intersections gives \(x\in B\) exactly when \(x\in C\). If \(x\notin A\), equality of unions gives the same equivalence, because membership in either union must then come from \(B\) or \(C\). Thus every element has identical membership in \(B\) and \(C\), so \(B=C\).

Every set can be decomposed into two disjoint parts relative to \(C\): the elements outside \(C\), namely \(A\setminus C\), and the elements inside \(C\), namely \(A\cap C\). The two corresponding parts of \(A\) and \(B\) are given equal. Their unions therefore are equal: \(A=(A\setminus C)\cup(A\cap C)=(B\setminus C)\cup(B\cap C)=B\). Hence option A must hold.

We have \(A\cap B=\{3,4\}\) and \(A\setminus B=\{1,2\}\). These two sets are disjoint. A set that belongs to both power sets must be a subset of both \(\{3,4\}\) and \(\{1,2\}\). The only common subset of disjoint sets is the empty set. Since the empty set is an element of every power set, the intersection of the two power sets is \(\{\varnothing\}\), not \(\varnothing\).