Complement of a Set and Its Properties

The equation Bᶜ = A means that B must be the complement of A in the given universal set U. Remove the elements 2, 4, and 6 from U = {1, 2, 3, 4, 5, 6, 7}; the remaining elements are 1, 3, 5, and 7. Hence B = Aᶜ = {1, 3, 5, 7}, making option A correct. Option B is A itself, not its complement.

The universal set contains a, b, c, d, e, and g. Set A contains a, d, and g, while the remaining elements b, c, and e form B. Thus B contains exactly the elements of U that are not in A, so B = U \ A = Aᶜ. The sets are disjoint and their union is U; therefore option A is correct.

The condition P ∩ Q = ∅ says that P and Q have no common elements, while P ∪ Q = U says that together they contain every element of the universal set. Thus, every element not in P must be in Q, and every element of Q is outside P. Therefore, Q is exactly the complement of P in U: Pᶜ = Q. Hence, option A is correct.

A set and its complement are disjoint, and their union is the universal set. Hence the elements of A and Aᶜ can be counted separately and added: n(U) = n(A ∪ Aᶜ) = n(A) + n(Aᶜ) = 27 + 33 = 60. The subtraction 33 − 27 = 6 is irrelevant here, so option A is correct.

The double-complement law states that taking the complement twice returns the original set, so (Aᶜ)ᶜ = A. The required expression therefore becomes A ∩ {3, 6, 9}. Since A = {1, 2, 3, 4, 5, 6}, the common elements are 3 and 6; 9 is not in A. Thus the intersection is {3, 6}, so option A is correct.

The complement A^c contains all elements of U that are not in A. Thus, A^c={1,2,4,5,7,8,10,11}, which has 8 elements. The elements 6 and 12 belong to A, so neither is already in A^c. Adding both of them to A^c gives {1,2,4,5,6,7,8,10,11,12}, containing 10 elements. Therefore, option A is correct.

The complement A^c consists precisely of the elements of the universal set U that are not in A. Therefore, no element can belong to both A and A^c, which gives A∩A^c=∅. At the same time, every element of U belongs to either A or A^c, so A∪A^c=U. Hence option A is the only universally true statement.

The digits in the universal set are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, so n(U) = 10. The prime digits are 2, 3, 5, and 7, giving n(A) = 4. The complement Aᶜ contains all digits in U that are not prime: {0, 1, 4, 6, 8, 9}. Therefore, n(Aᶜ) = n(U) − n(A) = 10 − 4 = 6. Hence, option A is correct.

A number divisible by both 2 and 5 must be divisible by their least common multiple, 10. Between 1 and 18, the only multiple of 10 is 10, so A={10}. The complement contains every element of U except 10. Hence A^c={1,2,3,4,5,6,7,8,9,11,12,13,14,15,16,17,18}, which is option A.

All complements are taken with respect to U. Since C=A^c, taking the complement again gives C^c=(A^c)^c=A, by the double-complement law. Therefore, C^c∩{1,2,3,4}=A∩{1,2,3,4}. The elements common to A={2,4,6,8,10} and the second set are 2 and 4, so the answer is {2,4}, option A.

A partition of U requires two essential conditions: its parts must be pairwise disjoint, and their union must be the whole universal set. A and A^c are disjoint because no element can belong to both, so A∩A^c=∅. Together they contain every element of U, so A∪A^c=U. Therefore, option A proves that they form a partition.

A set and its complement are disjoint, so A ∩ Aᶜ = ∅. If A = Aᶜ, replacing Aᶜ by A would give A ∩ A = ∅, which means A = ∅. But then Aᶜ = U, and the equality A = Aᶜ would require ∅ = U. This contradicts U ≠ ∅. Equivalently, no nonempty universal set can be equal to its own complement. Therefore, option A is correct.

A complement is defined relative to a particular universal set: A^c=U−A. If U is enlarged while A stays unchanged, any newly included elements that are not members of A will belong to the new complement. Thus the complement may gain new elements. It is not necessarily empty, equal to A, or unchanged. Therefore, option A is correct.

The complement depends on the universal set. Relative to U1, A^c=U1−A={1,3,4}. Relative to U2, the complement is A^c=U2−A={1,3,4,6,7}. Comparing these two complements, the elements that appear only in the second one are 6 and 7. Thus the extra elements are {6,7}, making option A correct.

All complements are taken with respect to U. First, Aᶜ = U − A = {5, 6, 7, 8, 9, 10, 11, 12}. Intersecting this with B = {3, 4, 5, 6, 7} gives Aᶜ ∩ B = {5, 6, 7}. Now take the complement of this result in U: (Aᶜ ∩ B)ᶜ = U − {5, 6, 7} = {1, 2, 3, 4, 8, 9, 10, 11, 12}. Thus, option A is correct.

The complement of A with respect to U is Aᶜ = {1,3,5,7,9}. Taking the union with B gives Aᶜ ∪ B = {1,2,3,4,5,7,9}. The elements of U not present in this union are {6,8,10}, so (Aᶜ ∪ B)ᶜ = {6,8,10}. This also follows from De Morgan’s law: (Aᶜ ∪ B)ᶜ = A ∩ Bᶜ.

Among the numbers 1 to 24, six numbers are divisible by 4: 4, 8, 12, 16, 20, and 24. Four are divisible by 6: 6, 12, 18, and 24. The common elements are 12 and 24, so |A ∪ B| = 6 + 4 − 2 = 8. By De Morgan’s law, Aᶜ ∩ Bᶜ = (A ∪ B)ᶜ, and its size is 24 − 8 = 16.

De Morgan’s first law states that the complement of a union is the intersection of the complements: (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ. An element is outside A ∪ B precisely when it is outside A and also outside B. Option B incorrectly retains the union sign, option C uses the wrong operation, and option D is false because (Aᶜ)ᶜ = A.

The complement A^c contains the elements of U that are not in A, while B^c contains the elements of U that are not in B. By De Morgan’s law, A^c ∩ B^c = (A ∪ B)^c. Here, A ∪ B = {1,2,4,6,7,8,10}. Removing these elements from U={1,2,3,...,11} leaves {3,5,9,11}. Therefore, option A is correct.

The complement Aᶜ consists of all elements of the universal set U that are not in A. Every element of U is therefore either in A or in Aᶜ, so their union is U: A ∪ Aᶜ = U. They have no common element, which means A ∩ Aᶜ = ∅. Thus options B, C, and D are not generally true.