Complement of a Set and Its Properties

De Morgan’s law gives \(A^c\cup B^c=(A\cap B)^c\). Thus every element of the universal set is included except the 22 elements in \(A\cap B\). Consequently, \(n(A^c\cup B^c)=n(U)-n(A\cap B)=120-22=98\). Option D is correct. The other regional counts are consistent but unnecessary for this particular identity.

De Morgan’s law gives Aᶜ∪Bᶜ=(A∩B)ᶜ. Thus, the requested set consists of every universal-set element except those in the intersection A∩B. Since U has 150 elements and A∩B has 29 elements, n(Aᶜ∪Bᶜ)=150−29=121. The difference-region values are consistent with the diagram but are not needed for this calculation.

First find the sizes of A and B from their complements: n(A) = 80 − 35 = 45 and n(B) = 80 − 42 = 38. Then n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 45 + 38 − 25 = 58. Since A′ ∩ B′ = (A ∪ B)′ by De Morgan’s law, n(A′ ∩ B′) = 80 − 58 = 22. Therefore, option A is correct.

Since A ∩ B = ∅, the sets are disjoint and have no common elements. Therefore, n(A ∪ B) = n(A) + n(B) = 57 + 46 = 103. The complement of A ∪ B contains all universal-set elements outside that union. Hence n((A ∪ B)′) = n(U) − n(A ∪ B) = 125 − 103 = 22, so option A is correct.

There are floor(120/8) = 15 multiples of 8 and floor(120/12) = 10 multiples of 12. Numbers counted in both sets are multiples of lcm(8, 12) = 24, and there are floor(120/24) = 5 of them. Thus n(A ∪ B) = 15 + 10 − 5 = 20. The complement within U therefore has 120 − 20 = 100 elements. Hence option A is correct.

Use the inclusion-exclusion principle. The set A has 15 multiples of 8, while B has 10 multiples of 12. Their overlap consists of multiples of lcm(8, 12) = 24, giving 5 common elements. Therefore n(A ∪ B) = 15 + 10 − 5 = 20. Since the universal set has 120 elements, n((A ∪ B)') = 120 − 20 = 100. Thus option A is unambiguously correct.

Mutually disjoint sets have no common elements, so their union has size equal to the sum of their sizes: n(A∪B∪C)=27+34+41=102. The complement of this union contains the elements of U outside all three sets. Using n(X′)=n(U)−n(X), we get 125−102=23. Thus option A is correct; 102 is the union size, not its complement.

The statement is De Morgan’s law for the complement of a union. An element is outside A∪B exactly when it is not in A and also not in B. The conditions “not in A” and “not in B” describe A′ and B′, and their simultaneous occurrence is A′∩B′. Thus both the assertion and its reason are true, and the reason explains the assertion.

First calculate the intersection: A ∩ B = {2, 5}, because 2 and 5 are the only elements common to A and B. The complement of a set is taken with respect to the universal set U, so remove 2 and 5 from U. This gives (A ∩ B)′ = U − {2, 5} = {1, 3, 4, 6, 7, 8, 9, 10}. Option A is the intersection itself, option C is A ∪ B, and option D omits several elements that belong to the complement.

The governing concept is that a complement is always taken with respect to the stated universal set. First combine all distinct elements: A ∪ B = {1, 2, 3, 5, 7, 9, 11}. Now remove these elements from U = {1, 2, ..., 12}. The elements left are 4, 6, 8, 10, and 12, so (A ∪ B)′ = {4, 6, 8, 10, 12}. Option B lists the union itself, while option C is only the intersection.

By De Morgan’s law, Aᶜ ∩ Bᶜ = (A ∪ B)ᶜ. The union A ∪ B is {2, 3, 4, 6, 8, 9, 10, 12}. Removing these elements from the universal set U leaves {1, 5, 7, 11}. Therefore, Aᶜ ∩ Bᶜ = {1, 5, 7, 11}. Notice that elements such as 6 and 12 are excluded because they belong to both A and B.

Apply De Morgan’s law: Aᶜ ∪ Bᶜ = (A ∩ B)ᶜ. The common elements of A and B are A ∩ B = {4,5}. The complement of {4,5} relative to U contains every other element of U, namely {1,2,3,6,7,8,9,10}. Hence option A is correct. This also shows why the intersection, not the union, is needed first.

The universal set contains the 20 natural numbers from 1 through 20. The primes in this range are 2, 3, 5, 7, 11, 13, 17, and 19, so n(A) = 8. Since the complement contains all non-prime members of U, n(Aᶜ) = n(U) - n(A) = 20 - 8 = 12. Number 1 is not prime and is correctly included in the complement.

The condition x² ≤ 4 is equivalent to -2 ≤ x ≤ 2. Because x must be an integer, A = {-2, -1, 0, 1, 2}. The universal set contains every integer from -4 through 4: {-4, -3, -2, -1, 0, 1, 2, 3, 4}. The elements of U that are absent from A are -4, -3, 3, and 4. Hence Aᶜ = {-4, -3, 3, 4}.

The complement is formed relative to U = (-5,5]. In A = [-1,3), the endpoint -1 is included, so it must be excluded from Aᶜ. The endpoint 3 is excluded from A, so it must be included in Aᶜ. Thus the part before A is (-5,-1), and the part after A is [3,5]. Therefore Aᶜ = (-5,-1) ∪ [3,5].

Since A is a subset of B, every element of A is also in B. Taking complements with respect to the same universal set reverses the direction of inclusion: if X ⊆ Y, then Yᶜ ⊆ Xᶜ. Therefore, A ⊆ B implies Bᶜ ⊆ Aᶜ. Equality is not guaranteed unless A and B are equal, and the other options contradict complement laws.

For complements taken with respect to the same universal set, inclusion reverses when complements are taken. The statement Aᶜ ⊆ Bᶜ can be viewed as Yᶜ ⊆ Xᶜ, which implies X ⊆ Y in reverse notation; hence B ⊆ A. The original inclusion A ⊆ B is not logically forced, and neither equality nor disjointness follows from the given condition.

First apply the inclusion–exclusion formula to the complements: \(n(A^c\cup B^c)=n(A^c)+n(B^c)-n(A^c\cap B^c)=40+55-20=75\). By De Morgan’s law, \(A^c\cup B^c=(A\cap B)^c\). Hence \(n((A\cap B)^c)=75\), and therefore \(n(A\cap B)=n(U)-75=100-75=25\).

With respect to \(U\), \(A^c=\{5,6,7,8,9\}\) and \(B^c=\{1,2,3,7,8,9\}\). The difference \(A^c-B^c\) retains elements in \(A^c\) that are not in \(B^c\), giving \(\{5,6\}\). Equivalently, \(X-Y=X\cap Y^c\), so \(A^c-B^c=A^c\cap B\). Thus option A is correct.

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.

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.

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.

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.