Operations on Sets (Union, Intersection, Difference)

A \ B contains the elements of A that are not in B. Because B = (2, 4) is open, it contains every number strictly between 2 and 4 but does not contain 2 or 4. Both endpoints belong to A, so they remain after removing B. Therefore, A \ B = [0, 2] ∪ [4, 6], which is option A.

First solve the equation defining each set. From \(x^2=9\), we obtain \(x=3\) or \(x=-3\), so \(A=\{-3,3\}\). From \(x^2-4=0\), we get \(x^2=4\), hence \(x=2\) or \(x=-2\), so \(B=\{-2,2\}\). The union contains every element that belongs to either set, without repeating any element. Therefore, \(A\cup B=\{-3,-2,2,3\}\), which is option A. Option B lists only A, option C lists only B, and option D omits 2.

Factor the first quadratic: \(x^2-5x+6=(x-2)(x-3)\), so its roots are 2 and 3 and \(A=\{2,3\}\). Factor the second quadratic: \(x^2-3x+2=(x-1)(x-2)\), so its roots are 1 and 2 and \(B=\{1,2\}\). The intersection consists only of elements common to both sets. The only common element is 2; therefore, \(A\cap B=\{2\}\), which is option B. Option A combines elements from both sets rather than finding common elements, option C includes 3 although it is not in B, and option D is incorrect because the intersection is not empty.

A − (A ∩ B) consists of elements that are in A but not in the intersection A ∩ B. An element of A is excluded from A ∩ B precisely when it is not in B. Therefore, the remaining elements are those in A but outside B, which is A − B. Equivalently, A − (A ∩ B) = A ∩ Bᶜ = A − B. Option A is correct.

The union A ∪ B contains all elements from A and B. Removing A eliminates every element that belongs to A, including the common elements A ∩ B. What remains are precisely the elements that belong to B but not to A, which is B − A. Therefore, (A ∪ B) − A = B − A, so option B is correct.

The intersection A ∩ B contains only the elements that are present in both A and B. Set A contains 1, 2, and 3, while set B contains 4 and 5. There is no common element between the two sets. Therefore, A and B are disjoint sets, and their intersection is the empty set, written as ∅. Hence, option C is correct.

For any two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Since A ∩ B = ∅, its cardinality is zero. Thus n(A ∪ B) = 9 + 11 − 0 = 20. Equivalently, because the sets are disjoint, no element is counted twice, so their cardinalities can be added directly. Therefore, option B is correct.

The complement A′ is taken with respect to the universal set U. Removing the elements of A from U gives A′ = {1, 3, 5, 7, 9}. We then find the intersection of A′ with B = {1, 2, 3, 4}. The elements common to both sets are 1 and 3, so A′ ∩ B = {1, 3}. Therefore, option A is correct.

First calculate the union and intersection. A ∪ B = {1, 2, 3, 4, 5, 6}, while A ∩ B = {2, 4}. Subtracting the intersection from the union removes the common elements 2 and 4, leaving {1, 3, 5, 6}. This operation gives the elements belonging to exactly one of the two sets, also called their symmetric difference. Hence, option B is correct.

The union A ∪ B consists of three mutually separate regions: elements in A but not B, elements in B but not A, and elements common to both sets. These are respectively A − B = {1, 5}, B − A = {7}, and A ∩ B = {2, 3}. Combining all distinct elements gives A ∪ B = {1, 2, 3, 5, 7}. Therefore, option A is correct.

The vowels in the English alphabet are A = {a, e, i, o, u}. Set B contains {a, b, c, d, e}. The intersection contains only letters present in both sets. The common letters are a and e; i, o, and u are vowels but do not belong to B. Therefore, A ∩ B = {a, e}, making option A correct.

The prime numbers less than 15 are A = {2, 3, 5, 7, 11, 13}. The odd numbers less than 15 are B = {1, 3, 5, 7, 9, 11, 13}. A − B contains elements in A that are not in B. Every prime in A except 2 is odd and therefore belongs to B. Removing those common elements leaves only 2, so option A is correct.

The union contains every distinct element appearing in either set. Combining A and B gives A ∪ B = {2, 3, 5, 7, 11, 13}, which has six elements. The same result follows from n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 4 + 4 − 2 = 6, because 5 and 7 are common and must not be counted twice. Hence, option B is correct.

Every element of B is also an element of A, so B is a subset of A. Removing B’s elements 2, 4, and 6 from A leaves A − B = {1, 3, 5}. Also, B − A would be empty, A ∩ B would equal B = {2, 4, 6}, and A ∪ B would equal A, not B. Therefore, only statement A is correct.

Let \(x\) be any element of \(B\). By the definition of union, \(x\in A\cup B\). Since \(A\cup B=A\), this means \(x\in A\). Thus every element of \(B\) belongs to \(A\), so \(B\subseteq A\). The reverse inclusion or equality is not necessary; for example, \(A=\{1,2\}\) and \(B=\{1\}\) satisfy the condition but are not equal.

The complement \(A'\) consists of all elements of the universal set \(U\) that are not in \(A\). Therefore, every element of \(U\) lies either in \(A\) or in \(A'\), and hence belongs to their union. Consequently, \(A\cup A'=U\). This is one of De Morgan’s basic complement laws; in contrast, \(A\cap A'=\varnothing\).

By definition, \(A'\) contains exactly those elements of \(U\) that are not in \(A\). Thus no element can belong to both \(A\) and \(A'\) simultaneously. Their common part is therefore empty, giving \(A\cap A'=\varnothing\). This conclusion remains valid for every set \(A\) relative to the same universal set \(U\), including when \(A\) itself is empty or equal to \(U\).

First evaluate the expression inside parentheses: \(B\cup C=\{2,3,4,6,7\}\). Next find the elements common to this set and \(A=\{1,2,3,4,5\}\). The common elements are 2, 3, and 4, so \(A\cap(B\cup C)=\{2,3,4\}\). Option D is only the union and includes 6 and 7, which are not in \(A\); option B contains elements excluded from the union.

The parentheses require the intersection first. The elements common to \(A=\{1,2,4,8\}\) and \(B=\{2,4,6\}\) are 2 and 4, so \(A\cap B=\{2,4\}\). Taking the union with \(C=\{4,6,8\}\) gives \(\{2,4\}\cup\{4,6,8\}=\{2,4,6,8\}\). Repeated elements such as 4 are written only once in a set.

Multiples of 3 from 1 through 20 are \(\{3,6,9,12,15,18\}\), while multiples of 5 are \(\{5,10,15,20\}\). The only number appearing in both lists is 15. Equivalently, the least common multiple of 3 and 5 is 15, and the next common multiple is 30, which exceeds 20. Therefore \(A\cap B=\{15\}\).