Operations on Sets (Union, Intersection, Difference)

For a real number x, the inequality x² > 0 holds exactly when x is nonzero. If x = 0, then x² = 0, and if x ≠ 0, then x² is positive. Thus A and B describe exactly the same set, namely R − {0}. The symmetric difference contains elements belonging to exactly one set, so equal sets have symmetric difference ∅. Option A is correct.

For two finite sets, the inclusion-exclusion formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). The given equality has no subtraction term, so n(A ∩ B) must be zero. A finite set with zero elements is empty; therefore A ∩ B = ∅. This means A and B are disjoint, but it does not imply equality or either subset relation. Hence option A is correct.

The condition \(x\ge2\) gives the interval \([2,\infty)\), while \(x<6\) gives \((-infty,6)\). Their intersection is therefore \([2,6)\), including 2 but excluding 6. The set \(C\) contains only the number 4. Taking the difference \((A\cap B)-C\) removes 4 from \([2,6)\), leaving \([2,4)\cup(4,6)\). Thus, option A is correct; 4 is excluded from both resulting intervals.

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.

A is the interval [−3, 5), while B is (1, 8]. These intervals overlap, so their union covers every real number from −3 through 8, namely A ∪ B = [−3, 8]. Intersecting this union with C = (−∞, 2] keeps only numbers up to and including 2. The left endpoint −3 remains included, giving [−3, 2]. Therefore, option A is correct.

Use the inclusion–exclusion formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 64 + 57 − 29 = 92. The complement of A ∪ B contains all elements of U outside the union. Therefore, n((A ∪ B)′) = n(U) − n(A ∪ B) = 110 − 92 = 18. Thus option A is correct; 92 is the size of the union, not its complement.

Distribute the intersection over the union: (A ∪ C) ∩ B = (A ∩ B) ∪ (C ∩ B). Since A ⊆ B, every element of A is already in B, so A ∩ B = A. The condition C ∩ B = ∅ removes the second part. Consequently, the expression becomes A ∪ ∅ = A. Hence option A is the only correct answer.

For two finite sets, the inclusion–exclusion formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 60 = 42 + 35 − n(A ∩ B), so 60 = 77 − n(A ∩ B). Therefore, n(A ∩ B) = 77 − 60 = 17. The common elements must be subtracted once because they were counted in both A and B.

Every element of A belongs either to A − B or to A ∩ B. These two parts are disjoint: an element cannot be outside B and simultaneously inside B. Thus A is partitioned into the two given parts, so n(A) = n(A − B) + n(A ∩ B) = 28 + 16 = 44. Therefore, option A is correct.

The condition defining \(A\) gives all integers from \(-3\) through \(5\), so \(A=\{-3,-2,-1,0,1,2,3,4,5\}\). For \(B\), the inequality \(x^2\le9\) is equivalent to \(|x|\le3\), or \(-3\le x\le3\). Therefore, \(B=\{-3,-2,-1,0,1,2,3\}\). The difference \(A\setminus B\) consists only of elements in \(A\) that are not in \(B\). Removing the seven elements of \(B\) from \(A\) leaves \(\{4,5\}\), so option A is correct. Option C is wrong because \(-3\in B\).

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 determine the intersection of A and B. The elements common to both sets are 6 and 12, so A ∩ B = {6, 12}. Now subtract C = {6, 12, 18}. Both elements of {6, 12} occur in C, so both are removed. No element remains, and therefore (A ∩ B) − C = ∅. Hence option A is correct.

The even elements of U are \(A=\{2,4,6,8,10,12,14,16,18,20\}\), so \(n(A)=10\). The multiples of 3 are \(B=\{3,6,9,12,15,18\}\), so \(n(B)=6\). Their common elements are \(\{6,12,18\}\), hence \(n(A\cap B)=3\). By inclusion-exclusion, \(n(A\cup B)=10+6-3=13\). Therefore, the complement has \(20-13=7\) elements, so option A is correct.

Because \(A\subseteq B\), every element of A is already an element of B. The difference \(B\setminus A\) contains precisely those elements of B that are not in A. Thus, A and \(B\setminus A\) are disjoint parts whose union contains every element of B exactly once. Therefore, \(A\cup(B\setminus A)=B\). This is a standard partition identity for a subset and its remainder.

The equality \(A\cap B=A\) means that taking only the elements common to A and B leaves all of A unchanged. Hence every element of A must belong to B, which is exactly \(A\subseteq B\). The second equality, \(A\cup B=B\), expresses the same containment: adding A to B does not introduce any new element. Therefore option A is the necessary conclusion; the other statements do not follow.

The set \(A\setminus B\) consists of elements in A but not in B. Therefore, the elements common to A and B number \(n(A\cap B)=n(A)-n(A\setminus B)=30-18=12\). Applying the inclusion-exclusion formula gives \(n(A\cup B)=n(A)+n(B)-n(A\cap B)=30+24-12=42\). We subtract the intersection once because those 12 elements were counted in both 30 and 24.

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.

Since the natural numbers begin at 1 and A is restricted by \(x\le 12\), we have \(A=\{1,2,3,4,5,6,7,8,9,10,11,12\}\). The primes in this range are \(B=\{2,3,5,7,11\}\). Removing these from A leaves \(\{1,4,6,8,9,10,12\}\). Number 1 is included because it is neither prime nor composite; it has exactly one positive divisor.