Complement of a Set and Its Properties

By De Morgan’s law, A′ ∪ B′ = (A ∩ B)′. An element belongs to A ∩ B when it is divisible by both 3 and 5, so it must be divisible by 15. The multiples of 15 in U are 15, 30, and 45, giving n(A ∩ B) = 3. Since U has 45 elements, n((A ∩ B)′) = 45 − 3 = 42. Therefore, option A is correct.

The union A ∪ B consists of the interval [−5, 1) together with (3, 8]. Its complement in the real numbers contains values less than −5, the complete interval from 1 to 3, and values greater than 8. The endpoint signs are important: −5 and 8 are included in the original union, so they are excluded from the complement; 1 and 3 are excluded from the original union, so they are included in the complement. Thus option A is correct.

The odd elements of U are B = {1, 3, 5, 7, 9, 11, 13, 15}. The square numbers in U are {1, 4, 9, 16}; among these, the odd square numbers are 1 and 9. A′ contains numbers that are not squares, so intersecting A′ with B removes 1 and 9 from the odd set. The result is {3, 5, 7, 11, 13, 15}, which is option A.

For every set A in a universal set U, A and its complement A′ are disjoint, so A ∩ A′ = ∅. If A = A′, then this would imply A ∩ A = ∅, hence A = ∅. But the complement of the empty set is U, so A′ = U, and A = A′ would require ∅ = U, which is impossible for a nonempty universal set. Therefore no such set exists, and option D is correct.

The prime numbers from 1 through 25 are 2, 3, 5, 7, 11, 13, 17, 19, and 23, so there are 9 primes. Set A contains the multiples of 5, including 5 itself. Among the primes, only 5 belongs to A; every other listed prime belongs to A′. Therefore the number of primes in A′ is 9 − 1 = 8. Hence option C, not option A, is correct.

A contains the 35 even numbers from 1 to 70. The set A − B contains even numbers that are not divisible by 7. An even number divisible by 7 must be divisible by 14; the multiples of 14 in U are 14, 28, 42, 56, and 70, so there are 5 such numbers. Therefore n(A − B) = 35 − 5 = 30. Its complement in a 70-element universal set has 70 − 30 = 40 elements. Option A is correct.

Factor the quadratic as x² − 6x + 8 = (x − 2)(x − 4). Since the parabola opens upward, the expression is positive outside its roots: A = (−∞, 2) ∪ (4, ∞). The points 2 and 4 are not in A because the expression equals zero there. Consequently, the complement in ℝ includes both endpoints and the interval between them, giving A′ = [2, 4]. Therefore option A is correct.

Apply De Morgan’s law: A′ ∪ B′ = (A ∩ B)′. The elements common to A and B are 6, 12, and 18, so A ∩ B = {6, 12, 18}. Taking the complement relative to U means removing these three elements from {1, ..., 18}. The remaining elements are {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17}. Hence option A is correct; option C would represent A′ ∩ B′ instead.

A′ ∩ B′ consists of elements that are neither even nor square. Thus they must be odd, non-square natural numbers not exceeding 40. There are 20 odd numbers from 1 through 40. The odd square numbers in this range are 1, 9, and 25; these must be excluded. Therefore n(A′ ∩ B′) = 20 − 3 = 17, so option A is correct. The square 49 is outside U and is not considered.

The complement A' contains the elements of U that are not in A. Therefore, A = U − A' = {p, r, t, v}. Set B is {p, q, t, v}. The elements common to A and B are p, t, and v, so A ∩ B = {p, t, v}. Element q belongs to B but not to A, so it must not be included in the intersection. Thus, option A is correct.

A contains all real numbers greater than −3, while B contains all real numbers less than or equal to 5. Hence, A ∩ B = (−3, 5]. Taking the complement relative to ℝ gives all real numbers at most −3 or greater than 5. Therefore, (A ∩ B)' = (−∞, −3] ∪ (5, ∞). The endpoint −3 is included because it was excluded from the intersection, whereas 5 is excluded because it belonged to the intersection.

An element belongs to A ∩ B only when it is divisible by both 6 and 10. Such numbers are multiples of lcm(6, 10) = 30. In U = {1, 2, …, 90}, the multiples of 30 are 30, 60, and 90, so n(A ∩ B) = 3. Since U has 90 elements, the complement has 90 − 3 = 87 elements. Therefore, option A is correct.

The complement A' with respect to U contains every element of U except 1, 4, 7, and 10. The second set is {2, 4, 6, 8, 10, 12}. Removing the elements 4 and 10, which are not in A', leaves 2, 6, 8, and 12. Hence A' ∩ {2, 4, 6, 8, 10, 12} = {2, 6, 8, 12}, so option A is correct.

If A ⊆ B, every element of A is also an element of B. Consider any element x in B'. It is not in B. Since every element of A must lie in B, x cannot be in A either; therefore x belongs to A'. This proves B' ⊆ A'. Complementation reverses the direction of subset inclusion, so option B is correct. Equality is possible only in the special case A = B, not in general.

Factor the equation as x² − 9x + 20 = (x − 4)(x − 5) = 0. Thus, A = {4, 5}. The elements of U that are less than 5 are {0, 1, 2, 3, 4}. Because 4 belongs to A, it is excluded from A'. Therefore, the elements of A' less than 5 are {0, 1, 2, 3}, which has 4 elements. Hence, option A is correct.

The elements of A are the positive multiples of 8 not exceeding 100. Their number is floor(100/8) = 12, since the multiples run from 8 through 96. The universal set U contains 100 elements. The complement A' therefore contains all elements of U that are not divisible by 8, so n(A') = 100 − 12 = 88. Thus, option A is correct.

Solve the absolute-value inequality: |x − 2| < 5 means −5 < x − 2 < 5. Adding 2 gives −3 < x < 7, so A = (−3, 7). Since the complement is taken in U = ℝ, it contains every real number outside this open interval. The boundary points −3 and 7 are included in the complement, giving A' = (−∞, −3] ∪ [7, ∞). Therefore, option A is correct.

Within the universal set U, the numbers that are not even are precisely the odd numbers, so A is the set of odd numbers from 1 to 50. The complement A' consists of all elements of U that are not in A. These are exactly the even numbers: {2, 4, 6, …, 50}. Therefore, A' is the set of even numbers in U, making option A correct.

A' ∩ B consists of numbers that are divisible by 3 but not divisible by 2. From 1 to 36, there are floor(36/3) = 12 multiples of 3. Among them, the even ones are multiples of 6, and there are floor(36/6) = 6 such numbers. Removing these 6 even multiples from the 12 multiples of 3 leaves 12 − 6 = 6 numbers. Hence, option A is correct.

The condition A ∪ B = U says that together the two sets contain every element of the universal set. The condition A ∩ B = ∅ says that they have no common element. Therefore, every element outside A must belong to B, and every element outside B must belong to A. Hence B is exactly the complement of A, so B = A′.