Operations on Sets (Union, Intersection, Difference)

Assuming N denotes the positive natural numbers, the condition x ≤ 5 gives A = {1, 2, 3, 4, 5}. The union A ∪ B contains every distinct element belonging to either set. Combining A with B = {3, 4, 5, 6} gives {1, 2, 3, 4, 5, 6}; repeated elements are written only once. Therefore, option A is correct.

A union contains every distinct element that occurs in at least one of the given sets. Begin with A = {1, 2, 3}, add the new element 4 from B, and then add the new element 5 from C. The elements 3 and 4 are repeated in different sets, but a set does not list duplicates. Thus A ∪ B ∪ C = {1, 2, 3, 4, 5}.

First calculate the expression inside the parentheses: B ∪ C = {2, 4, 5, 6}. Now A − (B ∪ C) contains the elements of A that are not in this union. From A = {1, 2, 3, 4, 5}, remove 2, 4, and 5; the element 6 is irrelevant because it is not in A. The result is {1, 3}, so option A is correct.

Use the parentheses first. The union B ∪ C combines all distinct elements from B and C: {b, c, e} ∪ {c, d, e} = {b, c, d, e}. Next intersect this result with A = {a, b, c, d}. The common elements are b, c, and d; a is absent from the union and e is absent from A. Hence A ∩ (B ∪ C) = {b, c, d}.

The parentheses require us to find A ∩ B first. The elements common to A = {1, 2, 3, 4} and B = {3, 4, 5} are {3, 4}. Now take the union with C = {4, 5, 6}: {3, 4} ∪ {4, 5, 6} = {3, 4, 5, 6}. Element 4 is written once, and 1 and 2 are not included because they are not in the intermediate intersection.

The union contains all distinct elements from both sets. Combining A and B gives A ∪ B = {1, 2, 3, 4, 5, 6, 8, 10}. The elements 2 and 4 occur in both sets, but each is counted only once in a set union. Consequently, the union has 8 elements. Equivalently, |A ∪ B| = |A| + |B| − |A ∩ B| = 5 + 5 − 2 = 8.

The intersection A ∩ B contains only the elements that occur in both sets. Comparing the two sets, 20 appears in A and B, and 40 also appears in A and B. The elements 10 and 30 occur only in A, while 60 occurs only in B. Therefore, A ∩ B = {20, 40}, which has 2 elements. Hence option B is correct.

The vowels in the English alphabet are A = {a, e, i, o, u}. The difference A − B means that we retain elements of A and remove every element that is also present in B. Since B contains a, e, and i, these three vowels are removed from A. The remaining vowels are o and u, so A − B = {o, u}. Therefore, option A is correct.

The union A ∪ B contains every distinct element that belongs to A or to B, including elements common to both. Here A ∪ B = {2, 4, 6, 8, 10}. Among the choices, 8 is a member of B, so it must be included in the union. The other numbers, 3, 12, and 1, belong to neither set. Therefore, option B is correct.

The difference A − B consists of elements that are in A but not in B. The elements 2 and 4 occur in both sets, so they are excluded from A − B. The element 3 is in A and does not occur in B, while 6 is not in A at all. Thus A − B = {1, 3}, and 3 is the only listed element belonging to this difference. Option C is correct.

An intersection contains only the elements common to both sets. Set A contains red and blue, while set B contains blue and green. The only name appearing in both sets is blue. Therefore, A ∩ B = {blue}. Option A lists an element found only in A, option C lists an element found only in B, and option D incorrectly combines all elements as a union. Hence option B is correct.

The union A ∪ B contains every distinct element from both sets. Combining A and B gives cat, dog, cow, and goat. The repeated element dog is written only once because a set does not list duplicate elements. Thus A ∪ B = {cat, dog, cow, goat}. Option B is only the intersection, while options C and D omit elements. Therefore, option A is correct.

The union A ∪ B is formed by collecting every element from A and every element from B, without repeating any element. Here A contributes 1, 2, and 3, while B contributes 4 and 5. Since the sets are disjoint, none of these elements is repeated or removed. Therefore, A ∪ B = {1, 2, 3, 4, 5}. Option A is correct.

First calculate the two differences separately. A − B contains the elements in A that are absent from B, so A − B = {2, 3}. Similarly, B − A contains the elements in B that are absent from A, so B − A = {6, 7}. Taking their union gives {2, 3} ∪ {6, 7} = {2, 3, 6, 7}. The common elements 4 and 5 are excluded, so option A is correct.

First form the union: A ∪ B = {1, 2, 3, 4, 5, 6}. The difference (A ∪ B) − A means that every element belonging to A must be removed from the union. Removing 1, 2, 3, and 4 leaves {5, 6}. This also follows from the identity (A ∪ B) − A = B − A. The elements 3 and 4 are not retained because they are already in A.

The union of the two sets is A ∪ B = {a, b, c, d, e}. Subtracting B means removing c, d, and e from this union. The elements left are a and b, so the answer is {a, b}. Equivalently, the identity (A ∪ B) − B = A − B can be used. Option B gives only the common element, while option D ignores the subtraction operation.

To find A − B, remove from A every element that also occurs in B. Thus, 2 and 4 are removed, while 1, 3, and 5 remain; therefore A − B = {1, 3, 5}. Since A − B is already a subset of A, intersecting it with A does not change it. Hence A ∩ (A − B) = {1, 3, 5}.

The union A ∪ B contains every element in either set, so A ∪ B = {1, 2, 3, 4}. Taking the intersection with A selects only elements common to A and this union. Because every element of A is automatically in A ∪ B, the intersection is A itself: A ∩ (A ∪ B) = A = {1, 2, 3}. This is the absorption law.

The common elements of A and B are found first: A ∩ B = {4}. Therefore, A ∪ (A ∩ B) = {2, 4} ∪ {4}. A union retains every element from either set, and 4 is already present in A, so no new element is added. The result is {2, 4}. This illustrates the absorption law A ∪ (A ∩ B) = A.

The difference B − C contains elements that are in B but not in C. Set B is {2, 4, 6}; among these, 2 also belongs to C and must be removed. The elements 4 and 6 are not in C, so they remain. Therefore B − C = {4, 6}. The set A is extra information here and does not affect the requested difference.