Operations on Sets (Union, Intersection, Difference)

The condition 2 ≤ x ≤ 7 for natural numbers gives A = {2, 3, 4, 5, 6, 7}. The difference A − B contains elements of A that are not in B. The elements 4 and 6 are common to A and B, so remove them from A. The remaining set is {2, 3, 5, 7}; hence option A is correct.

First find the union: A ∪ B = {0, 1, 2, 3, 4, 5, 6}. Next find the intersection: A ∩ B = {2, 3, 4}. Subtracting the intersection from the union removes the elements common to both sets and leaves the elements appearing in only one set: {0, 1, 5, 6}. Therefore, option A is correct.

Union is commutative: A ∪ B = B ∪ A for every pair of sets because an element belongs to the union when it belongs to A or B, and changing the order does not change that condition. The other statements are not always true. In general A − B differs from B − A, A ∩ ∅ = ∅, and A ∪ A = A. Hence option A is correct.

The intersection of several sets contains only elements common to every set. Comparing A and B first gives A ∩ B = {3, 5}. The set C contains 3 but does not contain 5. Thus, after checking all three sets, only 3 remains, so A ∩ B ∩ C = {3}. Option B is not correct because 5 is absent from C, and option C includes that same incorrect element. Therefore, option A is correct.

First calculate the union inside the parentheses: B ∪ C = {2, 3, 5, 6, 7}. Now A − (B ∪ C) means retain elements of A that are absent from this union. Removing 2, 3, 5, and 6 from A leaves {1, 4}. Although 7 belongs to the union, it is not in A and cannot appear in the difference. Thus option A is correct.

Use the order of operations for sets: calculate the union inside the parentheses first. B ∪ C = {b, d, e, f}. Now intersect this set with A = {a, b, c, d}; only elements appearing in both sets are retained. Those common elements are b and d. Hence A ∩ (B ∪ C) = {b, d}, so option A is correct. The other options either omit common elements or include elements not in A.

Follow the parentheses and calculate the intersection first. A ∩ B contains the elements common to A and B, so A ∩ B = {3, 5}. Now take the union with C = {5, 7, 11}; include every element appearing in either set, without repeating 5. The result is {3, 5, 7, 11}, so option A is correct.

The union contains every distinct element from both sets, with common elements counted only once. Here A ∩ B = {2, 4}, so there are two repeated elements. Using the formula n(A ∪ B) = n(A) + n(B) − n(A ∩ B), we get 4 + 5 − 2 = 7. Indeed, A ∪ B = {1, 2, 3, 4, 5, 6, 8}, which has seven elements. Option C is correct.

The intersection A ∩ B consists only of elements that occur in both A and B. Comparing the sets, 24 appears in both and 48 also appears in both. The elements 12 and 36 occur only in A, while 60 occurs only in B. Therefore, A ∩ B = {24, 48}, which has two elements. Hence option B is correct.

The set difference \(A-B\) contains every element that belongs to \(A\) but does not belong to \(B\). In \(A\), the elements 8 and 64 are also present in \(B\), so they are removed. The elements 1 and 27 are not in \(B\), so they remain. Therefore, \(A-B=\{1,27\}\). Option B is the intersection, option C contains an element only from \(B\), and option D is the union.

The difference \(A-B\) keeps elements of \(A\) that are not elements of \(B\). The sets \(A\) and \(B\) are disjoint because they have no common element: 1 through 5 occur only in \(A\), while 6 through 8 occur only in \(B\). Thus no element is removed from \(A\), and \(A-B=A=\{1,2,3,4,5\}\).

The intersection \(A\cap B\) consists only of elements that occur in both sets. The word “book” appears in \(A\) and also in \(B\), while “pen” and “bag” occur only in \(A\), and “desk” occurs only in \(B\). Therefore, \(A\cap B=\{\text{book}\}\). Option D represents the union, not the intersection.

The union \(A\cup B\) contains every distinct element that belongs to either \(A\), \(B\), or both. The elements are rose, lily, lotus, and jasmine. Because lily is common to both sets, it is written only once; repetition is not used in a set. Hence \(A\cup B=\{\text{rose},\text{lily},\text{lotus},\text{jasmine}\}\).

To find \(A-B\), remove from \(A\) every element that is also in \(B\). The elements east and west occur in both sets, so they are deleted from \(A\). The remaining elements are north and south. Therefore, \(A-B=\{\text{north},\text{south}\}\). This is not the intersection, empty set, or complete original set.

An element belongs to \(A\cap B\) only when it is present in both \(A\) and \(B\). The number 4 appears in both sets, and 8 also appears in both, although 8 is not offered as an option. The number 2 and 6 occur only in \(A\), while 12 occurs only in \(B\). Therefore, among the given choices, 4 is the correct answer.

The union \(A\cup B\) contains every element that belongs to at least one of the two sets. Since 6 is an element of \(A\), it must be included in the union. In fact, \(A\cup B=\{3,6,9,12,15\}\). The numbers 18, 21, and 0 are not present in either set, so they cannot belong to the union.

The set difference \(A-B\) includes elements of \(A\) that are absent from \(B\). Here 2 and 10 occur in both sets, so they are excluded. The number 1 occurs in \(A\) but not in \(B\), so it belongs to the difference; 5 also belongs to the difference, although it is not offered. The number 4 is only in \(B\).

The union combines all distinct elements from \(A\) and \(B\), without adding elements that belong only to the universal set \(U\). Combining \(\{1,2,7\}\) and \(\{2,4,6\}\) gives \(\{1,2,4,6,7\}\); the common element 2 is written once. Elements 3, 5, and 8 are in \(U\) but in neither \(A\) nor \(B\).

The intersection \(A\cap B\) contains only elements common to both sets. Comparing \(A=\{a,c,e\}\) with \(B=\{b,c,e,f\}\), the common elements are c and e. Therefore, \(A\cap B=\{c,e\}\). The universal set is not automatically the answer; a letter belongs to the intersection only when it is listed in both A and B.

The difference A − B contains the elements that are present in A but absent from B. Set A contains 10, 20, and 40, while set B contains 20 and 30. Therefore, remove 20 from A; 10 and 40 remain. Hence, A − B = {10, 40}. The universal set U is not needed for this particular difference operation.