Operations on Sets (Union, Intersection, Difference)

The inequality \(|x|\le4\) gives \(-4\le x\le4\), so \(A=\{-4,-3,-2,-1,0,1,2,3,4\}\). The condition \(x^2\le9\) is equivalent to \(|x|\le3\), giving \(B=\{-3,-2,-1,0,1,2,3\}\). Removing all elements of \(B\) from \(A\) leaves only \(-4\) and 4. Therefore \(A\setminus B=\{-4,4\}\).

The sets A ∩ B and A \ B are always disjoint: an element cannot be both in B and outside B. If two disjoint sets are equal, their common set must contain no elements. Therefore A ∩ B = A \ B = ∅. Since A \ B = ∅ alone does not always imply A = ∅, the equality with A ∩ B is essential here, and option A is the only necessary conclusion.

Every element of B belongs to A ∪ B, so the left side contains all elements of B. However, A \ B is defined as the set of elements in A that are not in B; it contains no element of B. If the two sides are equal, B can have no elements at all. Hence B = ∅. The other options are not forced by the given equality.

The union A ∪ B is divided into three pairwise disjoint regions: elements only in A, counted by |A \ B|; elements only in B, counted by |B \ A|; and common elements, counted by |A ∩ B|. Hence 70 = 25 + 30 + |A ∩ B|. Solving gives |A ∩ B| = 70 − 55 = 15, so option A is correct.

Every element of A is either outside B or inside B. Therefore A can be partitioned as A = (A \ B) ∪ (A ∩ B). Both parts are given to be subsets of C, and the union of subsets of C is also a subset of C. Consequently, A ⊆ C. No condition controls elements in B \ A, so B ⊆ C and A ∪ B ⊆ C do not necessarily follow.

An element belongs to the left side exactly when it is in A but not in B, and also in C. Thus it is simultaneously in A and C and outside B. The right side describes precisely the same condition: it is in A and in C \ B, meaning it belongs to C but not to B. Since both sides contain exactly the same elements, the identity is true for all sets A, B, and C.

The difference A \ B contains all elements that belong to A but do not belong to B. Set A includes every real number from -2 through 5, including both endpoints. Set B contains numbers greater than 1 and up to 7, but it does not include 1. Therefore, the part of A removed is (1, 5], while -2 through 1 remains. Hence A \ B = [-2, 1]. The closed endpoint at 1 is important because 1 is not in B.

The elements of A are the multiples of 6 from 1 to 90, so |A| = floor(90/6) = 15. The elements of B are the multiples of 15, so |B| = floor(90/15) = 6. Elements in both sets are multiples of lcm(6, 15) = 30; there are floor(90/30) = 3 such elements. By inclusion-exclusion, |A ∪ B| = |A| + |B| − |A ∩ B| = 15 + 6 − 3 = 18.

To find the number of students in at least one of the two sets, use the inclusion-exclusion formula: |M ∪ P| = |M| + |P| − |M ∩ P|. Thus |M ∪ P| = 72 + 64 − 28 = 108. The remaining students belong to neither set, so subtract the union from the total: 120 − 108 = 12. Therefore, 12 students are in neither mathematics nor physics.

There are floor(84/7) = 12 multiples of 7 in U, so |A| = 12. There are floor(84/12) = 7 multiples of 12, so |B| = 7. A number belongs to both sets when it is a multiple of lcm(7, 12). Since 7 and 12 are relatively prime, their least common multiple is 84, giving one common element. Therefore, |A ∪ B| = 12 + 7 − 1 = 18.

To find A \ B, retain elements of A that are not in B. Set B begins just greater than −1, so −1 itself is not in B and remains in the difference. Every number greater than −1 and below 6 belongs to B and must be removed. The left endpoint −4 is included in A, giving A \ B = [-4, −1].

The union A ∪ B is divided into three mutually disjoint regions: elements that are only in A, elements that are only in B, and elements common to both sets. These regions have sizes |A \ B| = 17, |B \ A| = 23, and |A ∩ B| = 14. Since they do not overlap with one another, add them directly: |A ∪ B| = 17 + 23 + 14 = 54. Thus option A is correct.

