Complement of a Set and Its Properties

A number divisible by both 2 and 3 must be divisible by their least common multiple, 6. Between 1 and 30, the multiples of 6 are 6, 12, 18, 24, and 30, so n(A) = 5. The universal set has 30 elements, and Aᶜ contains all remaining elements. Therefore n(Aᶜ) = 30 − 5 = 25, so option B is correct.

The prime numbers in U are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29, so |A| = 10. The multiples of 3 from 1 to 30 are 3, 6, 9, 12, 15, 18, 21, 24, 27 and 30, so |B| = 10. Their only common element is 3, hence |A ∩ B| = 1. By inclusion-exclusion, |A ∪ B| = 10 + 10 − 1 = 19. Therefore, |(A ∪ B)'| = |U| − |A ∪ B| = 30 − 19 = 11. Thus, option A is correct.

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.

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.

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 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}.

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.

Set A contains all real numbers less than 2 together with all real numbers at least 8. Its complement must contain numbers satisfying neither condition: they are not less than 2 and are less than 8. Thus \(2\le x<8\), represented by \([2,8)\). The endpoint 2 is included because it is excluded from A, while 8 is excluded because it belongs to A.

The inequality \(x^2<16\) is equivalent to \(|x|<4\), which means \(-4<x<4\). Therefore, \(A=(-4,4)\). Since the universal set is all real numbers, the complement consists of the two outside intervals, including the boundary points where equality holds: \(x\le-4\) or \(x\ge4\). Hence \(A'=(-\infty,-4]\cup[4,\infty)\), option A.

The universal set contains the integers from 1 through 25. Since A is the set of odd numbers, its complement A' within U is the set of even numbers. The perfect squares in U are {1, 4, 9, 16, 25}. Among them, the even squares are 4 and 16. Therefore, A' ∩ B = {4, 16}, and its cardinality is 2.

The condition (A ∪ B)' = ∅ means that A ∪ B = U, because the only set whose complement is empty is the universal set itself. The condition A ∩ B = ∅ says that A and B are disjoint. Thus A and B partition U, so every element outside A must belong to B. Hence B = A'.

The complement A' consists of elements in U that are not in A. The multiples of 3 from 1 to 20 are 3, 6, 9, 12, 15 and 18, so |A| = 6. Since U has 20 elements, the complement-count rule gives |A'| = |U| − |A| = 20 − 6 = 14. Hence option B is correct. The other numerical choices result from miscounting the multiples or subtracting incorrectly.

A complement must be determined relative to the stated universal set, not relative to all integers. The set U contains the integers −3 through 6. Within U, the condition x ≥ 2 gives A = {2, 3, 4, 5, 6}. Therefore the elements left outside A are −3, −2, −1, 0 and 1. Thus A' = {−3, −2, −1, 0, 1}, which is option A. Option B is A itself, while C and D omit or include boundary elements incorrectly.

Use inclusion–exclusion because numbers divisible by both 2 and 3 are counted in both groups. There are 30/2 = 15 multiples of 2 and 30/3 = 10 multiples of 3. Their common multiples are multiples of 6, numbering 30/6 = 5. Hence |A| = 15 + 10 − 5 = 20, and |A'| = 30 − 20 = 10. Therefore option C is correct; adding 15 and 10 without subtracting the overlap would give the wrong count.

The set difference B − A means the elements that are in B but not in A. By definition, “not in A” is represented by the complement A' relative to U. Intersecting B with A' keeps exactly those members of B outside A, so B − A = B ∩ A'. For these sets, the actual elements are {6, 10, 12, 14}, which confirms the identity. The union or the other intersections select different regions.

The universal set has |U| = 100 elements. The perfect squares in this range are 1², 2², 3², ..., 10², ending at 100, so A contains exactly 10 elements. The complement contains every member of U that is not a perfect square. Therefore |A'| = |U| − |A| = 100 − 10 = 90. Option B is correct. Counting 0 or a square above 100 would be inappropriate because neither belongs to the stated universal set.

First find the intersection of A and B. The only common element is 4, so A ∩ B = {4} and |A ∩ B| = 1. The complement is taken with respect to U, which contains 15 elements. Therefore, |(A ∩ B)′| = |U| − |A ∩ B| = 15 − 1 = 14. Hence, option B is correct.

A ⊆ B means that every element of A is also an element of B. Consider any element x in B′. Since x is not in B, it cannot be in A either; otherwise A ⊆ B would force x to be in B. Therefore x belongs to A′, proving B′ ⊆ A′. Taking complements reverses the direction of inclusion, so option A is correct.