Complement of a Set and Its Properties

The complement \(A'\) contains precisely those elements of the universal set \(U\) that are not in \(A\). Therefore, to recover \(A\), remove \(b,d,f\) from \(U\): \(A=U\setminus A'=\{a,b,c,d,e,f\}\setminus\{b,d,f\}=\{a,c,e\}\). Hence option A is correct. Option B is the complement itself, while option C incorrectly includes every element of \(U\).

The governing idea is that the complement reverses the stated condition within U. Numbers in U divisible by neither 2 nor 5 are 1, 3, 7, and 9, so A = {1, 3, 7, 9}. Consequently A′ contains every element divisible by 2 or by 5: the even numbers 2, 4, 6, 8, 10 together with 5. Thus A′ = {2, 4, 5, 6, 8, 10}, making option A correct; option C wrongly omits multiples of 2 other than 10.

The universal set consists of the natural numbers from 1 through 15. Set A contains all non-prime numbers, including 1, because 1 is neither prime nor composite. Therefore, the complement A′ contains exactly the prime numbers in U: 2, 3, 5, 7, 11, and 13. Thus option A is correct.

Because the universal set is all integers, A contains every integer from −2 through 3, including both endpoints: {−2,−1,0,1,2,3}. Its complement therefore contains all integers outside this closed interval. In set-builder form, A′ = {x ∈ Z : x < −2 or x > 3}; the endpoints are excluded from the complement.

First form the union: A ∪ B = {a, c, g} ∪ {b, c, h} = {a, b, c, g, h}. The complement is taken relative to the stated universal set U, so we select the elements of U that are absent from this union. The remaining elements are d and e. Therefore, (A ∪ B)' = {d, e}, making option A correct.

The complement of A relative to U is A' = {5, 6, 7, 8, 9}, because these are the elements of U not belonging to A. Now intersect A' with B = {4, 5, 6}. The common elements are 5 and 6; 4 is excluded because it belongs to A. Hence, A' ∩ B = {5, 6}, so option A is correct.

The common elements of A and B are 2 and 4, so A∩B={2,4}. The complement is taken relative to the stated universal set U. Therefore, remove 2 and 4 from U: (A∩B)'=U−{2,4}={1,3,5,6,7,8}. Hence option A is correct. Option B is the intersection itself, while options C and D omit or include elements incorrectly. This also agrees with De Morgan’s law: (A∩B)'=A'∪B'.

The universal set is U={1,2,...,20}, which has 20 elements. The perfect squares in this range are 1, 4, 9, and 16, so A has four elements. The complement A' contains every element of U that is not a perfect square. Therefore, |A'|=|U|−|A|=20−4=16. The complement must be taken relative to U, not to all natural numbers. Hence option C is correct.

Because U contains only nonnegative integers, the condition x^2<25 is satisfied by x=0,1,2,3,4. The value x=5 is excluded because 5^2=25, and the inequality is strict. Thus A={0,1,2,3,4}. The relative complement contains the elements of U outside A, namely A'=U−A={5,6,7,8,9}. Therefore, option A is correct. Checking the boundary value 5 prevents confusing <25 with ≤25.

The universal set U contains the integers −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, and 5, so |U| = 11. The condition |x| ≤ 2 gives A = {−2, −1, 0, 1, 2}, which has 5 elements. The complement relative to U therefore has |A′| = |U| − |A| = 11 − 5 = 6 elements. Hence option C is correct.

The complement is taken with respect to U = ℝ. The interval A = (−∞,2] contains every real number less than 2 and also 2 itself. Consequently, the complement contains exactly the real numbers greater than 2. Since 2 already belongs to A, it must not be included in A′, so the correct interval is (2,∞).

The complements are taken with respect to U. Thus A' = {e, f, g, h} and B' = {a, b, g, h}. The symmetric difference contains elements belonging to exactly one of the two sets, so A' Δ B' = {a, b, e, f}. Therefore, it has 4 elements, making option B correct. The common elements g and h are excluded.

The positive divisors of 12 that lie in U are A = {1, 2, 3, 4, 6, 12}. The complement A' contains every element of U that is not in A. Removing these divisors from {1, 2, ..., 12} leaves {5, 7, 8, 9, 10, 11}. Therefore option A is correct. The universal set is essential when finding a complement.