First form the union A ∪ B by listing every element that occurs in either A or B, without repeating any element. This gives {2, 3, 4, 6, 8, 9, 10, 12, 15}. Now intersect this set with C = {6, 12, 18}; intersection keeps only elements common to both sets. The common elements are 6 and 12, while 18 is absent from A ∪ B. Therefore, (A ∪ B) ∩ C = {6, 12}.

The set A \ B contains elements of A that are not in B. Since |A| = 41 and |A \ B| = 19, the elements common to A and B number |A ∩ B| = 41 − 19 = 22. Apply the inclusion–exclusion formula: |A ∪ B| = |A| + |B| − |A ∩ B|. Hence |A ∪ B| = 41 + 36 − 22 = 55. Subtracting the intersection prevents the common elements from being counted twice.

To prove the first inclusion, take any element x in A ∪ C. Then x belongs either to A or to C. If x ∈ A, the condition A ⊆ B gives x ∈ B, so x ∈ B ∪ D. If x ∈ C, the condition C ⊆ D gives x ∈ D, so again x ∈ B ∪ D. Thus every element of A ∪ C is in B ∪ D, proving A ∪ C ⊆ B ∪ D. The reverse and difference/intersection statements are not guaranteed.

The equality A ∪ B = A means that every element of B is already in A, so B ⊆ A. The condition A \ B = ∅ means that no element of A lies outside B; therefore A ⊆ B. Since each set is a subset of the other, the two sets are equal. Hence A = B, and no claim that either set is empty is justified.

The positive perfect squares not exceeding 40 are 1, 4, 9, 16, 25, and 36. Among these, the even numbers are 4, 16, and 36. The intersection requires both conditions simultaneously: being a perfect square and being even. Therefore A ∩ B = {4, 16, 36}. Option C contains odd squares as well and is therefore only A, not the intersection.

For any sets X and Y, X \ Y = ∅ exactly when every element of X belongs to Y, which is equivalent to X ⊆ Y. Applying this fact with X = A and Y = B ∩ C, the given condition implies A ⊆ B ∩ C. It does not require B ∩ C to be contained in A, nor does it imply disjointness or the stated union equality.

First calculate B ∩ C. The elements common to B = {2, 4, 6, 8} and C = {1, 4, 6, 9} are 4 and 6, so B ∩ C = {4, 6}. Set difference A \ (B ∩ C) keeps the elements of A that are not in {4, 6}. Removing 4 and 6 from A leaves {1, 2, 3, 5}. Therefore option A is correct.

The set A \ B contains elements of A that are outside B, whereas A ∩ B contains elements of A that are inside B. An element cannot simultaneously be outside B and inside B. Therefore (A \ B) ∩ (A ∩ B) = ∅, so the two parts are disjoint. Their union is A because every element of A is either in B or not in B.

The union A ∪ B consists of the elements in A together with the elements of B that are outside A. Therefore, |A ∪ B| = |A| + |B \ A|. Substituting the given values gives 65 = 39 + |B \ A|, so |B \ A| = 26. The intersection value 18 is consistent but is not needed for this direct calculation.

Solve the first equation: x² − 9 = (x − 3)(x + 3) = 0, so A = {−3, 3}. The second equation is x² − 6x + 9 = (x − 3)² = 0, so B = {3}. A union contains every distinct element appearing in either set; because 3 is repeated, it is written only once. Hence A ∪ B = {−3, 3}.

Because A ⊆ B, every element of A is also an element of B. However, C \ B contains only those elements of C that are not in B. Therefore no element can belong to both A and C \ B. Their intersection is consequently the empty set, ∅, regardless of the particular sets C and B.

Set A consists of the even numbers from 1 through 50, so it has 50 ÷ 2 = 25 elements. Elements in both A and B must be divisible by both 2 and 5, hence divisible by 10. There are 50 ÷ 10 = 5 such elements. Therefore, A \ B contains the even numbers that are not divisible by 5, and its size is 25 − 5 = 20.

First calculate the intersection A ∩ C. The elements common to A and C are a and d, so A ∩ C = {a,d}. Next calculate B \ A, which contains elements present in B but absent from A. Since b and d are in A, they are removed, leaving {f,g}. Taking the union gives {a,d} ∪ {f,g} = {a,d,f,g}. Therefore, option A is correct.