Operations on Sets (Union, Intersection, Difference)

To determine A, factor x² − 5x + 6 as (x − 2)(x − 3) = 0, so A = {2, 3}. To determine B, factor x² − 4x + 3 as (x − 1)(x − 3) = 0, so B = {1, 3}. The intersection contains only elements common to both sets. The only common element is 3; therefore, A ∩ B = {3}.

The set A contains all real numbers greater than -2 and up to and including 5. The set B contains every number from 1, including 1, up to but not including 8. Therefore, the portion of A removed by B begins at 1, and 1 must also be removed because it belongs to B. The remaining part is \((-2,1)\); -2 is excluded from A and 1 is excluded from the difference.

The set A is divided into two disjoint parts: A ∩ B, containing elements of A that are also in B, and A \ B, containing elements of A that are not in B. Every element of A belongs to exactly one of these parts. Therefore, their union reconstructs the entire set A. Here A ∩ B = (0, 2] and A \ B = (-∞, 0], whose union is (-∞, 2] = A.

Use the inclusion–exclusion formula n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Thus, 31 = 18 + 22 − n(A ∩ B), so n(A ∩ B) = 9. The difference A \ B contains the elements of A that are not in B, so n(A \ B) = n(A) − n(A ∩ B) = 18 − 9 = 9. Therefore, option A is correct.

The symmetric difference contains elements that belong to exactly one of the two sets. From A, removing the common elements 2 and 4 gives A \ B = {1, 6}. From B, removing 2 and 4 gives B \ A = {3, 5}. Their union is {1, 6} ∪ {3, 5} = {1, 3, 5, 6}. The common elements are deliberately excluded.

The equality A ∪ B = A means that adding every element of B to A produces no new element. Therefore, every element already in B must also belong to A. This is precisely the definition of B ⊆ A. The other statements are not necessary: A and B may overlap, neither set must be empty, and A need not be a subset of B.

An element belongs to A \ (B ∪ C) when it is in A but not in B ∪ C. Not being in B ∪ C means that it is neither in B nor in C. Thus the element must belong simultaneously to A \ B and A \ C. Consequently, A \ (B ∪ C) = (A \ B) ∩ (A \ C). This is the set-difference form of De Morgan’s law.

An element is in A \ (B ∩ C) if it belongs to A and does not belong to both B and C at the same time. Therefore, at least one of the conditions ‘not in B’ or ‘not in C’ must hold. The element is consequently in A \ B or in A \ C, giving A \ (B ∩ C) = (A \ B) ∪ (A \ C). The union is essential because either condition is sufficient.

First form the union A ∪ B by listing every element appearing in either set: A ∪ B = {1, 2, 3, 4, 5, 6}. Next intersect this result with C = {1, 4, 7}. The elements common to both sets are 1 and 4; 7 is absent from A ∪ B. Therefore, (A ∪ B) ∩ C = {1, 4}.

First find the difference A \ B: remove from A every element that is also in B. Since b and d are common to A and B, A \ B = {a, c}. Next find B ∩ C, the elements common to both B and C: B ∩ C = {d, e}. Taking the union combines all distinct elements from these two sets, giving {a, c, d, e}. Therefore, option A is correct.

Since A ⊆ C and B ⊆ C, every element of A and every element of B is already in C. Therefore, every element of A ∪ B is also in C, which means A ∪ B ⊆ C. The difference (A ∪ B) \ C asks for elements in A ∪ B that are outside C. There are none, so the result is the empty set ∅.

For any two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Here A ∩ B is the empty set, so it has zero elements: n(A ∩ B) = 0. Substituting the given values gives n(A ∪ B) = p + q − 0 = p + q. Because the sets are disjoint, no common element is counted twice.

The number choosing at least one subject is 40 − 7 = 33, because 7 students chose neither subject. By the inclusion–exclusion principle, n(M ∪ P) = n(M) + n(P) − n(M ∩ P). Therefore, 33 = 24 + 18 − n(M ∩ P), so n(M ∩ P) = 9. Hence, 9 students chose both Mathematics and Physics.

The primes in U are {2, 3, 5, 7, 11, 13, 17, 19}. The difference A \ B keeps primes that are not odd. Every prime other than 2 is odd, so 3, 5, 7, 11, 13, 17, and 19 are removed. The number 2 is prime and even, hence it remains. Therefore A \ B = {2}, making option A correct; 1 is not prime, so option D is also incorrect.

For integer x with −3 ≤ x < 4, A = {−3, −2, −1, 0, 1, 2, 3}. The inequality x² ≤ 4 gives −2 ≤ x ≤ 2, so B = {−2, −1, 0, 1, 2}. Subtracting B from A removes the five central elements and leaves the endpoints −3 and 3. Hence A \ B = {−3, 3}, so option A is correct.

A contains every real number less than 2, and B contains every real number greater than −1. If a real number is less than 2, it belongs to A. If it is not less than 2, then it is at least 2 and is certainly greater than −1, so it belongs to B. Thus every real number is covered, and A ∪ B = R.

By definition, A \ B consists of elements that are in A but not in B. Relative to a universal set U, the complement B′ contains precisely the elements of U that are not in B. Intersecting A with B′ therefore retains elements in A that are outside B, so A \ B = A ∩ B′. The universal set must be specified for the complement.

Set difference is not commutative. A \ B contains elements of A that are not in B, whereas B \ A contains elements of B that are not in A; these sets can be different. For example, if A = {1, 2} and B = {2, 3}, then A \ B = {1}, while B \ A = {3}. The other three statements are standard identities or inclusion properties.

The set A \ B is obtained by removing from A every element that also belongs to B. If the result is still exactly A, no element of A could have been removed. Therefore A and B have no common element, which means A ∩ B = ∅. This does not imply that B is empty or that A and B are equal; B may contain elements outside A.

First list the ordered pairs allowed by x < y: (1,2), (1,3), and (2,3). Now test the second condition x + y = 4. Only (1,3) has sum 4. The pair (3,1) also has sum 4, but it fails x < y, while (2,2) fails the strict inequality. Therefore the intersection contains only {(1,3)}.