Complement of a Set and Its Properties

Apply the distributive law \((X\cup Z)\cap(Y\cup Z)=Z\cup(X\cap Y)\), with \(X=A\), \(Y=A'\), and \(Z=B'\). The expression becomes \(B'\cup(A\cap A')\). A set and its complement are disjoint, so \(A\cap A'=\varnothing\). Therefore the result is \(B'\cup\varnothing=B'\).

Factor the quadratic: x² + x − 6 = (x + 3)(x − 2). Since the parabola opens upward, the expression is nonnegative outside the roots, so A = (−∞, −3] ∪ [2, ∞). The complement in R consists of the numbers strictly between the roots. The endpoints −3 and 2 belong to A because equality is allowed, so they are excluded from A'. Therefore A' = (−3, 2), option A.

For any set, \(A\cap A'=\varnothing\). If \(A=A'\), then this says \(A\cap A=\varnothing\), so \(A=\varnothing\). Also, \(A\cup A'=U\); replacing \(A'\) by A gives \(A\cup A=A=U\). Thus A must equal both the empty set and U, which is possible only when \(U=\varnothing\).

The inequality \(x^2\ge 9\) is equivalent to \(|x|\ge 3\), so \(A=(-\infty,-3]\cup[3,\infty)\). Its complement in \(\mathbb{R}\) consists of all real numbers that do not satisfy \(|x|\ge3\), namely those satisfying \(|x|<3\). Hence \(-3<x<3\), and \(A'=(-3,3)\). The endpoints are excluded because \((-3)^2=3^2=9\), so both belong to A. Therefore option A is correct.

A union B = U means every element of U is in A, in B, or in both. If an element is in A', it is not in A. To still belong to U = A union B, it must therefore be in B. Hence every element of A' belongs to B, so A' is a subset of B. Equality is not guaranteed because B may contain elements that are also in A.

Assuming ℕ begins at 1, U has 48 elements. There are 12 multiples of 4 and 8 multiples of 6 up to 48. Their common elements are multiples of lcm(4,6) = 12, namely 4 numbers: 12, 24, 36, and 48. Hence n(A ∪ B) = 12 + 8 − 4 = 16. Therefore n((A ∪ B)′) = 48 − 16 = 32.

First solve the inequality (x − 1)(x + 2) ≥ 0. Its critical values are −2 and 1, so the product is non-negative when x ≤ −2 or x ≥ 1. Restricting to U gives A = {−4, −3, −2, 1, 2, 3, 4, 5, 6}. Therefore, the elements of U not in A are −1 and 0. Hence A′ = {−1, 0}.

De Morgan’s law states that (A∩B)′=A′∪B′. The inclusion–exclusion formula gives n(A′∪B′)=n(A′)+n(B′)−n(A′∩B′). Substituting the given values, we obtain 76+89−31=134. The intersection must be subtracted once because it is counted in both sets. Therefore, option A is correct.

There are 13 even numbers from 1 through 27: 2, 4, 6, ..., 26, since floor(27/2) = 13. The even numbers that are divisible by 3 are 6, 12, 18, and 24; these belong to A and therefore are excluded from A′. Consequently, the number of even elements in A′ is 13 − 4 = 9. Option A is correct. Option B counts all even numbers without removing those in A, while option C counts only the excluded even multiples of 3.

By De Morgan’s law, A′ ∩ B′ = (A ∪ B)′. The set A contains the multiples of 6 up to 84, so n(A) = 84/6 = 14. The set B contains the multiples of 14, so n(B) = 84/14 = 6. Their common elements are multiples of lcm(6,14) = 42, giving n(A ∩ B) = 84/42 = 2. Therefore n(A ∪ B) = 14 + 6 − 2 = 18, and n(A′ ∩ B′) = 84 − 18 = 66.

Using De Morgan’s law, A′ ∩ B′ = (A ∪ B)′. There are 96/8 = 12 multiples of 8 and 96/12 = 8 multiples of 12 in U. Numbers counted in both sets are multiples of lcm(8,12) = 24; there are 96/24 = 4 of them. Thus n(A ∪ B) = 12 + 8 − 4 = 16. Since U has 96 elements, n(A′ ∩ B′) = 96 − 16 = 80.

Factor the quadratic inequality: x² − 9x + 18 = (x − 3)(x − 6). Since the parabola opens upward, the inequality is non-positive for 3 ≤ x ≤ 6. The integer elements of A are therefore 3, 4, 5, and 6, so n(A) = 4. The universal set contains all integers from −12 through 12, giving 25 elements. Hence n(A′) = 25 − 4 = 21, so option D is correct.

