Operations on Sets (Union, Intersection, Difference)

The difference A \ B contains elements that belong to A but not B, whereas B \ A contains elements that belong to B but not A. These two differences are always disjoint. They therefore cannot be equal to the same non-empty set {2, 5}, because that would make 2 and 5 belong to both differences simultaneously. The stated condition is impossible.

A number belongs to A ∩ B precisely when it is divisible by both 4 and 6. Such numbers are multiples of lcm(4, 6) = 12. The positive multiples of 12 not exceeding 50 are 12, 24, 36, and 48. There are four numbers, so n(A ∩ B) = 4. The definition of N as positive natural numbers removes any ambiguity about including zero.

The condition |x| ≤ 3, with x an integer, gives A = {−3, −2, −1, 0, 1, 2, 3}. For B, x² − 1 = 0 means x² = 1, so x = −1 or x = 1; hence B = {−1, 1}. The difference A \ B contains the elements of A that are not in B. Removing −1 and 1 from A leaves {−3, −2, 0, 2, 3}. Therefore option A is correct.

Separate the elements of A ∪ B into three disjoint regions: A \ B, A ∩ B, and B \ A. Subtracting A \ B removes only the elements that belong exclusively to A. The remaining two regions, A ∩ B and B \ A, together contain every element of B and no element outside B. Therefore, (A ∪ B) \ (A \ B) = B.

The union A ∪ B consists of three mutually disjoint regions: the elements only in A, the elements only in B, and the elements common to both sets. Therefore, n(A ∪ B) = n(A \ B) + n(B \ A) + n(A ∩ B). Substituting the values gives 54 = 17 + 21 + n(A ∩ B), so n(A ∩ B) = 54 − 38 = 16. Hence, option A is correct.

There are floor(60/2) = 30 multiples of 2 and floor(60/3) = 20 multiples of 3 in U. Elements divisible by both 2 and 3 are multiples of 6, and there are floor(60/6) = 10 of them. By inclusion–exclusion, |A ∪ B| = |A| + |B| − |A ∩ B| = 30 + 20 − 10 = 40.

For every pair of sets, A ∩ B is contained in A ∪ B. If the two are equal, then every element that belongs to either set must belong to both sets. Thus every element of A belongs to B, so A ⊆ B; similarly, every element of B belongs to A, so B ⊆ A. Mutual inclusion proves A = B. The other options describe only special cases or contradict the condition.

Use the cardinality formula |A ∪ B| = |A| + |B| − |A ∩ B|. Thus, 50 = 28 + 35 − |A ∩ B|, so |A ∩ B| = 13. The set difference A \ B contains the elements that are in A but not in B, so |A \ B| = |A| − |A ∩ B| = 28 − 13 = 15. Therefore, option A is correct.

The difference A \ B contains the elements of A that are not in B. If removing all elements of B from A leaves A unchanged, then no element of A can have belonged to B. Equivalently, there is no common element between the sets, so A ∩ B = ∅. The condition does not require B to be contained in A, nor does it imply that A is contained in B or that their union equals A.

First find the union of A and B by listing every element that occurs in either set: A ∪ B = {1, 2, 3, 4, 5, 6, 8}. Next, intersect this result with C, which means retain only elements also present in C = {1, 4, 7, 8}. The common elements are 1, 4, and 8. Therefore, (A ∪ B) ∩ C = {1, 4, 8}, so option A is correct.

Because A is a subset of B, every element of B is either already in A or belongs to B but not to A. Thus the two disjoint parts A and B \ A together partition B. Their union contains every element of B and contains nothing outside B. Consequently, A ∪ (B \ A) = B. This identity depends on the given subset condition; without A ⊆ B, it would not generally hold.

First list the integers in A: A = {−3, −2, −1, 0, 1, 2, 3, 4}. For B, the condition x² < 10 means −√10 < x < √10, so the integer elements are B = {−3, −2, −1, 0, 1, 2, 3}. The set difference A \ B keeps elements of A that are not in B. Therefore, only 4 remains, and A \ B = {4}.

The equality A ∩ B = A means that intersecting A with B does not remove any element of A. Thus every element of A must already belong to B, which is precisely the statement A ⊆ B. The equality A ∪ B = B expresses the same fact: adding A to B contributes no new elements. Equality of A and B is not required.

