Operations on Sets (Union, Intersection, Difference)

Remove from A every element that belongs to B. Since B contains 2, 4, and 6, the difference is \(A\setminus B=\{1,3,5\}\), which has 3 elements. A set with n elements has exactly \(2^n\) subsets in its power set, because each element may either be selected or not selected. Therefore, \(n(\mathcal{P}(A\\setminus B))=2^3=8\). The number 3 is the size of the difference set, not of its power set.

Although B − C has 6 elements that also belong to A, none of these elements can belong to A − B, because every element of A − B is specifically outside B, whereas every element of B − C is inside B. Thus (A − B) and (B − C) are disjoint. Therefore, n((A − B) ∪ (B − C)) = 17 + 22 = 39. The stated common elements with A do not create an intersection between the two sets in the union.

Because A is a subset of B, every element of B − A is also an element of C − A. Starting with C − A means retaining elements of C that are not in A. Removing B − A then removes all elements that are in B but not in A. The elements left are precisely those in C that are outside B, namely C − B. Hence option A is correct.

The conditions \(A\cap B=\varnothing\), \(B\cap C=\varnothing\), and \(C\cap A=\varnothing\) show that the three sets are pairwise disjoint. Therefore, no element is repeated in their union. We can add their cardinalities directly: \(n(A\cup B\cup C)=n(A)+n(B)+n(C)=12+15+18=45\). Hence, option A is correct. If the sets were not disjoint, common elements would have to be subtracted using the inclusion–exclusion principle.

The natural-number perfect squares not exceeding 30 are \(1,4,9,16,25\), so \(A=\{1,4,9,16,25\}\). The set \(B\) contains numbers up to 30 divisible by 3. Among the elements of \(A\), only 9 is divisible by 3, because \(9=3\times3\). Set difference \(A-B\) means retaining elements of \(A\) that are not in \(B\). Thus, \(A-B=\{1,4,16,25\}\), making option A correct.

An element belongs to A \ (B ∪ C) precisely when it belongs to A and belongs to neither B nor C. The condition of being outside the union B ∪ C therefore means being outside both B and C. This is exactly the membership condition for (A \ B) ∩ (A \ C). Thus option A is De Morgan’s difference identity and is true for all sets.

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.

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.

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.

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.

The condition \(A\setminus B=\varnothing\) says that A contains no element outside B, so every element of A belongs to B; hence \(A\subseteq B\). Similarly, \(B\setminus A=\varnothing\) implies \(B\subseteq A\). When each set is a subset of the other, the two sets have exactly the same elements. Therefore \(A=B\), not a proper-subset relation and not necessarily disjoint.

For two finite sets, the inclusion-exclusion formula is \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\). Substitute the given values: \(75=46+n(B)-18\). Rearranging gives \(n(B)=75-46+18=47\). The intersection is added back when solving for B because it had been subtracted from the sum of the two set sizes. Thus option A, 47, is the only correct value.

Students studying only Mathematics are 38 − 20 = 18, because the 20 students studying both subjects must be excluded. Students studying only Physics are 32 − 20 = 12. Therefore, the number studying exactly one subject is 18 + 12 = 30. Equivalently, this is 38 + 32 − 2(20) = 30.

Let T be the set of people who like tea and C the set of people who like coffee. By inclusion–exclusion, n(T ∪ C) = n(T) + n(C) − n(T ∩ C) = 52 + 47 − 19 = 80. Therefore, 80 people like at least one beverage. The number who like neither is 90 − 80 = 10, so option A is correct.

Set A contains the even numbers from 1 to 30, so n(A) = 30/2 = 15. The elements that belong to both A and B must be divisible by both 2 and 5, hence they are multiples of 10: 10, 20, and 30. Thus n(A ∩ B) = 3. Removing these from A gives n(A \ B) = 15 − 3 = 12. Therefore, option A is correct.

Factor the first quadratic: x² − 5x + 6 = (x − 2)(x − 3), so A = {2, 3}. Factor the second: x² − 4x + 3 = (x − 1)(x − 3), so B = {1, 3}. The intersection contains only values present in both sets. The only common value is 3, hence A ∩ B = {3}. Option A is correct.