Complement of a Set and Its Properties

The difference \(A-B\) contains elements of \(A\) that do not belong to \(B\). Since 1 and 7 are absent from \(B\), while 3 and 5 occur in \(B\), \(A-B=\{1,7\}\). Taking the complement relative to \(U\) removes 1 and 7 from \(U\), leaving \((A-B)^c=\{2,3,4,5,6,8\}\). Hence option A is correct.

Among the integers from 1 through 30, 15 are divisible by 2 and 10 are divisible by 3. The 5 numbers divisible by both 2 and 3 were counted twice, so \(n(A)=15+10-5=20\). Since \(U\) has 30 elements, the complement contains \(n(A^c)=30-20=10\) elements. Therefore, option A is correct.

The universal set contains the 50 natural numbers from 1 to 50. Its elements divisible by 5 are \(5,10,15,20,25,30,35,40,45,50\), so \(n(A)=10\). A and its complement partition U, meaning \(n(U)=n(A)+n(A^c)\). Therefore, \(n(A^c)=50-10=40\), making option A correct.

Factor the condition: \(x^2-5x+6=(x-2)(x-3)=0\). Thus the solutions that lie in U are \(x=2\) and \(x=3\), so \(A=\{2,3\}\). The complement relative to U is \(U\setminus A\), obtained by removing 2 and 3 from U. Hence \(A^c=\{1,4,5,6,7,8,9,10\}\), which is option A.

The integers from -5 through 5 give 11 elements in U. Solving \(x^2=9\) gives \(x=3\) or \(x=-3\), both of which belong to U; therefore \(A=\{-3,3\}\) and \(n(A)=2\). The complement has the remaining elements, so \(n(A^c)=n(U)-n(A)=11-2=9\). Thus option A is correct.

Let H be the set of students studying Hindi and E be the set studying English. By the inclusion–exclusion principle, n(H ∪ E) = n(H) + n(E) − n(H ∩ E) = 72 + 55 − 30 = 97. Students studying neither subject belong to the complement of H ∪ E. Therefore, their number is n(U) − n(H ∪ E) = 120 − 97 = 23. Hence, option A is correct.

The complement Aᶜ contains all students in the universal set who do not belong to A. Since A and Aᶜ are disjoint and together make U, their cardinalities satisfy n(A) + n(Aᶜ) = n(U). Substituting the given values gives n(A) + 34 = 90, so n(A) = 90 − 34 = 56. Therefore, 56 students like science, making option A correct.

The perfect squares in U from 1 through 25 are 1, 4, 9, 16, and 25. Thus A has 5 elements. The complement Aᶜ consists of every element of U that is not a perfect square. Because U has 25 elements, n(Aᶜ) = n(U) − n(A) = 25 − 5 = 20. Hence, option A is the only correct answer.

Since A is the set of odd numbers in U, its complement Aᶜ is the set of even numbers: {2, 4, 6, 8, 10, 12, 14}. The other set contains multiples of 3 in U: {3, 6, 9, 12, 15}. An intersection keeps only elements common to both sets. The common elements are 6 and 12, so Aᶜ ∩ {multiples of 3} = {6, 12}. Therefore, option A is correct.

The complement of A relative to U is obtained by removing the elements of A from U. Therefore, Aᶜ = U − A = {5, 6, 7, 8, 9}. The condition Aᶜ ⊆ B means that B must contain every one of these five elements. To make B as small as possible, we include no additional elements. Thus the minimum choice is B = Aᶜ = {5, 6, 7, 8, 9}, which is option A.

Aᶜ is defined relative to the stated universal set U. It contains every element of U that is not an element of A. Starting with U = {1, 2, 3, 4, 5, 6} and removing A = {1, 2, 3} leaves {4, 5, 6}. Therefore, Aᶜ = {4, 5, 6}, so the set B satisfying B = Aᶜ is option A. The universal set must always be used when determining a complement.

The complement Aᶜ contains all elements of U that are absent from A. Since A = {a, c, e}, removing these elements from U = {a, b, c, d, e, f} leaves {b, d, f}. This remaining set is exactly B. Hence Aᶜ = {b, d, f} = B, so statement A is correct. In fact, A and B are disjoint and their union is U, which confirms that they are complements.

The complement Aᶜ of A in the universal set U is the set of all elements of U that are not in A. The condition A ∩ B = ∅ says that A and B have no common elements, while A ∪ B = U says that together they contain every element of U. Thus B contains exactly those elements of U that are outside A. Therefore, B = Aᶜ, so option A is correct.

A set and its complement are disjoint, as shown by A ∩ Aᶜ = ∅, and their union is the entire universal set, as shown by A ∪ Aᶜ = U. For two disjoint finite sets, the cardinality of their union equals the sum of their cardinalities. Therefore, n(A ∪ Aᶜ) = n(A) + n(Aᶜ). Since A ∪ Aᶜ = U, the result is n(A) + n(Aᶜ) = n(U).

First, Aᶜ relative to U is {6, 7, 8}. The double-complement law states that taking the complement twice returns the original set, so (Aᶜ)ᶜ = A = {1, 2, 3, 4, 5}. Now intersect this set with {2, 4, 6, 8}. The common elements are only 2 and 4, because 6 and 8 are not in A. Therefore, the required set is {2, 4}, which is option A.

The complement Aᶜ contains all elements of U that are not in A. Therefore, Aᶜ = {1,3,4,6,7,9,10}, which has 7 elements. Since 5 is not in Aᶜ, taking the union with {5} adds one new element. Thus Aᶜ ∪ {5} has 7 + 1 = 8 elements, so option A is correct.

The complement of \(A\) is formed by taking all elements of the universal set that are not in \(A\). Therefore, \(A^c=U\setminus A=\{1,3,5,7,8,9\}\). The expression then asks us to remove \(\{1,3,5\}\) from this complement. The elements left are \(\{7,8,9\}\), so option A is correct. Option B stops after finding the complement and does not perform the set difference.

The digits are U = {0,1,2,3,4,5,6,7,8,9}. The even digits are A = {0,2,4,6,8}. The complement Aᶜ therefore contains the digits that are not even, namely the odd digits {1,3,5,7,9}. This set has 5 elements, so option A is correct. The digit 0 is even and is correctly included in A, not in its complement.

A number divisible by both 2 and 3 must be divisible by their least common multiple, 6. Within U = {1,2,...,12}, the numbers divisible by 6 are 6 and 12, so A = {6,12}. The complement contains every other element of U: {1,2,3,4,5,7,8,9,10,11}. Hence option A is correct.

Since \(C=A^c\), first find the complement of \(A\) in the universal set: \(C=\{2,4,6,8\}\). Taking the complement of \(C\) again gives all elements of \(U\) that are not in \(C\), namely \(\{1,3,5,7\}\). Thus \(C^c=(A^c)^c=A\), so option A is correct. This illustrates the double-complement identity: the complement of a complement is the original set.