Complement of a Set and Its Properties

All complements are taken relative to \(U\). Thus \(A'=U\setminus A=\{5,6,7,8\}\) and \(B'=U\setminus B=\{1,2,7,8\}\). Their union contains every element appearing in either complement: \(A'\cup B'=\{1,2,5,6,7,8\}\). This also agrees with De Morgan’s law, \(A'\cup B'=(A\cap B)'\), since \(A\cap B=\{3,4\}\). Therefore option A is correct.

By De Morgan’s law, A ∩ B = (A′ ∪ B′)′. The union of the given complements is A′ ∪ B′ = {1,2,4,7,8}. Taking its complement in U leaves the elements of U not in this union: {3,5,6}. Therefore A ∩ B = {3,5,6}. Option D is only A, not the required intersection.

The distinct prime factors of 18 are 2 and 3, so A = {2, 3} and n(A) = 2. The universal set U contains 18 elements, and therefore the complement A' contains 18 − 2 = 16 elements. For any finite set with k elements, its power set has 2^k elements. Thus n(P(A')) = 2^16 = 65,536, making option A correct.

By De Morgan’s law, Aᶜ∪Bᶜ=(A∩B)ᶜ. The complement of A∩B in the universal set must therefore contain 210−52=158 elements. However, the question states that n(Aᶜ∪Bᶜ)=170. Because 158 and 170 are different, the two supplied values cannot describe the same sets; hence the data are inconsistent.

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.

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.

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 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.

For any two subsets of the same universal set, taking complements does not change which elements belong to exactly one of the sets. Algebraically, \(A'\Delta B'=(A'\cap B)\cup(B'\cap A)=A\Delta B\). Here \(A\Delta B=\{1,2,5,6\}\), since these elements occur in exactly one of A and B. Therefore, A is correct.

The sets A and A' partition the universal set: every element belongs to exactly one of them. Consequently, \(B=(A\cap B)\cup(A'\cap B)\), and similarly \(C=(A\cap C)\cup(A'\cap C)\). The two given equalities make the corresponding parts equal, so their unions are equal. Therefore, \(B=C\), and option A is correct.

The interval A begins at -2 and includes -2, while B extends through 7 and includes 7. Together, the intervals overlap and form A ∪ B = [-2, 7]. Since the universal set is R, the complement contains real numbers strictly less than -2 or strictly greater than 7. Thus (A ∪ B)' = (-∞, -2) ∪ (7, ∞).

Apply De Morgan’s law: A' ∩ B' = (A ∪ B)'. The intervals A = (−3, 4] and B = [1, 6) overlap, so their union is (−3, 6). The endpoint −3 is excluded because A excludes it, and 6 is excluded because B excludes it. The complement in R therefore includes both endpoints: (−∞, −3] ∪ [6, ∞). Hence option A is correct; option B incorrectly excludes the boundary points.

There are \(72/4=18\) multiples of 4 and \(72/9=8\) multiples of 9. Numbers counted in both sets are multiples of \(\operatorname{lcm}(4,9)=36\); these are 36 and 72, so there are 2. Thus \(|A\cup B|=18+8-2=24\). The complement within a 72-element universal set has \(72-24=48\) elements.

Solve the inequality by factoring: \(x^2-4x=x(x-4)\le0\). Therefore, the real solution interval is \([0,4]\). Since the universal set contains only integers from -8 through 8, \(A=\{0,1,2,3,4\}\). Removing these five integers from \(U\) leaves \(A'=\{-8,-7,-6,-5,-4,-3,-2,-1,5,6,7,8\}\), which is option A.

Use inclusion–exclusion. There are 45 multiples of 2, 30 of 3, and 18 of 5. Pairwise overlaps contain 15 multiples of 6, 9 of 10, and 6 of 15; the triple overlap contains 3 multiples of 30. Thus |A ∪ B ∪ C| = 45 + 30 + 18 − 15 − 9 − 6 + 3 = 66. Therefore, the complement has 90 − 66 = 24 elements, so option B is correct.

Factor the inequality as x² − 9 = (x−3)(x+3). The product is positive when x < −3 or x > 3, so A = (−∞,−3) ∪ (3,∞). The points −3 and 3 make the expression zero and are not in A. Consequently, the complement in R contains every number from −3 through 3, including both endpoints: A′ = [−3,3].

The symmetric difference A △ B consists of elements belonging to exactly one of A and B. Its complement therefore consists of elements that belong to both sets or to neither set. The ‘both’ part is A ∩ B, and the ‘neither’ part is A′ ∩ B′. Taking their union gives (A △ B)′ = (A ∩ B) ∪ (A′ ∩ B′).

By De Morgan’s law, \(A'\cup B'=(A\cap B)'\). The common part of \(A=[-1,5)\) and \(B=(2,8]\) is \((2,5)\); both endpoints are excluded because 2 is not in \(B\) and 5 is not in \(A\). The real-number complement of \((2,5)\) is \((-infty,2]\cup[5,\infty)\).

Every multiple of 4 is also a multiple of 2, so \(C\subseteq A\) and \(A\cup B\cup C=A\cup B\). In \(1\) through \(24\), \(|A|=12\), \(|B|=8\), and the common multiples of 2 and 3 are the multiples of 6, of which there are 4. Hence \(|A\cup B|=12+8-4=16\), and the complement has \(24-16=8\) elements.

By De Morgan's law, A' ∩ B' = (A ∪ B)'. In U, the multiples of 5 are counted by floor(55/5) = 11, and the multiples of 11 by floor(55/11) = 5. The common multiples are multiples of 55; only 55 lies in U, so inclusion–exclusion gives |A ∪ B| = 11 + 5 − 1 = 15. Therefore |A' ∩ B'| = 55 − 15 = 40, so option C is correct.

Factor the quadratic: x² − 2x − 8 = (x − 4)(x + 2). Since the parabola opens upward, the inequality (x − 4)(x + 2) ≤ 0 holds between and at the roots, so A = [−2,4]. The complement in ℝ contains numbers strictly outside this closed interval. Hence A' = (−∞,−2) ∪ (4,∞). The roots are excluded from the complement because they belong to A.

Factor the quadratic: x² − 9x + 20 = (x − 4)(x − 5). Since the parabola opens upward, the inequality is at most zero for 4 ≤ x ≤ 5. Because the universe contains integers, A = {4, 5}. Removing these two elements from U = {−10, …, 10} gives A′ = {−10, …, 3, 6, …, 10}, listed in option A.

Use inclusion–exclusion. There are 30 multiples of 4, 20 of 6, and 12 of 10. Pairwise intersections contain 10 multiples of 12, 6 multiples of 20, and 4 multiples of 30; the triple intersection has 2 multiples of 60. Thus |A ∪ B ∪ C| = 30 + 20 + 12 − 10 − 6 − 4 + 2 = 44. Therefore the complement has 120 − 44 = 76 elements, so option C is correct.

The universal set is partitioned into four mutually disjoint regions: A ∩ B, A′ ∩ B, A′ ∩ B′, and A ∩ B′. Their cardinalities must add to |U|. Therefore |A ∩ B′| = 90 − 24 − 17 − 29 = 20. Hence option B is correct. This partition prevents any element from being counted twice or omitted.