Operations on Sets (Union, Intersection, Difference)

For two subsets of a universal set, n(A ∪ B) = n(A) + n(B) − n(A ∩ B), and n(A ∪ B) cannot exceed n(U). To make the intersection as small as possible, the union must be as large as possible, namely 90. Hence 90 ≥ 58 + 49 − n(A ∩ B), which gives n(A ∩ B) ≥ 17. This bound is attainable, so the minimum is 17.

The intersection A ∩ B contains only elements that belong to both sets, so it cannot contain more elements than either A or B. Consequently, n(A ∩ B) ≤ min(n(A), n(B)) = min(82, 63) = 63. This maximum is possible when every element of B is also an element of A, meaning B is a subset of A. Therefore option A is correct.

The set A is the disjoint union of A − B and A ∩ B. Thus n(A ∩ B) = n(A) − n(A − B) = 69 − 28 = 41. Now apply the union formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 69 + 74 − 41 = 102. The value 143 incorrectly counts the common elements twice, while 41 is only the intersection. Therefore option A is correct.

The union consists of three disjoint parts: the elements only in A, the elements only in B, and the common elements in A ∩ B. Therefore, n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Hence the required value is 115 − 42 = 73. Thus option A is correct. This is also the cardinality of the symmetric difference A △ B.

The union A ∪ B is the set of all elements that belong to A or to B or to both. Combining A = {1, 2, 3} and B = {3, 4, 5} gives 1, 2, 3, 4, and 5. The common element 3 is written only once because a set does not repeat elements. Therefore, A ∪ B = {1, 2, 3, 4, 5}.

The intersection A ∩ B contains only those elements that are present in both A and B. Checking the elements of A, 2 occurs in B and 4 also occurs in B, whereas 6 and 8 do not. Therefore, the common-element set is A ∩ B = {2, 4}. The other options represent unrelated combinations or an empty intersection.

The difference A − B consists of elements that belong to A but do not belong to B. In A = {a, b, c, d}, the elements b and d are also in B, so they are removed. The elements a and c are not in B and remain. Hence A − B = {a, c}. Notice that set difference is order-sensitive; B − A would give a different result.

First convert P from set-builder form into roster form. Taking N as the positive natural numbers, x < 5 gives P = {1, 2, 3, 4}. The union includes every distinct element of P and Q. Since 2 and 4 are already present in P, they are not repeated; adding 6 gives P ∪ Q = {1, 2, 3, 4, 6}.

The intersection contains elements common to both sets. The elements of A are 5, 10, and 15, while the elements of B are 20 and 25. There is no number appearing in both lists, so the sets are disjoint. Therefore, their intersection is the empty set: A ∩ B = ∅. The union would contain all five elements, but that is not being asked.

The union of sets contains every distinct element that belongs to at least one of the sets. Starting with \(A\cup B\), we get \(\{1,2,3\}\). Taking the union of this result with \(C=\{3,4\}\) adds only 4, because 3 is already present. Therefore, \(A\cup B\cup C=\{1,2,3,4\}\). Repeated elements are written only once in a set, so option D is not acceptable.

An intersection contains only elements common to every set involved. The elements 3 and 5 occur in \(A\), in \(B\), and in \(C\). The elements 2 and 7 occur only in \(A\), while 1 and 9 occur only in \(B\), and 11 occurs only in \(C\). Hence the common intersection is \(A\cap B\cap C=\{3,5\}\), making option A correct.

The difference \(B\setminus A\) consists of elements that belong to \(B\) but do not belong to \(A\). Here, both elements of \(B\), namely 2 and 4, are already elements of \(A\). Therefore, no element remains after removing from \(B\) the elements common with \(A\), so \(B\setminus A=\varnothing\). Option A is correct.

The union \(A\cup B\) contains every distinct element found in either \(A\) or \(B\). Set \(A\) contributes 1, 2, and 3; set \(B\) contributes 3 and 4. Since 3 is common, it is listed only once, giving \(A\cup B=\{1,2,3,4\}\). The universal set \(U\) provides the surrounding context but does not mean that all elements of \(U\) belong to the union.

The set difference \(A\setminus B\) keeps elements that are in \(A\) but not in \(B\). From \(A\), the elements 4 and 8 are also present in \(B\), so they must be removed. The elements 2, 6, and 10 are not in \(B\), while 12 is not even in \(A\). Hence \(A\setminus B=\{2,6,10\}\), so option A is correct.

The empty set \(\varnothing\) contains no elements. Therefore, taking the union of \(A\) with the empty set adds nothing to \(A\), and the result remains unchanged: \(A\cup\varnothing=A\). Since \(A=\{1,2,3\}\), the answer is \(\{1,2,3\}\). This is called the identity property of union; it should not be confused with intersection with the empty set, which is empty.

An intersection contains elements common to both sets. The empty set \(\varnothing\) has no elements at all, so there cannot be any element common to \(A=\{4,5,6\}\) and \(\varnothing\). Therefore, \(A\cap\varnothing=\varnothing\). Notice that this differs from union with the empty set: \(A\cup\varnothing=A\), whereas intersection with it always produces the empty set.

The difference \(A\setminus A\) consists of elements that belong to the first copy of \(A\) but do not belong to the second copy. Since both sets are identical, every element \(p\), \(q\), and \(r\) is removed. No element can remain in one copy while being absent from the other, so \(A\setminus A=\varnothing\). This is a general identity for every set.

First, form the union: \(A\cup B=\{1,2,3,4,5,6\}\). Next, find the intersection: \(A\cap B=\{3,4\}\). Subtracting the intersection from the union removes the elements common to both sets, leaving \(\{1,2,5,6\}\). Thus option A is correct. This result is also called the symmetric difference because it contains elements belonging to exactly one of the two sets.

The vowels in English are \(A=\{a,e,i,o,u\}\). The set \(B=\{a,e,i\}\) is a subset of \(A\), so every element of \(B\) is already present in \(A\). Taking the union adds no new element; therefore \(A\cup B=A=\{a,e,i,o,u\}\). Option B is only the smaller subset, not the union.

Assuming the standard school convention \(\mathbb{N}=\{1,2,3,\ldots\}\), we have \(A=\{1,2,3,4,5,6\}\). The even natural numbers not exceeding 8 are \(B=\{2,4,6,8\}\). The common elements are 2, 4, and 6, so \(A\cap B=\{2,4,6\}\). The element 8 is excluded because it is not in A.