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.

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.

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.

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 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.

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.

First evaluate the difference \(A-B\). Remove every element of \(B\), namely 2, 4, and 6, from \(A\); this gives \(A-B=\{1,3,5\}\). Now intersect this result with \(C\). Since \(C=\{1,3,5\}\), every element of \(A-B\) is also in \(C\). Hence \((A-B)\cap C=\{1,3,5\}\). Difference removes the second set’s elements, while intersection retains only common elements.

The set \(A\) contains all integers from \(-6\) through 6. For \(B\), solve \(x^2<10\). Since \(3^2=9<10\) but \(4^2=16>10\), the integer solutions are \(-3,-2,-1,0,1,2,3\). Thus \(B=\{-3,-2,-1,0,1,2,3\}\). Removing these elements from \(A\) leaves \(\{-6,-5,-4,4,5,6\}\), so option A is correct. The endpoints \(-3\) and 3 must be removed because their squares are 9.

A − B = ∅ means that no element of A lies outside B, so A ⊆ B. Likewise, B − C = ∅ means B ⊆ C. Subset inclusion is transitive: if every element of A is in B and every element of B is in C, then every element of A is in C. Therefore A ⊆ C must be true. Equality or disjointness is not forced.

A is the closed interval [1, 9], while B is the open interval (3, 7). To find A − B, remove every real number strictly between 3 and 7 from A. The endpoints 3 and 7 are not elements of B because B uses strict inequalities, so they remain. The result is [1, 3] ∪ [7, 9].

The condition \(A\cap B=\varnothing\) means that A and B have no common elements. The union \(A\cup B\) contains every element of both sets. When all elements belonging to B are removed from this union, no element of A is removed because A and B are disjoint. Therefore, only A remains, so \((A\cup B)\setminus B=A\). This also follows from the identity \((A\cup B)\setminus B=A\setminus B\), together with \(A\setminus B=A\) for disjoint sets.

Use inclusion–exclusion to calculate the size of the union: \(n(A\cup B)=n(A)+n(B)-n(A\cap B)=70+65-40=95\). De Morgan’s law gives \(A'\cap B'=(A\cup B)'\), so the required set consists of elements in the universal set that belong to neither A nor B. Its size is therefore \(n(U)-n(A\cup B)=120-95=25\). Thus option A is correct; 95 is the union size, not the required complement size.

Compare the sets pair by pair. The elements common to A and B are only 1, so \(A\cap B=\{1\}\). The elements common to B and C are also only 1, so \(B\cap C=\{1\}\). Finally, the elements common to C and A are only 1, so \(C\cap A=\{1\}\). Taking the union of these three pairwise intersections still gives \(\{1\}\), because repeating an element does not create a new element in a set. Therefore option A is correct.

The set A contains every integer from −3 through 8, while B contains 0, 1, 2, 3, 4, and 5. Removing B from A leaves \(A\setminus B=\{-3,-2,-1,6,7,8\}\). The set C contains all even integers, so we retain only the even members of this difference set. Those are −2, 6, and 8. Hence \((A\setminus B)\cap C=\{-2,6,8\}\). Option B stops before applying the even-number condition.

Within the natural numbers up to 10, the odd numbers are \(1,3,5,7,9\). The odd prime numbers in this range are 3, 5, and 7. Subtracting C from B removes these primes and leaves \(B\setminus C=\{1,9\}\). Both 1 and 9 satisfy the definition of A, so intersecting with A does not remove either element. Therefore, \(A\cap(B\setminus C)=\{1,9\}\). Remember that 1 is neither prime nor composite.

The inequality \(x^2\le 9\) is equivalent to \(-3\le x\le 3\), so \(A=[-3,3]\). Similarly, \(x^2<1\) gives \(-1<x<1\), so \(B=(-1,1)\). The difference \(A\setminus B\) removes every point strictly between −1 and 1 from A. The boundary points −1 and 1 remain because they are not members of B. Therefore, \(A\setminus B=[-3,-1]\cup[1,3]\), making option A correct.

First form the union by listing each distinct element only once: \(A\cup B=\{1,2,3,4,5\}\). Thus the union has 5 elements. For any finite set with n elements, its power set contains every possible subset, including the empty set and the full set, and its cardinality is \(2^n\). Hence \(|\mathcal{P}(A\cup B)|=2^5=32\). The repeated elements 2 and 3 are counted only once in the union, so option A is correct.