Operations on Sets (Union, Intersection, Difference)

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.

The symmetric difference A △ B contains the elements that belong to exactly one of A or B, while A ∩ B contains the elements common to both. These two parts are disjoint and together make up A ∪ B. Therefore, n(A ∪ B) = n(A △ B) + n(A ∩ B) = 26 + 9 = 35. Hence option A is correct.

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.

To prove the result, consider any element x. If x belongs to A, the equality of intersections tells us that x belongs to B exactly when it belongs to C. If x does not belong to A, the equality of unions tells us that x belongs to B exactly when it belongs to C. Thus every element has the same membership status in B and C, so B and C contain precisely the same elements. Therefore \(B=C\). The other options are not forced by the two given equalities.

Factor the quadratic: \(x^2-4x+3=(x-1)(x-3)\). Because the quadratic opens upward, it is positive outside its roots, so A is \((-infty,1)\cup(3,infty)\). Set B contains numbers greater than 1, represented by \((1,infty)\). Their common part is therefore only the interval \((3,infty)\). The endpoints 1 and 3 are excluded because the inequality is strict and the roots make the expression zero. Hence option A is correct.

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.

Remove from A every element that belongs to B. Since B contains 2, 4, and 6, the difference is \(A\setminus B=\{1,3,5\}\), which has 3 elements. A set with n elements has exactly \(2^n\) subsets in its power set, because each element may either be selected or not selected. Therefore, \(n(\mathcal{P}(A\\setminus B))=2^3=8\). The number 3 is the size of the difference set, not of its power set.

Although B − C has 6 elements that also belong to A, none of these elements can belong to A − B, because every element of A − B is specifically outside B, whereas every element of B − C is inside B. Thus (A − B) and (B − C) are disjoint. Therefore, n((A − B) ∪ (B − C)) = 17 + 22 = 39. The stated common elements with A do not create an intersection between the two sets in the union.

Because A is a subset of B, every element of B − A is also an element of C − A. Starting with C − A means retaining elements of C that are not in A. Removing B − A then removes all elements that are in B but not in A. The elements left are precisely those in C that are outside B, namely C − B. Hence option A is correct.

The conditions \(A\cap B=\varnothing\), \(B\cap C=\varnothing\), and \(C\cap A=\varnothing\) show that the three sets are pairwise disjoint. Therefore, no element is repeated in their union. We can add their cardinalities directly: \(n(A\cup B\cup C)=n(A)+n(B)+n(C)=12+15+18=45\). Hence, option A is correct. If the sets were not disjoint, common elements would have to be subtracted using the inclusion–exclusion principle.

The natural-number perfect squares not exceeding 30 are \(1,4,9,16,25\), so \(A=\{1,4,9,16,25\}\). The set \(B\) contains numbers up to 30 divisible by 3. Among the elements of \(A\), only 9 is divisible by 3, because \(9=3\times3\). Set difference \(A-B\) means retaining elements of \(A\) that are not in \(B\). Thus, \(A-B=\{1,4,16,25\}\), making option A correct.

An element belongs to A \ (B ∪ C) precisely when it belongs to A and belongs to neither B nor C. The condition of being outside the union B ∪ C therefore means being outside both B and C. This is exactly the membership condition for (A \ B) ∩ (A \ C). Thus option A is De Morgan’s difference identity and is true for all sets.