Complement of a Set and Its Properties

A and Aᶜ are disjoint and together contain every element of U, so n(A) + n(Aᶜ) = n(U) = 84. Let n(Aᶜ) = x. The given relation gives n(A) = 2x. Hence 2x + x = 84, so 3x = 84 and x = 28. Therefore, n(Aᶜ) is 28. The finite-set complement formula is essential here.

Let n(A)=x. For a finite universal set, A and its complement A^c are disjoint and together contain every element of U. Therefore, n(A)+n(A^c)=n(U). Substituting n(A^c)=3n(A), n(U)=72, and n(A)=x gives x+3x=72, so 4x=72 and x=18. Thus, option A is the only correct answer.

For every element x of U, the law of excluded middle says that either x ∈ A or x ∉ A. If x ∈ A, it is included in A; if x ∉ A, the definition of the complement places it in Aᶜ. Thus every element of U belongs to A or Aᶜ, and both sets are subsets of U. Consequently, A ∪ Aᶜ = U.

The complement of A relative to U contains every element of U that is absent from A. Therefore Aᶜ = U \ A = {1,4,6,7,8,9}. Option B lists exactly these six elements, and options C and D express the same set using difference notation and set-builder notation. Option A omits 9, so it is not the complement.

The positive multiples of 8 from 1 through 80 are 8, 16, 24, 32, 40, 48, 56, 64, 72, and 80, so A has 10 elements. The complement contains all remaining elements of U. Therefore n(Aᶜ) = n(U) − n(A) = 80 − 10 = 70. Option B counts A itself, not its complement.

Because \(B=A^c\), set \(B\) contains every element of the universal set that is not in \(A\). Here, \(B=\{1,4,6,8,9,10,12\}\). Together, \(A\) and \(B\) contain every element from 1 through 12, so \(A\cup B=U\). This illustrates the identity \(A\cup A^c=U\).

The complement \(B=A^c\) contains exactly those elements of the universal set \(U\) that are not members of \(A\). Therefore, no element can belong to both \(A\) and \(B\) at the same time. Hence, \(A\cap B=A\cap A^c=\varnothing\). The correct answer is option A. This is a standard complement identity, and it is true for every set relative to its universal set.

The union combines every element that belongs to either set. Here, \(A\cup B=\{1,2,3,4,5,6,7,8\}=U\), because the two sets together contain all elements of the universal set. Therefore, \((A\cup B)^c=U^c\). The complement of the universal set relative to itself contains no elements, so \(U^c=\varnothing\). Thus, option A is the only correct answer.

Set \(A\) contains only odd numbers from the universal set, whereas \(B\) contains only even numbers. Hence they have no common element and \(A\cap B=\varnothing\). The complement of the empty set relative to \(U\) is the whole universal set, so \((A\cap B)^c=\varnothing^c=U\).

The complement \(A^c\) consists of every element of the universal set \(U\) that is not an element of \(A\). Removing 2, 4, and 6 from \(U=\{1,2,3,4,5,6\}\) leaves \(\{1,3,5\}\). Therefore, option A is correct. The complement always depends on the stated universal set.

The complement is found by removing the elements of \(B\) from the universal set: \(B^c=U\setminus B=\{b,c,d\}\). This set has three elements, so option B is correct. The answer is not 5, because 5 is the cardinality of the entire universal set, not the number of elements outside \(B\).

The empty set contains no elements, so there is no element of \(U\) to remove when forming its complement. By definition, \(A^c=U\setminus A\). With \(A=\varnothing\), this becomes \(\varnothing^c=U\setminus\varnothing=U\). Thus, the entire universal set is the complement of the empty set.

The complement of \(A\) contains the elements of the universal set that are not in \(A\). If \(A=U\), every element of the universal set is already in \(A\), so no element remains outside it. Therefore, \(A^c=U\setminus U=\varnothing\). Notice that \(\{U\}\) would be a singleton containing the set \(U\), not the empty set.

First form the union by listing each element that appears in either set: \(A\cup B=\{1,2,3,5,7,8\}\). To find its complement, select the elements of \(U\) that are absent from this union. The only such elements are 4 and 6. Therefore, \((A\cup B)^c=\{4,6\}\), making option A correct. Option C is the intersection \(A\cap B\), not the complement.

If every element of \(A\) is also in \(B\), then any element outside \(B\) must certainly be outside \(A\). Thus, the elements of \(B^c\) are a subset of the elements of \(A^c\), giving \(B^c\subseteq A^c\). Complementation reverses the direction of a subset relation; this is called order reversal.

The governing cardinality rule for a finite universal set is n(Aᶜ) = n(U) − n(A), because U is partitioned into A and its complement. Substituting the given values gives n(Aᶜ) = 40 − 18 = 22. Thus 22 elements are outside A but still inside U, so option B is correct. Option A repeats n(A), option C repeats the size of U, and option D incorrectly adds the two quantities.

Treat the complete class as the universal set, with 60 students in total. The cricket players form a subset containing 35 students, so the non-players form its complement. Assuming each student is counted once, the complement has 60 − 35 = 25 students. Therefore option A is correct. Option B is the number who play cricket, option C is the total class size, and option D results from an incorrect addition.

Using the usual school convention ℕ = {1,2,3,...}, the universal set is U = {1,2,3,4,5,6,7,8,9,10}. The even elements form A = {2,4,6,8,10}. The complement relative to U consists of the remaining elements, which are precisely the odd numbers {1,3,5,7,9}. Hence option B is correct; option A is A itself.

The universal set is the set of all integers. A contains the positive integers, namely integers greater than zero. The complement must therefore contain every integer that is not positive: zero and all negative integers. This set is described by x ≤ 0, so option B is correct.

The interval A = (2,5) contains real numbers strictly greater than 2 and strictly less than 5; its endpoints are excluded. Therefore, in the universal set ℝ, the complement contains every real number at most 2 and every real number at least 5. In interval notation this is (−∞,2] ∪ [5,∞). Option A wrongly excludes the endpoints, while option D is A itself.