Operations on Sets (Union, Intersection, Difference)

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 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 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 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.

First determine the intersection. The elements common to A = {a, b, c, d} and B = {c, d, e} are c and d, so A ∩ B = {c, d} and has 2 elements. A set with n elements has 2ⁿ subsets in its power set. Therefore, P(A ∩ B) has 2² = 4 elements, making option A correct.

Rewrite the difference as an intersection with a complement: \(A\setminus B=A\cap B'\). Taking the complement with respect to \(U\) and applying De Morgan’s law gives \((A\setminus B)'=(A\cap B')'=A'\cup(B')'=A'\cup B\). Thus the required complement contains everything outside \(A\), together with all elements of \(B\), and option A is correct.

The set \(A\setminus B\) consists of all elements of \(A\) except those that are also in \(B\). The elements removed from \(A\) are precisely the elements in \(A\cap B\). Therefore, subtracting the size of the common part from the size of \(A\) gives \(n(A\setminus B)=n(A)-n(A\cap B)\). Option C is numerically equivalent, because \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\), but the requested direct formula is option A.

The difference \(A\setminus B\) contains elements that are in \(A\) but not in \(B\). Removing 2, 3, and 5 from \(A\) gives \(\{1,4,6\}\). Now intersect this result with \(C=\{3,4,5,8\}\). The only common element is 4, so \((A\setminus B)\cap C=\{4\}\). Option C stops after the difference operation, and option D incorrectly includes 8.

The identity \(A\cap(B\cup C)=(A\cap B)\cup(A\cap C)\) states that intersection distributes over union. The operation outside the parentheses, intersection, is applied separately to both terms inside the union. This is different from the commutative law, which merely changes order, and the associative law, which changes grouping. Therefore option A is the only correct answer.

This identity shows union distributing over intersection: \(A\cup(B\cap C)=(A\cup B)\cap(A\cup C)\). The union with \(A\) is distributed across both members of the intersection. It is not De Morgan’s law because no complements appear, and it is not absorption because neither side reduces simply to \(A\). Thus option A is correct. The two distributive identities should be distinguished by the operation outside the parentheses.

The identity \(A\cup(A\cap B)=A\) is the absorption law. Since \(A\cap B\) is always a subset of \(A\), taking its union with \(A\) cannot add any new element. Consequently, the result remains \(A\). The distributive law expands an operation across parentheses, while De Morgan’s laws involve complements. Therefore option A is unambiguously correct. The companion absorption identity is \(A\cap(A\cup B)=A\).

The expression \(A\cap(A\cup B)=A\) is the second standard form of the absorption law. Every element of \(A\) is automatically in \(A\cup B\), so intersecting \(A\cup B\) with \(A\) leaves exactly \(A\). The commutative law changes the order of operations or sets, and the associative law changes grouping; neither explains this reduction. Hence option A is correct.

The complement \(B'\), relative to the universal set \(U\), contains every element that is not in \(B\). Therefore \(A\cap B'\) consists precisely of elements that belong to \(A\) and do not belong to \(B\). This is the definition of the difference \(A\setminus B\). Option B reverses the sets, while option D describes elements in \(B\) but outside \(A\). Hence option A is correct.

The condition \(A\subseteq B\) means every element of \(A\) is already an element of \(B\). On the other hand, \(C\setminus B\) contains elements of \(C\) that are specifically not in \(B\). No element can therefore belong to both \(A\) and \(C\setminus B\). Their intersection is empty, so \(A\cap(C\setminus B)=\varnothing\). Hence option A is correct.

The equality \(A\setminus B=\varnothing\) means that no element of \(A\) lies outside \(B\). By definition, this is equivalent to \(A\subseteq B\). The additional condition \(A\cap B\ne\varnothing\) says that the two sets share at least one element. Since every element of \(A\) is in \(B\), this also confirms that \(A\) is non-empty. Therefore option A is the correct conclusion.

Use the cardinality formula \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\). Substituting the given values gives \(25=18+25-n(A\cap B)\), so \(n(A\cap B)=18\). Thus every one of the 18 elements of \(A\) is also in \(B\), meaning \(A\subseteq B\). Therefore \(A\setminus B\) is empty and its cardinality is 0. Option A is correct.

The first interval is \(A=[0,5)\), while the second is \(B=(3,8]\). Their union covers every real number from 0 through 8, so \(A\cup B=[0,8]\). Their intersection consists of numbers common to both, namely \((3,5)\). Removing this common part from the union leaves the numbers in exactly one set: \([0,3]\cup[5,8]\). The endpoints 3 and 5 are included because 3 belongs to A but not B, and 5 belongs to B but not A.

Use the inclusion–exclusion formula for three finite sets: \(n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C)\). Substitution gives \(86=38+34+31-12-10-9+x=72+x\). Hence \(x=14\), so \(n(A\cap B\cap C)=14\). The triple intersection is added at the end because elements common to all three sets were counted three times initially and then subtracted three times through the pairwise terms.

There are \(20\) multiples of 2, \(13\) multiples of 3, and \(8\) multiples of 5 in \(\{1,\ldots,40\}\). The pairwise overlaps contain multiples of 6, 10, and 15, giving \(6,4,2\) elements respectively. The common overlap consists of multiples of 30, so it has 1 element. Therefore, by inclusion–exclusion, \(n(A\cup B\cup C)=20+13+8-6-4-2+1=30\).

The union \(A\cup B\) is partitioned into three mutually disjoint regions: elements belonging only to A, elements belonging only to B, and elements belonging to both sets. Thus, \(n(A\cup B)=n(A\setminus B)+n(B\setminus A)+n(A\cap B)\). Substituting the values gives \(70=22+31+n(A\cap B)\), so \(n(A\cap B)=70-53=17\).