Operations on Sets (Union, Intersection, Difference)

The set A contains the 15 multiples of 2 from 1 to 30, and B contains the 10 multiples of 3. Numbers counted in both sets are multiples of 6; there are 5 of them: 6, 12, 18, 24, and 30. Therefore, inclusion–exclusion gives \(n(A\cup B)=n(A)+n(B)-n(A\cap B)=15+10-5=20\).

The condition \(A\subseteq B\) means every element of A is already in B. Consequently, the union is \(A\cup B=B\), while the intersection is \(A\cap B=A\). Substituting these identities into the expression gives \((A\cup B)\setminus(A\cap B)=B\setminus A\). This is precisely the part of B that is not contained in A.

The statement \(A\setminus B=\varnothing\) means that A has no element outside B, so \(A\subseteq B\). Similarly, \(B\setminus A=\varnothing\) implies \(B\subseteq A\). Since each set is a subset of the other, the two sets have exactly the same elements. Therefore, \(A=B\). The conditions do not imply that the sets are empty or disjoint.

For any two sets, \(A\cap B\subseteq A\cup B\) always holds. Equality can occur only when no element belongs to one set without also belonging to the other. Equivalently, every element of A is in B and every element of B is in A. Hence both mutual containments hold, which proves \(A=B\). The condition does not require the sets to be empty.

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.

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 ∅.

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.

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.

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)}.

Set A contains all integers divisible by 2, and set B contains all integers divisible by 3. An element of their intersection must satisfy both divisibility conditions at the same time. Since 2 and 3 are coprime, every integer divisible by both is divisible by their least common multiple, 6. Thus A ∩ B is the set of all integer multiples of 6, including 0 and negative multiples.

The union A ∪ B is divided into three mutually disjoint regions: elements belonging only to A, represented by A \ B; elements belonging only to B, represented by B \ A; and elements common to both, represented by A ∩ B. Therefore no element is counted twice when these three numbers are added: n(A ∪ B) = 5 + 7 + 4 = 16.

First calculate the number of elements in A ∪ B. Because the 11 elements in A ∩ B are included in both A and B, they would be counted twice in n(A) + n(B), so subtract them once: n(A ∪ B) = 32 + 27 − 11 = 48. The complement (A ∪ B)′ contains all elements of U that are not in the union. Therefore, n((A ∪ B)′) = n(U) − n(A ∪ B) = 60 − 48 = 12. Thus, option A is correct.

For A \ B to contain exactly 2 elements, we must choose exactly 2 elements of A to be excluded from B. Once those two elements are chosen, the other two elements must belong to B, so each choice determines exactly one valid subset B. The number of choices is the combination 4 choose 2, equal to 4!/(2!2!) = 6. Hence six subsets satisfy the condition.

Factor the quadratic: x² − 4x + 3 = (x − 1)(x − 3). Since the parabola opens upward, the inequality (x − 1)(x − 3) ≤ 0 holds for 1 ≤ x ≤ 3. Thus B = [1, 3]. This interval is already contained in A = [0, 4], so their intersection is [1, 3]. Therefore, option A is correct.

The sets A ∩ B and A \ B represent two non-overlapping parts of A: the elements inside B and the elements outside B, respectively. Their intersection is always empty. If these two disjoint sets are equal, the common set must itself be empty. Their union is A, so A must also be empty. Hence option A is correct.