Complement of a Set and Its Properties

There are floor(60/4) = 15 multiples of 4 and floor(60/9) = 6 multiples of 9. Numbers in both sets must be multiples of lcm(4, 9) = 36; only 36 occurs up to 60. Thus n(A ∪ B) = 15 + 6 − 1 = 20 by inclusion–exclusion. The complement contains the remaining 60 − 20 = 40 elements.

The first part of A contains all real numbers up to and including −4, so the complement begins just after −4; therefore −4 is excluded. The second part contains numbers strictly between 2 and 6, so both 2 and 6 are excluded from A and included in its complement. Hence A′ = (−4, 2] ∪ [6, ∞).

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.

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.

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.

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.

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.

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.

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′.

The only common element of A and B is 5, so A ∩ B = {5}. The complement is taken relative to U, which contains all integers from 1 through 15. Removing 5 from U gives (A ∩ B)′ = {1,2,3,4,6,7,8,9,10,11,12,13,14,15}. Thus option A is correct; option C represents A ∪ B.

Because the universal set is the real numbers, the complement of A contains all real numbers that are not in (-∞,0) or (4,∞). The endpoints 0 and 4 are excluded from A because both defining intervals are open there. Therefore A′ = [0,4]. Since [0,4] is wholly contained in B = [-2,6], their intersection is [0,4].

By De Morgan’s law, A′ ∩ B′ = (A ∪ B)′. In U = {1, 2, ..., 54}, the number of multiples of 9 is ⌊54/9⌋ = 6, and the number of multiples of 6 is ⌊54/6⌋ = 9. Numbers counted in both groups are multiples of lcm(9, 6) = 18; there are ⌊54/18⌋ = 3. Hence n(A ∪ B) = 6 + 9 − 3 = 12, and n(A′ ∩ B′) = 54 − 12 = 42.

The set A contains every real number less than or equal to −1 and every real number greater than 3. Its complement therefore consists of real numbers greater than −1 and less than or equal to 3. The point −1 is excluded because it belongs to A, while 3 is included because the condition x > 3 excludes it from A. Hence A′ = (-1,3].

If A ⊆ B, every element of A is also in B. Consider an element of B′: it is not in B. Since every element of A would have to be in B, that element cannot be in A either; therefore it belongs to A′. Thus every element of B′ is in A′, giving B′ ⊆ A′. Taking complements reverses the direction of inclusion.

The universal set U has 14 elements. Set A contains the seven even numbers from 2 through 14, so its complement A′ contains the seven odd numbers {1,3,5,7,9,11,13}. The Cartesian product A′×A consists of ordered pairs whose first element is chosen from A′ and second from A. Therefore, n(A′×A)=n(A′)×n(A)=7×7=49. Hence, option A is correct.

With respect to U, A′ = {4, 5, 6, 7, 8, 9, 10}. Intersecting this set with B = {3, 4, 5} gives A′ ∩ B = {4, 5}. Now take the complement of {4, 5} in U. The remaining elements are {1, 2, 3, 6, 7, 8, 9, 10}. Thus option A is correct; option B is only the intermediate intersection, not its complement.

Complements are taken in the real-number universal set. Because A includes 1 and all larger numbers, A′=(−∞,1). Because B contains numbers strictly less than 4 but not 4 itself, B′=[4,∞). Their union is therefore (−∞,1)∪[4,∞). Endpoint brackets are essential: 1 is excluded from A′, while 4 is included in B′.

The numbers from 1 to 81 divisible by 9 are 9, 18, 27, 36, 45, 54, 63, 72, and 81, so there are 9 such numbers. Among them, 9 = 3², 36 = 6², and 81 = 9² are perfect squares and therefore belong to A. Removing these three from the nine multiples leaves 9 − 3 = 6 numbers in A′.