To construct B, test each element x of A and check whether x + 1 is also an element of A. For x = 2, x + 1 = 3, which belongs to A, so 2 ∈ B. For 3, 5, 7, and 11, the successors 4, 6, 8, and 12 are not in A. Hence B = {2}. Removing B from A leaves A \ B = {3, 5, 7, 11}.

The union B ∪ C contains every element that belongs to B, to C, or to both. The difference A \ (B ∪ C) therefore removes from A every element found in either B or C. What remains consists of elements that are in A, not in B, and not in C. Equivalently, it can be written as A ∩ Bᶜ ∩ Cᶜ, so option A gives the precise meaning.

First find the intersection A ∩ C, which contains elements common to both sets: A ∩ C = {4, 8, 16}. Next take the difference with B by removing elements that occur in B. Both 4 and 8 belong to B, but 16 does not. Consequently, only 16 remains, so (A ∩ C) \ B = {16}.

The distributive law of 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 in at least one of \(B\) or \(C\). This is precisely the condition represented on the right side. Therefore option A is correct; the other expressions use union, difference, or reversed membership conditions and are not generally equivalent.

An element in \(A\cap B\) must be divisible by both 4 and 10. Such numbers are multiples of \(\operatorname{lcm}(4,10)=20\). The multiples of 20 in \(\{1,\ldots,100\}\) are 20, 40, 60, 80, and 100. There are five numbers, so \(|A\cap B|=5\), making option A correct.

First find each difference separately. Elements in \(A\) but not \(B\) are \(a,c\), so \(A\setminus B=\{a,c\}\). Elements in \(B\) but not \(A\) are \(e\), so \(B\setminus A=\{e\}\). Their union is therefore \(\{a,c\}\cup\{e\}=\{a,c,e\}\). This is the symmetric difference, so option A is correct.

Every element of the union belongs to exactly one of three disjoint regions: \(A\setminus B\), \(A\cap B\), or \(B\setminus A\). Combining the given regions gives \(\{1,4\}\cup\{2,3\}\cup\{5,6,7\}=\{1,2,3,4,5,6,7\}\). Therefore option A is correct. The other options omit one of the three regions.

To find each set, solve its defining quadratic equation. For A, x² − 5x + 6 = (x − 2)(x − 3) = 0, so A = {2, 3}. For B, x² − 7x + 12 = (x − 3)(x − 4) = 0, so B = {3, 4}. The intersection contains only elements present in both sets. Since 3 is common to A and B, A ∩ B = {3}. The set {2, 3, 4} would be the union, not the intersection.

The sets A \ B, B \ A, and A ∩ B represent three mutually disjoint regions in the Venn diagram. Every element of A ∪ B belongs to exactly one of these regions. Therefore, |A ∪ B| = |A \ B| + |B \ A| + |A ∩ B| = 9 + 4 + 11 = 24. Equivalently, |A| = 9 + 11 = 20 and |B| = 4 + 11 = 15, so |A ∪ B| = |A| + |B| − |A ∩ B| = 20 + 15 − 11 = 24.

The equation \(A\cap B=B\) says that every element of \(B\) is also an element of \(A\), so \(B\subseteq A\). The equation \(A\cup B=A\) gives exactly the same conclusion: adding all elements of \(B\) to \(A\) does not enlarge \(A\). Thus option A is necessary. The reverse inclusion, disjointness, and complement relation do not necessarily follow.

First form the union: \(B\cup C=\{0,3,4,5,6,9\}\), because every element appearing in either set is included once. Now intersect this result with \(A=\{1,3,5,7,9\}\). The common elements are 3, 5, and 9, so \(A\cap(B\cup C)=\{3,5,9\}\). Therefore option A is correct; option D is only the union, not the final intersection.

From A ∩ B = A \ C, every element of A ∩ B belongs to A but does not belong to C. Hence no element can simultaneously belong to A ∩ B and C. Since A ∩ B ∩ C consists precisely of elements common to A, B, and C, it follows that A ∩ B ∩ C = ∅. The first given equality is consistent with this result but is not needed for the final conclusion.