Complement of a Set and Its Properties

Because x is an integer and x² ≤ 9, we have −3 ≤ x ≤ 3. Therefore, A = {−3, −2, −1, 0, 1, 2, 3}. The universal set is U = {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}. The complement A' consists of every element of U that is not in A. Removing the seven elements of A leaves {−5, −4, 4, 5}. Hence, option A is correct. The complement must always be taken with respect to the stated universal set U.

Taking complements reverses the direction of set inclusion. Since every element of A is in B, any element that is not in B is certainly not in A. Therefore every element of Bᶜ belongs to Aᶜ, giving Bᶜ ⊆ Aᶜ. Equality is not guaranteed unless A and B are equal, so option B is the only always-true statement.

Apply De Morgan’s law to the intersection: (X ∩ Y)' = X' ∪ Y'. Here, X = A and Y = B'. Thus, (A ∩ B')' = A' ∪ (B')'. The complement of a complement returns the original set, so (B')' = B. Therefore, (A ∩ B')' = A' ∪ B, which is option A. This result also agrees with the element-wise interpretation: an element is outside A ∩ B' whenever it is outside A or it belongs to B.

Aᶜ consists of numbers that are not even, and Bᶜ consists of numbers that are not multiples of 5. Their intersection therefore contains numbers satisfying both negative conditions: neither even nor a multiple of 5. In U these are {1,3,7,9,11,13,17,19}. This also follows from De Morgan’s law, Aᶜ ∩ Bᶜ = (A ∪ B)ᶜ, so option B is correct.

Use the inclusion-exclusion formula: |A ∪ B| = |A| + |B| − |A ∩ B|. Substituting the given values gives |A ∪ B| = 35 + 40 − 15 = 60. The complement contains all elements of U outside this union, so |(A ∪ B)ᶜ| = |U| − |A ∪ B| = 80 − 60 = 20. Therefore option A is correct.

Since A' contains 58 elements, A contains 100 − 58 = 42 elements. Similarly, B contains 100 − 47 = 53 elements. By inclusion–exclusion, |A ∪ B| = |A| + |B| − |A ∩ B| = 42 + 53 − 21 = 74. Therefore, |(A ∪ B)'| = 100 − 74 = 26. Thus, option C is correct.

Use the given identity A − B = A ∩ B'. Taking complements on both sides gives (A − B)' = (A ∩ B')'. Applying De Morgan's law, (X ∩ Y)' = X' ∪ Y', we obtain (A ∩ B')' = A' ∪ (B')' = A' ∪ B. Therefore, option A is correct. The key step is to rewrite set difference before taking the complement.

The complement contains all real numbers that are not in A = (2, 7]. The number 2 is excluded from A because the left endpoint is open, so 2 belongs to A′. The number 7 is included in A because the right endpoint is closed, so 7 does not belong to A′. Thus A′ = (−∞, 2] ∪ (7, ∞), making option A correct.

The intersection consists of numbers that are at least 1 and less than 3, so A ∩ B = [1, 3). Its complement in R contains numbers less than 1 and numbers greater than or equal to 3. Hence, (A ∩ B)' = (−∞, 1) ∪ [3, ∞), making option A correct. Notice that 1 belongs to the intersection, while 3 does not.

By De Morgan's law, A' ∩ B' = (A ∪ B)'. The condition says that the complement of A ∪ B is empty. A set has an empty complement in U only when it contains every element of U. Therefore, A ∪ B = U, so option B is correct. Equivalently, every element of the universal set belongs to A or to B.

De Morgan’s law always gives (A ∩ B)′ = A′ ∪ B′. The stated equality therefore requires A′ ∪ B′ = A′ ∩ B′. For two sets, a union equals their intersection exactly when the two sets are equal, so A′ = B′. Taking complements on both sides gives A = B. Consequently, A ∪ B = A ∩ B, which is option A.

There are 25 multiples of 2 from 1 to 50 and 10 multiples of 5. The numbers counted in both sets are multiples of 10, of which there are 5. Thus |A ∪ B| = 25 + 10 − 5 = 30. By De Morgan's law, A' ∩ B' = (A ∪ B)', so |A' ∩ B'| = 50 − 30 = 20. Hence, option A is correct.

The governing idea is that a complement is taken relative to the universal set U. Since A' = {b, d, f, h}, the elements of A are the remaining members: A = {a, c, e, g}. Also, B' = U − B = {e, f, g, h}. The intersection contains elements common to both sets, namely e and g. Thus A ∩ B' = {e, g}; option A is correct. Options B, C and D contain elements excluded from one of the required sets.

A set and its complement have two defining properties: their union is the universal set, and their intersection is empty. The equations A ∪ A′ = U and A ∩ A′ = ∅ state exactly these properties. Thus A′ contains every element of U that is not in A, so A′ is the complement of A. Therefore, option C is correct.

A number in A ∩ B must be divisible by both 4 and 6. Such numbers are multiples of lcm(4, 6) = 12. In the set {1, 2, …, 40}, these are 12, 24, and 36, so |A ∩ B| = 3. Therefore, the complement contains 40 − 3 = 37 elements. Hence, option B is correct.

Factor the quadratic as \(x^2-5x+6=(x-2)(x-3)\). Therefore, the solutions are \(x=2\) and \(x=3\), so \(A=\{2,3\}\). Since the universal set is all real numbers, the complement contains every real number except 2 and 3. Hence \(A'=\mathbb{R}\setminus\{2,3\}\), making option A correct.

All complements are taken with respect to the universal set U. By De Morgan’s law, (A′ ∪ B′)′ = A ∩ B. The intersection contains exactly those elements common to both A and B. Comparing the two sets, the common elements are 3, 5, 7, and 11. Therefore, the correct answer is option A: {3, 5, 7, 11}.

The complement operation is reversible because the complement of a complement is the original set. Starting with \(A'=B'\), take complements on both sides: \((A')'=(B')'\). Thus \(A=B\). The other statements do not necessarily follow from equality of the complements, so option C is correct.

There are 20 multiples of 3, 15 of 4, and 12 of 5 in the universal set. Their pairwise overlaps contain 5 multiples of 12, 4 of 15, and 3 of 20; the triple overlap contains 1 multiple of 60. Inclusion-exclusion gives \(|A\cup B\cup C|=20+15+12-5-4-3+1=36\). Therefore, the complement has \(60-36=24\) elements, so C is correct.

Let the mathematics and physics sets be M and P. By inclusion-exclusion, \(|M\cup P|=|M|+|P|-|M\cap P|=70+65-30=105\). Students studying neither subject lie outside this union, so their number is \(|U|-|M\cup P|=120-105=15\). Thus option B is correct.