Complement of a Set and Its Properties

Every element of the universal set U is either in A or not in A. The elements that are not in A form Aᶜ. Consequently, taking the union of A and Aᶜ includes every element of U and does not include anything outside U, so A ∪ Aᶜ = U. This is called the complement or union law. Option B describes the intersection law, A ∩ Aᶜ = ∅, not the union law.

The complement Aᶜ contains precisely those elements of the universal set that are not in A. Therefore, no element can belong to both A and Aᶜ at the same time. Their common part is empty, so A ∩ Aᶜ = ∅. This is called the complement or intersection law. Option B belongs to the union identity A ∪ Aᶜ = U, while options C and D incorrectly treat the intersection as one of the two sets.

Taking a complement means selecting all elements of the universal set that are outside the given set. The first complement Aᶜ contains the elements outside A. Taking the complement again selects the elements outside Aᶜ, which are exactly the original elements of A. Therefore, the double-complement law is (Aᶜ)ᶜ = A. Option B is only the first complement, and U or ∅ are not generally equal to the result.

When A is a subset of B, every element of A is also an element of B. Taking complements reverses the inclusion order: an element outside B is certainly outside A, because A is contained in B. Hence every element of Bᶜ belongs to Aᶜ, and Bᶜ ⊆ Aᶜ. Equality is not guaranteed unless A and B are equal. The intersection in option D is also not generally the universal set.

The complement Aᶜ consists of all elements of the universal set U that are not members of A. The set A contains the even numbers from 1 through 10. Removing 2, 4, 6, 8, and 10 from U leaves the odd numbers 1, 3, 5, 7, and 9. Therefore, Aᶜ = {1, 3, 5, 7, 9}, so option A is correct. Option B repeats A instead of finding its complement.

To find Aᶜ, list the elements in the universal set U and remove every element that belongs to A. In the English version, U contains a, e, i, k, and m, while A contains a, e, and i. The remaining elements are k and m, so Aᶜ = {k, m}. Thus option A is correct. The Hindi version expresses the same structure with corresponding letters.

The complement Aᶜ contains the elements of U that are not in A. Starting with U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, remove 2, 3, 5, and 7. The remaining elements are 1, 4, 6, 8, and 9, so Aᶜ = {1, 4, 6, 8, 9}. Notice that 1 is included because it belongs to U even though it is neither prime nor a member of A. Therefore, option A is correct.

An element belongs to (A ∪ B)ᶜ when it does not belong to the union A ∪ B. Not belonging to a union means that the element belongs to neither A nor B. Therefore, it must belong to both complements Aᶜ and Bᶜ. Hence De Morgan's law gives (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ. The union and intersection are interchanged when a complement is distributed. Option B instead represents (A ∩ B)ᶜ.

First calculate the union: \(A\cup B=\{1,2,3,4\}\), because the union contains every element appearing in either set, with repeated elements written only once. The complement is taken relative to the universal set \(U\). Removing 1, 2, 3, and 4 from \(U\) leaves 5 and 6. Hence \((A\cup B)^c=\{5,6\}\), so option A is correct. The other choices either give the union, the intersection, or an incomplete complement.

The intersection contains elements common to both sets. Since 3 is the only common element, \(A\cap B=\{3\}\). The complement is formed by selecting elements of \(U\) that are not in this intersection. Thus \(U\setminus\{3\}=\{1,2,4,5,6\}\). Option A is correct. Option B is the intersection itself, option C still includes 3, and option D incorrectly claims that no elements remain.

The complement of a set is determined relative to the stated universal set. Start with all elements of \(U\): 0, 1, 2, 3, and 4. Remove the elements belonging to \(A\), namely 0, 2, and 4. The elements left are 1 and 3, so \(A^c=\{1,3\}\). Therefore option A is correct. The presence of 0 in A does not change the procedure; zero is an ordinary element of the set.

The complement of A consists of elements in U that are not in A. Since \(A=\{1,4\}\) and \(U=\{1,2,3,4\}\), we obtain \(A^c=\{2,3\}\). Therefore 2 belongs to \(A^c\), making option A true. Option B is false because 1 belongs to A, option C is false because 4 belongs to A, and option D is false because 3 is not in A. Membership must always be checked relative to U.

By definition, \(A^c=U\setminus A\), meaning that the complement contains only elements taken from the universal set U. Therefore every element of \(A^c\) is necessarily an element of U, so \(A^c\subseteq U\) is always true. The other statements require special conditions: \(A^c=U\) only when A is empty, \(A^c=A\) is not generally true, and \(A^c=\varnothing\) only when A equals U. Hence option A is the universal statement.

The complement \(A^c\) contains all elements of U that are not in A. If this complement is empty, there is no element of U outside A. Consequently, A must contain every element of U, so \(A=U\). This also follows from the identity \(A\cup A^c=U\): when \(A^c=\varnothing\), the union becomes A, giving A=U. Therefore option A is correct; the other choices do not follow from the definition.

The complement \(A^c\) consists of all elements of the universal set \(U\) that are not in \(A\). If \(A^c=U\), every element of \(U\) lies outside \(A\). Therefore, \(A\) contains no element and must be the empty set, \(A=\varnothing\). This also agrees with the standard identity \(A\cup A^c=U\) and \(A\cap A^c=\varnothing\). Option B is incorrect because \(A=U\) would imply \(A^c=\varnothing\), not \(U\).

In a Venn diagram, the rectangle represents the universal set U, so only the region inside that rectangle is considered. The circle represents A. The part inside U but outside the circle contains all elements of U that are not in A; by definition, this is the complement \(A^c=U\setminus A\). Therefore option A is correct. The circle itself represents A, while \(A\cap U=A\) represents the circle, not the outside region.

A complement is always defined with respect to its universal set. Here U contains all alphabet letters, while A contains the vowels. Therefore \(A^c=U\setminus A\) contains every alphabet letter that is not a vowel. These letters are the consonants, so option A is correct. The answer depends on the given universe: it is not the empty set or all of U, and it is not A itself because A contains vowels.

The complement Aᶜ contains every element of the universal set U that is not an element of A. Starting with U, remove 3, 6, 9, and 12. The elements left are 1, 2, 4, 5, 7, 8, 10, and 11, so option A is correct. Option B repeats A rather than giving its complement; options C and D contain elements of A and therefore cannot be the required complement.

A and its complement Aᶜ are disjoint and together contain every element of U. Therefore, n(A) + n(Aᶜ) = n(U). Substituting the given values gives n(A) + 18 = 30, so n(A) = 30 − 18 = 12. Thus option A is correct. Option C is the size of the complement, option D is the size of U, and option B is impossible because a subset cannot have more elements than its universal set.

A and Aᶜ partition the universal set U, so n(A) + n(Aᶜ) = n(U). Since the two cardinalities are equal, let each one be x. Then x + x = 16, giving 2x = 16 and x = 8. Therefore n(A) = 8, so option A is correct. A value of 16 or 0 would make the two parts unequal, while 4 would account for only 8 elements altogether.