To find the complement in ℝ, reverse the inclusion of each interval and include the boundary points that were excluded from A. The point −5 is included in the complement, while 1 is excluded because it belongs to A. Similarly, 4 is excluded and 9 is included. Therefore A′ = [−5, 1) ∪ (4, 9].

The set A contains the multiples of 3 up to 42, so n(A) = 42/3 = 14. The elements removed in A − B are those that are also multiples of 7, namely the common multiples of 3 and 7. These are multiples of lcm(3,7) = 21; there are 42/21 = 2 such elements, 21 and 42. Therefore n(A − B) = 14 − 2 = 12. Since U has 42 elements, n((A − B)′) = 42 − 12 = 30. Thus option A is correct.

The inequality |x + 1| ≥ 4 means x + 1 ≤ -4 or x + 1 ≥ 4. Therefore A = (-∞, -5] ∪ [3, ∞). Since the universal set is R, its complement consists of the real numbers strictly between -5 and 3: A' = (-5, 3). The endpoints are excluded because they belong to A.

B and B' are disjoint because no element can simultaneously belong to a set and its complement. Consequently, A ∩ B and A ∩ B' are also disjoint. The question states that these two disjoint sets are equal. The only set equal to two disjoint copies of itself is the empty set, so A ∩ B = A ∩ B' = ∅, which forces A = ∅.

Use inclusion–exclusion. There are 18 multiples of 4, 12 of 6, and 8 of 9. Pairwise intersections have sizes 6, 2, and 4 because the relevant LCMs are 12, 36, and 18. The triple intersection has size 2, from multiples of 36. Thus n(A ∪ B ∪ C) = 18 + 12 + 8 - 6 - 2 - 4 + 2 = 28. Therefore the complement has 72 - 28 = 44 elements.

Using De Morgan's law, A' ∪ B' = (A ∩ B)'. The intersection of A = (-2, 6] and B = [0, 9) is [0, 6]. Taking its complement in ℝ excludes every number from 0 through 6, including both endpoints. Therefore the result is (-∞, 0) ∪ (6, ∞). The open endpoints are essential.

The integers from -5 through 15 inclusive number 15 - (-5) + 1 = 21, so n(U) = 21. The members congruent to 2 modulo 5 in this interval are -3, 2, 7, and 12, giving n(A) = 4. Since A' is the complement within U, n(A') = n(U) - n(A) = 21 - 4 = 17. Therefore option A is correct.

First use inclusion–exclusion for the complements: n(A′ ∩ B′) = n(A′) + n(B′) − n(A′ ∪ B′) = 120 + 150 − 210 = 60. De Morgan’s law gives (A ∪ B)′ = A′ ∩ B′, so the complement of A ∪ B has 60 elements. Therefore n(A ∪ B) = n(U) − n((A ∪ B)′) = 300 − 60 = 240. Option A is correct; 210 is the given union of complements, not the requested union.

Since \(A=(-\infty,2]\), its complement in \(\mathbb{R}\) is \(A'=(2,\infty)\); the endpoint 2 is excluded because it belongs to A. Since \(B=[-1,\infty)\), its complement is \(B'=(-\infty,-1)\); the endpoint −1 is excluded because it belongs to B. No real number can be both greater than 2 and less than −1, so \(A'\cap B'=\varnothing\).

The set A contains all real numbers less than 3 and all real numbers greater than or equal to 8. Its complement must therefore contain the real numbers that are not less than 3 and are not at least 8. These conditions are \(x\ge3\) and \(x<8\), giving \(A'=[3,8)\). The endpoint 3 is included, while 8 is excluded, so option A is correct.

By De Morgan's law, \(A'\cap B'=(A\cup B)'\), so we count numbers from 1 to 60 divisible by neither 2 nor 3. There are 30 multiples of 2, 20 multiples of 3, and 10 multiples of both 2 and 3. Thus \(n(A\cup B)=30+20-10=40\). Therefore, \(n(A'\cap B')=60-40=20\), so option A is correct.

Within U, the condition \(x+1\ge4\) gives \(x\ge3\), so \(A=\{3,4,5,6,7,8\}\). Also, \(B=\{-3,-2,-1,0,1,2,3,4,5\}\). Therefore, \(A'\) contains \(-3,-2,-1,0,1,2\), while \(B'\) contains \(6,7,8\). Their union is \(\{-3,-2,-1,0,1,2,6,7,8\}\), option A.

The universal set U has 32 elements. Exactly 16 of them are even, so A contains 16 elements. The complement A' therefore contains the 16 odd numbers from 1 to 32. For any finite sets X and Y, n(X × Y) = n(X)n(Y). Hence n(A' × A') = 16 × 16 = 256. The order matters in a Cartesian product, and every first component can be paired with every second component.