Operations on Sets (Union, Intersection, Difference)

The condition \(A\setminus B=\varnothing\) says that A contains no element outside B, so every element of A belongs to B; hence \(A\subseteq B\). Similarly, \(B\setminus A=\varnothing\) implies \(B\subseteq A\). When each set is a subset of the other, the two sets have exactly the same elements. Therefore \(A=B\), not a proper-subset relation and not necessarily disjoint.

Use the difference identity \(A\setminus X=A\cap X'\). Thus \(A\setminus(B\cap C)=A\cap(B\cap C)'\). By De Morgan's law, \((B\cap C)'=B'\cup C'\), so the expression becomes \(A\cap(B'\cup C')\). Distributing intersection over union gives \((A\cap B')\cup(A\cap C')\), which is exactly \((A\setminus B)\cup(A\setminus C)\). Therefore option A is correct.

An element belongs to \(A\setminus(B\cup C)\) exactly when it is in A and is not in the union \(B\cup C\). Not being in a union means that it is in neither B nor C. Therefore the element is simultaneously in \(A\setminus B\) and in \(A\setminus C\), giving \((A\setminus B)\cap(A\setminus C)\). Equivalently, use \(A\cap(B\cup C)'=A\cap(B'\cap C')\).

For two finite sets, the inclusion-exclusion formula is \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\). Substitute the given values: \(75=46+n(B)-18\). Rearranging gives \(n(B)=75-46+18=47\). The intersection is added back when solving for B because it had been subtracted from the sum of the two set sizes. Thus option A, 47, is the only correct value.

Students studying only Mathematics are 38 − 20 = 18, because the 20 students studying both subjects must be excluded. Students studying only Physics are 32 − 20 = 12. Therefore, the number studying exactly one subject is 18 + 12 = 30. Equivalently, this is 38 + 32 − 2(20) = 30.

Let T be the set of people who like tea and C the set of people who like coffee. By inclusion–exclusion, n(T ∪ C) = n(T) + n(C) − n(T ∩ C) = 52 + 47 − 19 = 80. Therefore, 80 people like at least one beverage. The number who like neither is 90 − 80 = 10, so option A is correct.

Set A contains the even numbers from 1 to 30, so n(A) = 30/2 = 15. The elements that belong to both A and B must be divisible by both 2 and 5, hence they are multiples of 10: 10, 20, and 30. Thus n(A ∩ B) = 3. Removing these from A gives n(A \ B) = 15 − 3 = 12. Therefore, option A is correct.

Factor the first quadratic: x² − 5x + 6 = (x − 2)(x − 3), so A = {2, 3}. Factor the second: x² − 4x + 3 = (x − 1)(x − 3), so B = {1, 3}. The intersection contains only values present in both sets. The only common value is 3, hence A ∩ B = {3}. Option A is correct.

The inequality x² ≤ 16 is equivalent to −4 ≤ x ≤ 4, so A = [−4, 4]. Set B contains numbers strictly greater than 1. To belong to the intersection, a number must satisfy both conditions, giving 1 < x ≤ 4. The lower endpoint 1 is excluded and the upper endpoint 4 is included, so A ∩ B = (1, 4]. Option A is correct.

Set A consists of all real numbers greater than −1 and at most 5. Set B contains every real number from 2 through 5, with 2 included. In A \ B, we retain elements of A that are not in B. Thus all values from −1 up to, but not including, 2 remain. The endpoint −1 was already excluded from A, and 2 is removed because it belongs to B. Therefore, A \ B = (−1, 2), option A.

First evaluate the expression inside the parentheses. B ∪ C = {1, 2, 4, 6, 7}, because union collects every distinct element from both sets. Now intersect this result with A = {1, 2, 3, 4, 5}. The common elements are 1, 2, and 4; numbers 6 and 7 are not in A. Hence A ∩ (B ∪ C) = {1, 2, 4}, so option A is correct.

First form the union A ∪ B by listing every distinct element from A and B: A ∪ B = {1, 3, 5, 6, 7, 9}. The difference (A ∪ B) \ C removes every element that occurs in C. Since C = {1, 2, 3, 4}, the elements 1 and 3 are removed; 2 and 4 were not present anyway. The remaining set is {5, 6, 7, 9}, so option A is correct.

The condition A ∩ B = ∅ means that A and B are disjoint, so they have no common elements. For any two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives n(A ∪ B) = 21 + 17 − 0 = 38. Equivalently, because no element is counted twice, we can simply add the two cardinalities. Therefore, option A is correct.

The union contains every element that belongs to A or B. The three given parts are disjoint: elements only in A are {2, 5}, elements only in B are {7}, and common elements are {1, 3}. Combining all these parts without repetition gives A ∪ B = {1, 2, 3, 5, 7}. Therefore, option A is correct.

The sets A and B can be divided into three non-overlapping regions: elements only in A, elements only in B, and elements in both sets. Their sizes are 12, 9, and 6 respectively. The union contains all three regions, so n(A ∪ B) = 12 + 9 + 6 = 27. Thus, option A is correct.

The equality \(A\cup B=A\) means that adding every element of \(B\) to \(A\) does not produce any new element. Therefore, every element of \(B\) is already an element of \(A\), so \(B\subseteq A\). The difference \(B\setminus A\) contains elements that belong to \(B\) but do not belong to \(A\). Since no such element exists, \(B\setminus A=\varnothing\). Hence option A is correct.

The condition A ∩ B = B means that every element of B belongs to A, so B ⊆ A. However, A may contain additional elements outside B, or A may equal B. Therefore, A \ B may be non-empty or empty, and its emptiness is not guaranteed. The other three statements are identities or direct consequences of set difference, so option A is correct.

An element belongs to \(A\cap B\) only when it is divisible by both 4 and 6. Such numbers are multiples of their least common multiple: \(\operatorname{lcm}(4,6)=12\). The multiples of 12 from 1 through 50 are \(12,24,36,48\). There are four such elements, so \(n(A\cap B)=4\). Therefore, option A is correct.

The condition |x| ≤ 4 gives A = {−4, −3, −2, −1, 0, 1, 2, 3, 4}, which has 9 elements. Factor the equation for B: x² − 2x − 3 = (x − 3)(x + 1) = 0, so B = {3, −1}. Both elements of B belong to A, and removing them leaves 9 − 2 = 7 elements. Therefore, option A is correct.

The symmetric difference A △ B contains elements that belong to exactly one of A and B. The union contains both exclusive elements and common elements, while the intersection contains only the common elements. Removing the intersection from the union leaves the symmetric difference. Thus n(A △ B) = n(A ∪ B) − n(A ∩ B) = 40 − 13 = 27. Option A is correct.