Complement of a Set and Its Properties

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.

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.

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.

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.

Because the universal set is all real numbers, the complement contains every real number outside the interval \([-2,4)\). The endpoint \(-2\) belongs to \(A\) because the left bracket is closed, so it is excluded from the complement. The endpoint \(4\) does not belong to \(A\) because the right parenthesis is open, so \(4\) belongs to the complement. Therefore, \(A'=(-\infty,-2)\cup[4,\infty)\), option A.

The condition \(A\cap B=\varnothing\) means that no element is common to both \(A\) and \(B\). Therefore, every element of \(B\) lies outside \(A\). By the definition of complement, all such elements belong to \(A'\), so \(B\subseteq A'\) must hold. The other statements need not always be true: \(B\) may be smaller than \(A'\), and the union need not equal \(U\).

By De Morgan’s law, \(A'\cap B'=(A\cup B)'\). Thus, we count numbers from 1 to 40 that are not divisible by 2 or 5. There are 20 multiples of 2 and 8 multiples of 5, while 4 numbers are multiples of both, namely multiples of 10. Therefore, \(n(A\cup B)=20+8-4=24\), and \(n(A'\cap B')=40-24=16\). Hence, option A is correct.

First solve the defining inequality: \(x+2>4\) gives \(x>2\). Since \(x\) must be an integer in \(U=\{-3,-2,-1,0,1,2,3,4,5,6,7\}\), we obtain \(A=\{3,4,5,6,7\}\). The complement consists of the remaining elements of \(U\), namely \(A'=\{-3,-2,-1,0,1,2\}\). Thus option A is correct. The endpoint 2 is included in the complement because the inequality is strict.

Use the given identity \(A-B=A\cap B'\). Taking complements on both sides gives \((A-B)'=(A\cap B')'\). De Morgan’s law changes the complement of an intersection into the union of the complements: \((A\cap B')'=A'\cup(B')'\). Since the complement of \(B'\) is \(B\), the result is \(A'\cup B\). Therefore, option A is correct.

A contains all real numbers less than 1, while B contains all real numbers greater than 4. Therefore, A union B excludes the endpoints 1 and 4 as well as every number between them. The complement in R consequently contains 1, 4, and all real numbers between them, so (A union B)' = [1, 4]. Thus option A is correct.

A complement is defined relative to a specified universal set U. The elements outside A may change when U changes, even if A itself remains the same. Hence A' is not independent of U. The other statements are standard complement identities: A union A' equals U, A intersection A' is empty, and A' equals U minus A. Therefore option A is false.

For a subset A of U, the relative difference U minus A is exactly the complement A'. Taking the complement once more gives (A')'. By the double-complement law, the complement of the complement of A is A itself. Therefore (U minus A)' = (A')' = A, making option A correct.

An element belongs to A intersection B exactly when it is divisible by both 3 and 4. Such numbers are multiples of lcm(3, 4) = 12. From 1 through 60, the multiples are 12, 24, 36, 48, and 60, so the intersection has 5 elements. Since U has 60 elements, its complement has 60 - 5 = 55 elements. Option A is correct.

Using De Morgan’s law, A' ∪ B' = (A ∩ B)'. The common elements of A = {1, 2, 5} and B = {2, 4, 6} are A ∩ B = {2}. Taking the complement of {2} in the universal set U = {1, 2, 3, 4, 5, 6, 7, 8} removes 2 and retains every other element, giving {1, 3, 4, 5, 6, 7, 8}. Therefore, option A is correct.

The interval A = (-1, 6] contains numbers strictly greater than -1 and includes 6. Therefore -1 is excluded from A and must be included in its complement, while 6 is included in A and must be excluded from the complement. All real numbers less than -1 and greater than 6 belong to A', giving (-infinity, -1] union (6, infinity).

The members of A whose last digit is 0 or 5 are 5, 10, 15, 20, 25, and 30, so n(A) = 6. The universal set contains 30 elements. Since A′ contains all elements of U not in A, the complement formula gives n(A′) = n(U) − n(A) = 30 − 6 = 24. Hence option A is correct.

When A ⊆ B, every element of A is also an element of B. Therefore, any element that is outside B must certainly be outside A. This means B′ ⊆ A′. Complementation reverses the direction of subset inclusion. The other statements are not always true: for example, A′ ∩ B′ equals (A ∪ B)′, not necessarily the empty set.

Within U, set A contains 1 through 8, while set B contains 14 through 20. Their union therefore contains the two outer ranges: {1, ..., 8, 14, ..., 20}. The elements of U that are missing from this union are exactly 9, 10, 11, 12, and 13. Hence (A ∪ B)′ = {9, 10, 11, 12, 13}.

Complements are taken relative to ℝ. Since A includes all real numbers up to and including 0, A′ = (0, ∞). Since B includes all real numbers from 2 onward, including 2, B′ = (−∞, 2). Their intersection is therefore the real interval (0, 2). Both endpoints are excluded: 0 is in A and 2 is in B.

First factor the quadratic: x² − 11x + 30 = (x − 5)(x − 6). Thus the roots are x = 5 and x = 6, and both belong to U, so A = {5, 6}. The complement A' in U contains every element of U except 5 and 6. Intersecting A' with {5, 6, 7, 8} removes 5 and 6 and leaves {7, 8}. Therefore, option A is correct.

Factor the quadratic: x² − 2x − 15 = (x − 5)(x + 3). The inequality (x − 5)(x + 3) ≤ 0 holds for −3 ≤ x ≤ 5. Therefore, A = {−3, −2, −1, 0, 1, 2, 3, 4, 5}, which has 9 integers. The universal set U contains all integers from −8 through 8, so n(U) = 17. Hence n(A′) = n(U) − n(A) = 17 − 9 = 8.

The complement contains every real number that is not in A. Because (−∞, −2) excludes −2, the point −2 belongs to A′. The interval [3, 7] includes both endpoints 3 and 7, so neither endpoint belongs to the complement. Numbers between −2 and 3, together with numbers greater than 7, are excluded from A and therefore form A′ = [−2, 3) ∪ (7, ∞).

A contains all even numbers in U, so A′ contains all odd numbers from 1 to 20. The prime numbers in U are 2, 3, 5, 7, 11, 13, 17, and 19. Intersecting this prime set with A′ removes 2 because it is even. The remaining odd primes are {3, 5, 7, 11, 13, 17, 19}, which is therefore A′ ∩ B.