Complement of a Set and Its Properties

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.

The equality \(A'\cup B=B\) means that adding every element of \(A'\) to \(B\) does not enlarge \(B\). Therefore, every element of \(A'\) must already be an element of \(B\), so \(A'\subseteq B\). The reverse inclusion, equality of the sets, and disjointness are not forced by the given condition.

Every multiple of 18 is also a multiple of 6, so \(B\subseteq A\). Hence \(A\cup B=A\). Applying De Morgan’s law gives \(A'\cap B'=(A\cup B)'=A'\). Notice that although \(A'\subseteq B'\), the intersection of the two complements is the smaller complement, namely \(A'\).

By De Morgan’s law, \(A'\cup B'=(A\cap B)'\). A number belonging to both \(A\) and \(B\) must be divisible by \(\operatorname{lcm}(6,9)=18\). The multiples of 18 from 1 through 54 are 18, 36, and 54, so \(|A\cap B|=3\). Therefore, \(|A'\cup B'|=54-3=51\).

The prime numbers in \(U\) are \(2,3,5,7,11,13,17,19\). The complement contains every element of the stated universal set that is not prime. Thus it contains 1 and all composite numbers from 4 to 20: \(A'=\{1,4,6,8,9,10,12,14,15,16,18,20\}\). Remember that 1 is neither prime nor composite, but it is still in the complement.

The universal set is U = {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}. The condition x² < 10 means |x| < √10. Since 3 < √10 < 4 and x must be an integer, the possible values are −3 through 3, so A = {−3, −2, −1, 0, 1, 2, 3}. The complement contains every element of U that is not in A. Therefore, A′ = U \ A = {−5, −4, 4, 5}.

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 \(U=\varnothing\).

Among the integers from 1 to 30, 15 are divisible by 2 and 10 are divisible by 3. The numbers divisible by both 2 and 3 are the multiples of 6, and there are 5 of them: 6, 12, 18, 24, and 30. By inclusion–exclusion, n(A ∪ B) = 15 + 10 − 5 = 20. The complement contains the remaining elements of U, so n((A ∪ B)′) = 30 − 20 = 10. Hence option A is correct.

Complements are taken relative to U. Thus \(A'=\{b,d,f\}\) and \(B'=\{a,d,e\}\). Their only common element is d, so \(A'\cap B'=\{d\}\). Equivalently, De Morgan’s law gives \(A'\cap B'=(A\cup B)'\); since \(A\cup B=\{a,b,c,e,f\}\), the only element of U outside the union is d.

First calculate the number of elements in the union using the inclusion–exclusion formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substitution gives n(A ∪ B) = 35 + 42 − 18 = 59. The universal set has 80 elements, so the complement of the union contains the elements outside both A and B. Therefore, n((A ∪ B)′) = n(U) − n(A ∪ B) = 80 − 59 = 21. Thus option A is correct.

The difference \(A-B\) contains the elements of \(A\) that are not in \(B\). Since 6 and 12 occur in both sets, \(A-B=\{2,4,8,10\}\). The complement is taken relative to the stated universal set \(U\), so remove these four elements from \(\{1,2,\ldots,12\}\). Therefore, \((A-B)'=\{1,3,5,6,7,9,11,12\}\), which is option A. Option B is the difference itself, not its complement.

A complement is determined by the universal set. Here the universal set is \(\mathbb{N}\), and \(A\) consists of all odd natural numbers. Every natural number is either odd or even, and no natural number is both. Therefore, the elements of \(\mathbb{N}\) that are not in \(A\) are precisely the even natural numbers: \(A'=\{x\in\mathbb{N}:x\text{ is even}\}\). Option B incorrectly changes the universe to \(\mathbb{Z}\), while C includes only primes.

Factor the quadratic equation: \(x^2-5x+6=(x-2)(x-3)=0\). Its roots are 2 and 3, and both belong to the specified universe, so \(A=\{2,3\}\). The universal set is \(U=\{0,1,2,3,4,5,6,7,8,9,10\}\). Hence the complement \(A'=U\setminus A\) contains every element except 2 and 3, giving \(\{0,1,4,5,6,7,8,9,10\}\). Thus option A is correct; option B is the original set \(A\).

De Morgan's law states that the complement of a union equals the intersection of the complements. Applying it to three sets gives \((A\cup B\cup C)'=A'\cap B'\cap C'\). An element is outside the union exactly when it is outside A, outside B, and outside C simultaneously. Therefore option A is correct. Option B fails to interchange union and intersection, while option C incorrectly gives the complement of an intersection.

The set \(A\) contains the positive multiples of 5 from 1 through 50: \(5,10,15,20,25,30,35,40,45,50\). There are \(50/5=10\) such multiples, so \(n(A)=10\). The universal set has 50 elements. Since \(A'\) contains all elements of \(U\) that are not in \(A\), its cardinality is \(n(A')=n(U)-n(A)=50-10=40\). Therefore option A is correct; 10 is the cardinality of A itself.

Take the complement of both sides of the given equality \(A'=B'\). Complementation is an involution, meaning that taking a complement twice returns the original set: \((A')'=A\) and \((B')'=B\). Thus \((A')'=(B')'\) implies \(A=B\). The other statements are not necessary consequences: equal sets need not be disjoint, their union need not be the universal set, and A need not be contained in \(B'\). Therefore option A is correct.

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 number belongs to both A and B exactly when it is divisible by both 4 and 6. Such numbers are multiples of their least common multiple, \(\operatorname{lcm}(4,6)=12\). The multiples of 12 from 1 to 100 are \(12,24,36,48,60,72,84,96\), so \(n(A\cap B)=\lfloor100/12\rfloor=8\). Since \(U\) has 100 elements, \(n((A\cap B)')=100-8=92\). Thus option A is correct.

The two given relations describe a complement: \(A\cup A'=U\) means that together A and \(A'\) contain every element of the universe, while \(A\cap A'=\varnothing\) means that they have no common element. Therefore \(A'\) consists exactly of the elements of \(U\) that do not belong to A, and it can be written as \(U\setminus A\). Option A states this definition. Option B is impossible for a subset A of U, and C and D do not express the complement.

Assuming the standard school convention \(\mathbb{N}=\{1,2,3,\ldots\}\), the universal set contains 25 elements. The square numbers not exceeding 25 are \(1,4,9,16,25\), so \(n(A)=5\). The complement contains all remaining natural numbers from 1 through 25. Therefore \(n(A')=n(U)-n(A)=25-5=20\), making option A correct. Option B counts the square numbers themselves, not the elements of their complement.