Complement of a Set and Its Properties

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,∞).

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 set A contains all prime numbers from 1 through 12: 2,3,5,7, and 11. The complement therefore contains every element of U that is not prime. Number 1 is neither prime nor composite, while 4,6,8,9,10, and 12 are composite. Hence A′ is {1,4,6,8,9,10,12}, described by option A.

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.

A complement is always determined relative to the stated universal set. Here U is the set of natural numbers, and A contains the even natural numbers. Removing all even natural numbers from N leaves exactly the odd natural numbers, such as 1, 3, 5, 7, and so on. Therefore, A′ represents all odd natural numbers.

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}.

The complement A' contains exactly those elements of U that are not in A. Therefore A is obtained by removing A' from the universal set: A=U\A'. Removing 2, 4, and 6 from {1,2,3,4,5,6} leaves {1,3,5}. Hence A={1,3,5}, so option A is correct. The universal set is essential because complements are always relative to it.

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.

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.

The complement \(A'\) contains all elements of the universal set \(U\) that are not in \(A\). Therefore, its cardinality is found by subtracting the number of elements in \(A\) from the number in \(U\): \(n(A')=n(U)-n(A)=30-11=19\). Thus, option B is correct. The other options either give the size of \(A\), the size of \(U\), or an incorrect sum.

De Morgan’s law states that the union of the complements of two sets equals the complement of their intersection: A' ∪ B' = (A ∩ B)'. An element belongs to the left side if it is outside A or outside B, which means it cannot be simultaneously inside both A and B. Therefore, option A is correct.

First find the union using inclusion–exclusion: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 20 + 16 − 6 = 30. The elements in neither A nor B form the complement of A ∪ B in U. Therefore, their number is n(U) − n(A ∪ B) = 50 − 30 = 20. Hence, option B is correct; 30 is the union, not the neither region.

The complement of A ∩ B contains every element of the universal set except the elements common to A and B. Therefore, use the complement formula n((A ∩ B)') = n(U) − n(A ∩ B). With the given values, the result is 32 − 9 = 23. Hence, option B is correct. The intersection itself has 9 elements, while its complement has the remaining 23.

The complement A′ is determined relative to the universal set U. It contains all elements of U that are not in A. From U = {1, 2, 3, 4, 5}, remove 2 and 5, the elements of A. The remaining elements are 1, 3, and 4. Therefore, A′ = {1, 3, 4}. Without specifying U, the complement cannot be determined uniquely.

The complement \(A'\) contains all elements of the universal set \(U\) that are not in \(A\). Thus, for a finite universal set, \(n(A')=n(U)-n(A)\). Substituting the given values gives \(n(A')=38-16=22\). Therefore, option B is correct. The number 16 is the size of A itself, 38 is the size of the whole universal set, and 54 incorrectly adds the two quantities instead of finding the elements outside A.

The complement A' is defined relative to the universal set U. It contains every element of U that is not a member of A. Starting with U = {1, 2, 3, 4, 5, 6} and removing 1, 4, and 6 leaves 2, 3, and 5. Therefore A' = {2, 3, 5}. The answer would be empty only if A were equal to U.

First form the union by listing every element that occurs in A or B: A ∪ B = {a, b, c, d}. The complement of this union contains the elements of U that are absent from it. In U, those remaining elements are e and f. Hence (A ∪ B)' = {e, f}. This also agrees with De Morgan’s law: (A ∪ B)' = A' ∩ B'.