Operations on Sets (Union, Intersection, Difference)

The symmetric difference contains elements that occur in exactly one of the two sets. From A, the elements not in B are A − B = {a, c}. From B, the elements not in A are B − A = {e, f}. Their union is therefore A △ B = {a, c, e, f}. The common elements b and d are excluded because they belong to both sets.

Consider any element x. If x belongs to A, the equality A ∩ B = A ∩ C forces x to belong to B exactly when it belongs to C. If x does not belong to A, the equality A ∪ B = A ∪ C forces the same conclusion. Thus every element belongs to B and C together or to neither, so B = C. The stronger claim A = B = C does not necessarily follow.

A contains every real number from 0, including 0, up to but not including 3. B contains every real number greater than 1 through 5, including 5. The intervals overlap on (1, 3), so there is no gap between them. Their union therefore begins at 0 and ends at 5, with both endpoints included. Hence A ∪ B = [0, 5].

The difference A − B consists of elements that are in A but not in B. Set A is the interval [−3, 2], while B contains all numbers strictly between −1 and 4. Thus, every point from just greater than −1 through 2 is removed from A. The point −1 remains because −1 belongs to A but is excluded from B. Therefore, A − B = [−3, −1].

The union is divided into three disjoint regions: elements only in A, counted by n(A − B) = 12; elements only in B, counted by n(B − A) = 9; and common elements, counted by n(A ∩ B) = 7. Therefore n(A ∪ B) = 12 + 9 + 7 = 28. The common elements must be included once in the union.

First apply the inclusion–exclusion formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 52 + 47 − 21 = 78. The complement (A ∪ B)' contains the elements of the universal set that are in neither A nor B. Therefore, n((A ∪ B)') = n(U) − n(A ∪ B) = 80 − 78 = 2. Hence option A is correct.

Disjoint sets have no common elements, so A ∩ B = ∅ and n(A ∩ B) = 0. The general formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives n(A ∪ B) = 15 + 19 − 0 = 34. Thus every element from both sets is counted exactly once in the union.

The equality A ∪ B = A means that adding every element of B to A introduces no new element. Therefore, each element of B must already belong to A, which is exactly the definition of B ⊆ A. The sets need not be equal, disjoint, or complements; B may be a proper subset of A or may equal A.

The intersection \(A\cap B\) contains elements common to both sets. If this intersection is exactly \(B\), then every element of \(B\) must also belong to \(A\). Therefore, \(B\) is a subset of \(A\), written as \(B\subseteq A\). The statement does not necessarily imply that every element of \(A\) belongs to \(B\), so \(A\subseteq B\) is not required.

A three-set intersection contains only elements that occur in all three sets. The elements 3, 5, and 9 appear in \(A\), \(B\), and \(C\). Every other listed element is absent from at least one set. Hence \(A\cap B\cap C=\{3,5,9\}\), making option A correct.

The primes not exceeding 20 are \(\{2,3,5,7,11,13,17,19\}\). Set \(B\) contains all odd natural numbers below 20, so it contains every odd prime in \(A\), including 19. The only element of \(A\) that is not in \(B\) is the even prime 2. Thus \(A-B=\{2\}\).

Set \(A\) is the set of all even integers, while set \(B\) is the set of all integers divisible by 4. Every number of the form \(4k\) can be written as \(2(2k)\), so it is also an element of \(A\). However, 2 belongs to \(A\) but not to \(B\). Therefore, \(B\subseteq A\), but \(A\not\subseteq B\).

The distinct letters in MATHEMATICS are \(\{M,A,T,H,E,I,C,S\}\), and those in STATISTICS are \(\{S,T,A,I,C\}\). The letters common to both sets are A, I, M, S, and T? On checking, M is not present in STATISTICS, so the correct common set is actually \(\{A,C,I,S,T\}\). Therefore option C is mathematically correct, and the original answer key A must be corrected.

The difference \(B-C\) consists of elements that belong to \(B\) but do not belong to \(C\). If its intersection with \(A\) is empty, then no element can simultaneously be in \(A\), in \(B\), and outside \(C\). Thus, no element of \(A\) is in \(B\) but not in \(C\), which is exactly option A.

The empty set has no elements. Therefore, taking the union of \(A\) with the empty set adds nothing, so \(A\cup\varnothing=A\). Their intersection has no common element, so \(A\cap\varnothing=\varnothing\). Subtracting the empty set removes nothing, giving \(A-\varnothing=A\). Hence option A is correct.

The condition \(A\cup B=U\) says that together the sets cover the entire universal set. The condition \(A\cap B=\varnothing\) says that they have no common elements. Consequently, every element of \(U\) that is not in \(A\) must be in \(B\), and no element of \(A\) can be in \(B\). Therefore, \(B\) is the complement of \(A\) in \(U\), so \(B=A'\).

First calculate each difference separately. From \(A\), remove the elements also in \(B\): \(A-B=\{1,3,5\}\). From \(A\), remove the elements also in \(C\): \(A-C=\{3,4,5,6\}\). Their union contains every distinct element appearing in either result, namely \(\{1,3,4,5,6\}\). Thus option A is correct.

Set \(B\) is formed by selecting from \(A\) only those elements strictly greater than 5. Testing the elements gives \(B=\{7,11,13\}\); the value 5 is excluded because 5 is not greater than 5. Since \(B\subseteq A\), the intersection \(A\cap B\) equals \(B\), namely \(\{7,11,13\}\).

First, \(|x|\le 3\) gives \(-3\le x\le 3\), so \(A=\{-3,-2,-1,0,1,2,3\}\). Next, \(x^2-2x\le0\) becomes \(x(x-2)\le0\), which is true for \(0\le x\le2\). Since \(x\) is an integer, \(B=\{0,1,2\}\). Every element of B is in A, hence \(A\cap B=\{0,1,2\}\).

The positive divisors of 36 are \(\{1,2,3,4,6,9,12,18,36\}\), while those of 48 are \(\{1,2,3,4,6,8,12,16,24,48\}\). The common elements are therefore \(1,2,3,4,6,12\). Equivalently, common divisors are the divisors of \(\gcd(36,48)=12\), giving option A.