Complement of a Set and Its Properties

The complement A′ contains those elements of the universal set U that are not members of A. Starting with U = {1,2,3,4,5,6,7,8} and removing A = {1,2,3,4}, we obtain A′ = {5,6,7,8}. The smallest element of this remaining set is 5, so option C is correct. The number 4 is excluded because it belongs to A.

The complement of A in U is A' = U \ A, meaning the elements of U that are not in A. Here A and U are exactly the same set, so there are no elements of U left after removing A. Therefore A' = ∅, and the empty set has zero elements. Hence the correct answer is 0.

The complement A′ in U consists of all elements of U that are not in A. From U = {black, white, red, yellow}, remove the elements black and yellow because they belong to A. The remaining elements are white and red, so A′ = {white, red}. Therefore option B is correct; a complement cannot contain any element already present in A.

The complement M' contains all elements of the universal set U that are not members of M. Removing 2, 4, 6, 8, and 10 from U gives M' = {1, 3, 5, 7, 9}. Among the choices, only 9 appears in this complement. Therefore, option C is correct; 2, 6, and 10 belong to M itself.

The complement is taken with respect to U. Removing the elements of A from U gives A' = U − A = {1, 4, 6}. Among these elements, only 1 is strictly less than 4; the number 4 itself is not less than 4. Therefore, exactly one element satisfies the condition, so option A is correct.

The complement of A in U is A' = U − A = {2, 3, 4}. Taking the complement once more gives (A')' = U − {2, 3, 4} = {1, 5}, which is the original set A. This illustrates the double-complement law: the complement of a complement returns the original set, provided the same universal set is used. Hence option A is correct.

The complement A' consists of all elements in the universal set U that are not in A. The digits are 0 through 9, while A contains all even digits: 0, 2, 4, 6, and 8. Removing them from U leaves the odd digits 1, 3, 5, 7, and 9. Therefore A' is {1, 3, 5, 7, 9}, so option A is correct.

The complement B = A' is taken with respect to the universal set U. Since A = {1,2,3}, its complement is B = {4,5,6}. Taking the complement once more gives B' = (A')' = A, by the double-complement law. Therefore B' = {1,2,3}. Option B is B itself, not B', while option C is the universal set and option D is the empty set.

The complement A' contains every element of the universal set U that is not an element of A. Starting with U = {2,4,6,8,10,12}, remove 4, 8, and 12 because they belong to A. The elements left are 2, 6, and 10. Therefore A' = {2,6,10}. Option B is A itself, and the other options omit or incorrectly include elements.

The complement of A relative to U is defined as A' = U \ A: all elements in U that are not in A. From U = {1,2,3,4,5,6,7,8}, remove 2, 4, and 8. The remaining elements are {1,3,5,6,7}, so option B is correct. Option A repeats A itself, while option C still contains elements of A and option D is not justified.

The complement C' contains elements of the universal set U that are not contained in C. The elements a, c, and f are already in C, so they cannot be in C'. Removing them from U = {a,b,c,d,f} leaves C' = {b,d}. Therefore b is in C'. Although d is also in the complement, it is not among the answer choices; among the given options, only b is correct.

The complement \(A'\) contains all elements of the universal set \(U\) that are not elements of \(A\). Removing 1, 3, and 5 from \(U\) gives \(A'=\{0,2,4,6\}\). This set has four elements, so \(n(A')=4\). The same result follows from the formula \(n(A')=n(U)-n(A)=7-3=4\). Thus option B is correct; 3 is the size of \(A\), while 7 is the size of \(U\).

The complement of a set is always defined relative to its universal set. Here, \(U\) contains every book in the library, while \(H\) contains only the Hindi books. Therefore, \(H'=U\setminus H\) consists of all books in the library that are not Hindi books, such as English or other-language books. It does not refer to authors, publication age, or books outside the library. Hence option B is correct.

The complement \(A'\) contains the elements of the universal set \(U\) that are not in \(A\). The intersection \(A'\cap A\) would require an element to be in \(A\) and not in \(A\) at the same time, which is impossible. Therefore, the two sets are disjoint and \(A'\cap A=\emptyset\). This is one of the fundamental complement identities; the related union identity is \(A\cup A'=U\). Hence option C is correct.

The complement \(A'\) is formed by selecting elements that belong to the universal set \(U\) but do not belong to \(A\). The elements 1, 3, 5, 7, and 9 are removed from \(U\), leaving 2, 4, and 6. Hence \(A'=\{2,4,6\}\), so option A is correct. Option C is wrong because 8 is not even in the universal set; option B repeats \(A\), and option D contains an element already in \(A\).

The complement \(A'\) contains the elements of the universal set \(U\) that are not in \(A\). Therefore, to recover \(A\), remove the elements of \(A'\) from \(U\): \(A=U\setminus A'=\{1,2,3,4,5,6\}\setminus\{2,6\}=\{1,3,4,5\}\). Hence option B is correct. Option A is the complement itself, while C and D contain incorrect elements or omit required ones.

The complement of \(A\) relative to \(U\) is \(A'=U\setminus A\), meaning all elements that are in the universal set but not in \(A\). Removing apple and mango from \(U=\{apple, banana, grapes, mango\}\) leaves banana and grapes. Therefore \(A'=\{banana, grapes\}\), represented by option B. Option A is the original set, option C incorrectly retains mango, and option D is not empty.

The complement A′ contains precisely those elements of the universal set U that are not elements of A. Removing 2, 3, 4, and 5 from U gives A′ = U − A = {6, 7, 8, 9}. The largest element of this set is 9, so option D is correct. The element 8 is in the complement, but it is not the largest element.

The complement of A with respect to U is defined as A′ = U − A, meaning the elements in U that are not in A. Here A and U are exactly the same set, so every element of U is already in A. No element remains after removing A from U; therefore A′ = ∅ and its cardinality is 0. Thus option A is correct.

The complement A′ contains all elements of the universal set U that are not in A. Starting with U = {blue, green, yellow, pink}, remove green and pink because both belong to A. The elements left are blue and yellow, so A′ = U − A = {blue, yellow}. Therefore option B is correct; options C and D include an element that belongs to A.

The complement T′ consists of elements in U that are not in T. Removing 3, 6, and 9 from U gives T′ = {1, 2, 4, 5, 7, 8}. Among the given options, 1, 2, and 8 belong to T′, whereas 6 belongs to T itself. Therefore 6 is the only element that is not in T′, so option C is correct.

The complement is formed relative to U: A' = U \ A = {2, 3, 5, 7}. Now apply the strict condition “less than 5.” The elements 2 and 3 satisfy it, while 5 does not because it is equal to 5. Therefore, exactly two elements of A' are less than 5. This also shows why counting 5 would produce the incorrect answer 3.

All complements are taken with respect to U. First, A' = U \ A = {4, 6, 8}. Taking the complement again gives (A')' = U \ {4, 6, 8} = {2, 10}. Thus the double-complement law, (A')' = A, is verified. Option B is only the first complement, and option C is U itself, not the double complement.

The complement A' contains every element of the universal set U that is not in A. Starting with U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, remove 2, 4, 6, and 8. The remaining elements are {1, 3, 5, 7, 9}. Zero cannot be included because it is not an element of the stated universal set, so option B is incorrect.

Because B = A', taking the complement of B gives B' = (A')'. The double-complement law states that the complement of a complement is the original set, so (A')' = A. Hence B' = A = {2, 5, 7}. In fact, B itself is {1, 3, 4, 6}; that is why option B represents the single complement and not B'.