Because U consists of the positive integers less than 10, U = {1, 2, 3, 4, 5, 6, 7, 8, 9}. The condition x² < 20 is true for x = 1, 2, 3, and 4, since 4² = 16, but false for x = 5 because 5² = 25. Thus A = {1, 2, 3, 4}, and its complement in U is {5, 6, 7, 8, 9}.

Within the universal set \(U\), the multiples of 2 are \(\{2,4,6,8,10\}\), and the multiples of 3 are \(\{3,6,9\}\). Taking their union gives \(A=\{2,3,4,6,8,9,10\}\). The complement \(A'\) contains precisely those elements of \(U\) that are not in \(A\). Removing these seven elements from \(U\) leaves \(A'=\{1,5,7\}\). Thus option A is correct; option B is the set \(A\) itself.

The intersection A ∩ B contains the elements common to both sets. Since A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, we get A ∩ B = {3, 4}. The complement is taken within the universal set U, so remove 3 and 4 from U = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Therefore, (A ∩ B)' = {1, 2, 5, 6, 7, 8, 9}.

Because x is an integer and −3 ≤ x ≤ 3, the universal set is U = {−3, −2, −1, 0, 1, 2, 3}. The condition x ≥ 0 selects A = {0, 1, 2, 3}. The complement A' consists of every element of U that is not in A. Hence A' = {−3, −2, −1}, so option A is correct.

First find the number of elements in the union using inclusion–exclusion: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 52 + 44 − 21 = 75. The complement contains all elements of the universal set that are not in the union. Hence n((A ∪ B)ᶜ) = n(U) − n(A ∪ B) = 100 − 75 = 25.

By De Morgan’s law, \(A^c\cap B^c=(A\cup B)^c\). First calculate the number of elements in the union: \(n(A\cup B)=n(A)+n(B)-n(A\cap B)=65+58-29=94\). The required region is outside both sets, so subtract the union from the universal set: \(n(A^c\cap B^c)=n(U)-n(A\cup B)=120-94=26\). Therefore, option A is correct. The value 94 represents the union, not its complement, while 29 represents only the intersection.

The complement of A∩B is taken with respect to the universal set U. It contains every element of U that is not in the common part A∩B. For a finite universal set, n(Xᶜ)=n(U)−n(X). Hence n((A∩B)ᶜ)=130−46=84. Therefore option B is correct; the intersection itself has 46 elements, not its complement.

The seven listed regions are disjoint and together form A∪B∪C. Their total is 23+27+21+13+11+9+7=111. The complement contains the universal-set elements outside this union. Hence n((A∪B∪C)ᶜ)=n(U)−n(A∪B∪C)=130−111=19. The triple intersection is counted once because it is given as its own region.

De Morgan’s law states that the complement of an intersection is the union of the individual complements. Therefore, \((A\cap B\cap C)^c=A^c\cup B^c\cup C^c\). An element is outside the intersection if it fails to belong to at least one of the three sets. Hence, option A is the only correct expression.

De Morgan’s law states that the complement of a union equals the intersection of the complements. Applying it successively gives (A∪B∪C)ᶜ = Aᶜ∩Bᶜ∩Cᶜ. In words, an element is outside the union exactly when it is outside A, outside B, and outside C simultaneously. Option A incorrectly uses a union of complements, while C and D omit the required complementation. Therefore, option B is correct.

De Morgan’s law states that \(A^c\cap B^c=(A\cup B)^c\). Thus the required set contains precisely the elements of \(U\) that are outside the union. Its cardinality is \(n(U)-n(A\cup B)=160-125=35\). Therefore option A is correct. The separate values of \(n(A)\) and \(n(B)\) are not needed once the union is known.

By De Morgan’s law, \(A^c\cup B^c=(A\cap B)^c\). Therefore its cardinality must be \(n(U)-n(A\cap B)=170-44=126\). The stated value is 146, which differs from 126 by 20. Since both values cannot describe the same complement in the same universal set, the supplied data are inconsistent. Hence option A is correct.