Operations on Sets (Union, Intersection, Difference)

Every element of A ∪ B belongs to exactly one of three disjoint regions: A − B, A ∩ B, or B − A. The union is {2,4,6,8,10,12,14}. Removing the elements belonging only to A, namely {2,10}, and those belonging only to B, namely {6,14}, leaves {4,8,12}. These remaining elements must belong to both sets, so A ∩ B = {4,8,12}.

The set A contains its exclusive part A − B and its common part A ∩ B. Therefore n(A ∩ B) = n(A) − n(A − B) = 58 − 23 = 35. Using the union formula, n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 58 + 44 − 35 = 67. The common elements must be subtracted once because they were counted twice.

Factor the quadratic: x² − 5x + 6 = (x − 2)(x − 3). Since the parabola opens upward, the inequality (x − 2)(x − 3) ≤ 0 holds for 2 ≤ x ≤ 3, so A = [2, 3]. Every number in [2, 3] is less than 4 and therefore belongs to B. Thus A is a subset of B, leaving no element in A − B; the answer is the empty set.

Solve the first inequality by removing the absolute value: |x − 2| ≤ 3 means −3 ≤ x − 2 ≤ 3. Adding 2 throughout gives −1 ≤ x ≤ 5, so A = [-1,5]. The second inequality x² ≤ 4 means −2 ≤ x ≤ 2, so B = [-2,2]. The common part of these two closed intervals starts at -1 and ends at 2. Therefore, A ∩ B = [-1,2].

Every element of A falls into exactly one of two cases: it is either outside B, in which case it belongs to A − B, or it is inside B, in which case it belongs to A ∩ B. These two parts are disjoint and together contain all elements of A. Hence (A − B) ∪ (A ∩ B) = A. The expression cannot generally equal B or A ∪ B because those may contain elements outside A.

First find the union of A and B by listing every element that occurs in either set: \(A\cup B=\{1,2,3,4,5,6,7,8\}\). Next, take the intersection with C, which means retain only the elements common to this union and C. Since 1, 4, 7, and 8 all belong to the union, the result is \(\{1,4,7,8\}\). Therefore, option A is correct; option B wrongly omits 8.

The elements of set \(A\) are divided into two disjoint parts: those belonging only to \(A\), represented by \(A-B\), and those common to both sets, represented by \(A\cap B\). Their sizes are 2 and 3 respectively. Thus \(n(A)=2+3=5\). The element in \(B-A\) belongs only to \(B\), so it is not counted.

Assume \(A\cap C=\varnothing\) as well. Since \(A\cap B=\varnothing\), every element of \(B\) lies outside \(A\). Equality of the unions \(A\cup B\) and \(A\cup C\) then forces every element of \(B\) to occur in \(C\); otherwise it would appear only in the first union. Hence \(B\subseteq C\), so option A is sufficient.

An element in \(A\cap B\) must be divisible by both 4 and 6. Such numbers are multiples of their least common multiple, \(\operatorname{lcm}(4,6)=12\). The positive multiples of 12 not exceeding 40 are 12, 24, and 36. Therefore, \(A\cap B=\{12,24,36\}\), making option A correct.

The intersection contains natural numbers that divide both 72 and 90. Therefore, its elements are exactly the positive divisors of \(\gcd(72,90)\). Since \(\gcd(72,90)=18\), the positive divisors are 1, 2, 3, 6, 9, and 18. Thus \(A\cap B=\{1,2,3,6,9,18\}\), so option A is correct.

First find each difference separately. In \(A-B\), retain elements of A absent from B, giving \(\{1,2\}\). In \(B-C\), retain elements of B absent from C, giving \(\{3,4\}\), because 5 and 6 belong to C. Their union is \(\{1,2\}\cup\{3,4\}=\{1,2,3,4\}\). Hence option A is correct.

The union \(A\cup B\) contains every element of A and every element of B. When A is removed from this union, all elements belonging to A disappear, including those that may also be in B. The elements left are precisely the elements that belong to B but do not belong to A. Therefore, \((A\cup B)-A=B-A\), so option A is correct. This is a standard identity involving union and set difference.

The equality \(A\cap B=A\) means every element of A is also in B, so \(A\subseteq B\). Similarly, \(B\cap C=B\) means every element of B is in C, so \(B\subseteq C\). Subset inclusion is transitive; therefore \(A\subseteq B\subseteq C\), which gives \(A\subseteq C\). Hence option A is correct.

To determine A, solve \(x^2-1=0\), which factors as \((x-1)(x+1)=0\). Thus, for integer x, \(x=1\) or \(x=-1\), so \(A=\{-1,1\}\). The condition defining B is already \(x^2=1\), giving the same set \(B=\{-1,1\}\). Since every element of A is also in B, no element remains after subtracting B from A. Hence \(A-B=\varnothing\), so option A is correct.

An element belongs to \(A-(B\cup C)\) exactly when it is in A and is in neither B nor C. The same condition means that it belongs to both \(A-B\) and \(A-C\). Therefore, \(A-(B\cup C)=(A-B)\cap(A-C)\). This is a difference form of De Morgan’s law, so option C is always true.

The distinct letters of ALGEBRA are {A, L, G, E, B, R}, while the distinct letters of GEOMETRY are {G, E, O, M, T, R, Y}. A union B contains every element appearing in either set, with repeated letters written only once. Therefore, A ∪ B = {A, B, E, G, L, M, O, R, T, Y}.

List the odd natural numbers not exceeding 25: \(B=\{1,3,5,7,9,11,13,15,17,19,21,23,25\}\). The primes among them are \(3,5,7,11,13,17,19,23\), which form \(A\). Set difference \(B-A\) keeps elements of \(B\) that are not in \(A\). Therefore, it contains 1 and the odd composite numbers 9, 15, 21, and 25, giving \(\{1,9,15,21,25\}\).

A contains the multiples of 2 in U, and B contains the multiples of 5. Thus A ∪ B contains every number divisible by 2 or by 5: {2, 4, 5, 6, 8, 10, 12, 14, 15, 16, 18, 20}. The complement contains the elements of U divisible by neither 2 nor 5, namely {1, 3, 7, 9, 11, 13, 17, 19}.

An intersection contains only numbers satisfying both conditions. The condition from A is x ≤ 1, and the condition from B is x > −2. Combining them gives −2 < x ≤ 1. In interval notation this is (−2, 1]. The left endpoint is open because −2 is excluded, while the right endpoint is closed because 1 is included.

The equality A ∪ B = B means that adding every element of A to B does not enlarge B, so every element of A is already in B; therefore A ⊆ B. Similarly, A ∩ C = C means every element of C belongs to A, so C ⊆ A. By transitivity of subset relation, C ⊆ A ⊆ B, and hence C ⊆ B.