Operations on Sets (Union, Intersection, Difference)

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.

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 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.

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 inequality \(x^2<9\) is equivalent to \(-3<x<3\), so \(A=(-3,3)\). Set B contains 1 and every real number greater than 1. The difference \(A-B\) retains elements of A that are not in B, so all numbers from -3 up to but not including 1 remain. Thus \(A-B=(-3,1)\).

Solving \(x^2\le4\) gives \(-2\le x\le2\), so \(B=[-2,2]\). Set A already contains every negative real number, including all numbers less than -2. Set B adds the interval from -2 through 2, including 2. Their union therefore contains every real number less than or equal to 2, namely \((-\infty,2]\).

The difference \(A-B\) removes from A every element that also belongs to B. If the result is still exactly A, no element of A can have been removed. Therefore A and B have no common element, which means \(A\cap B=\varnothing\). Notice that B may contain elements outside A; it need not itself be empty.

Suppose an element belongs to A but not B. It would then belong to \(A-B\), but not to \(B-A\), contradicting equality. Similarly, an element belonging to B but not A would occur only in \(B-A\). Hence neither set can have an element absent from the other, so every element is common and \(A=B\).

For any two sets, \(A\cap B\subseteq A\cup B\). If the union and intersection are equal, every element in the union must also lie in the intersection. Thus any element belonging to A must belong to B, giving \(A\subseteq B\); similarly, \(B\subseteq A\). Therefore the two sets are equal: \(A=B\).

The difference \(A-B\) contains elements that are in \(A\) but not in \(B\). Therefore, \(A-B=\{1,2\}\), which has 2 elements. If a finite set has \(n\) elements, its power set has exactly \(2^n\) subsets, including the empty set and the original set. Hence, \(|\mathcal{P}(A-B)|=2^2=4\). Option B gives only the cardinality of \(A-B\), not its power set.

The union \(A\cup B\) is partitioned into three mutually disjoint parts: \(A-B\), \(B-A\), and \(A\cap B\). Let \(n(A\cap B)=x\). Then \(75=28+31+x\). Hence \(x=75-59=16\). Therefore, the intersection contains 16 elements. The calculation also shows why the difference parts must not be counted as overlapping with each other.

First, \(A\cap B=\{2,4,6,8\}\), because these elements occur in both sets. Next, \(A\cup B=\{1,2,3,4,5,6,7,8,10\}\). Removing the intersection from the union leaves the elements that belong to exactly one of the two sets: \(\{1,3,5,7,10\}\). Thus the expression is the symmetric difference of \(A\) and \(B\), so option A is correct.

In general, \(A-(A-B)=A\cap B\). This is because \(A-B\) removes from \(A\) every element that is not in \(B\), leaving only the elements common to both sets. Since the question states that \(A\cap B=\varnothing\), the remaining set is empty. Equivalently, disjointness gives \(A-B=A\), so the expression becomes \(A-A=\varnothing\).

The distributive law for sets states that intersection distributes over union: \(A\cap(B\cup C)=(A\cap B)\cup(A\cap C)\). An element belongs to the left side exactly when it is in \(A\) and also in at least one of \(B\) or \(C\); this is precisely the condition described by the right side. Option B is the other distributive identity and is not generally equal to the given expression.

The second distributive law for sets is \(A\cup(B\cap C)=(A\cup B)\cap(A\cup C)\). To verify it elementwise, an element is on the left if it is in \(A\), or if it is in both \(B\) and \(C\). On the right, it must be in both unions, which gives exactly the same condition. Therefore, option A is correct.

Within the restriction \(x\le15\), the multiples of 3 are \(\{3,6,9,12,15\}\), and the multiples of 5 are \(\{5,10,15\}\). Their union contains every number divisible by 3 or 5: \(\{3,5,6,9,10,12,15\}\). Intersecting with \(A\) simply enforces the upper bound of 15. Thus option A is correct; options C and D omit one of the two groups.

An element belongs to \(A\cap B\) only if it is divisible by both 2 and 3 and is at most 30. Every number divisible by both 2 and 3 is divisible by their least common multiple, \(\operatorname{lcm}(2,3)=6\). The positive multiples of 6 not exceeding 30 are \(6,12,18,24,30\). Hence option A is correct.

To evaluate A − (B ∩ C), first calculate the operation inside the parentheses. The elements common to B = {2, 3, 6} and C = {3, 4, 7} form B ∩ C = {3}. Set difference A − {3} means retaining every element of A that is not 3. Removing 3 from A = {1, 2, 3, 4, 5} gives {1, 2, 4, 5}. Therefore, option A is correct. The order of operations matters: find the intersection first, then perform the difference.

First combine A and C. Since A covers all real numbers greater than -2 up to 6, and C covers numbers greater than 4 up to 10, their union is (-2,10]. Now subtract B = [1,8), which removes every number from 1 inclusive to 8 exclusive. The part below 1 remains (-2,1), while 8 remains because it is not in B, giving [8,10]. Therefore, the answer is (-2,1) ∪ [8,10].

By definition, B − C consists of elements that belong to B but do not belong to C. Intersecting this set with A retains precisely those elements that are simultaneously in A and B and are outside C. This is exactly the meaning of (A ∩ B) − C. The other expressions either include extra elements or remove the wrong condition.

The condition x² ≤ 16 gives -4 ≤ x ≤ 4 for integer x. This entire set lies inside A, because A contains every integer from -4 through 6. Hence A ∩ B = {-4,-3,-2,-1,0,1,2,3,4}. Set C contains the even integers, so removing C deletes -4,-2,0,2,4. The remaining odd integers are {-3,-1,1,3}, which is option A.