Complement of a Set and Its Properties

A′ contains the elements of U that are not in A. Therefore, A is obtained by removing 1, 5, 7, and 11 from U. The remaining elements are 2, 3, 4, 6, 8, 9, 10, and 12. Hence A = U − A′ = {2, 3, 4, 6, 8, 9, 10, 12}, so option A is correct. The answer depends on the stated universal set.

The set A contains all real numbers satisfying x ≤ −1 or x > 4. Its complement must satisfy the opposite of both conditions simultaneously: x > −1 and x ≤ 4. Therefore A′ = {x : −1 < x ≤ 4} = (−1, 4]. The endpoint −1 is excluded because it belongs to A, while 4 is included because 4 is not greater than 4. Thus option A is correct.

Let H be the set of students taking Hindi and E the set taking English. By the inclusion–exclusion principle, |H ∪ E| = |H| + |E| − |H ∩ E| = 120 + 90 − 50 = 160. Students taking neither language belong to (H ∪ E)′. Therefore their number is 200 − 160 = 40, so option A is correct.

The condition A ∩ B′ = ∅ says that no element of A lies outside B. If an element x belongs to A, it cannot belong to B′, because that would place x in the empty intersection. Therefore x must belong to B. Since every element of A is in B, we conclude A ⊆ B. The other statements are stronger and need not hold, so option A is correct.

Take the complement of both sides of A ∪ B′ = U. Using De Morgan’s law, (A ∪ B′)′ = A′ ∩ B, while U′ = ∅. Hence A′ ∩ B = ∅, which means that no element of B lies outside A. Therefore every element of B belongs to A, so B ⊆ A. Equality or disjointness is not required, making option A correct.

There are \(\lfloor36/4\rfloor=9\) multiples of 4 and \(\lfloor36/9\rfloor=4\) multiples of 9. Their common elements are multiples of \(\operatorname{lcm}(4,9)=36\), so only 36 is common. Therefore, \(|A\cup B|=9+4-1=12\). By De Morgan’s law, \(A'\cap B'=(A\cup B)'\\), hence \(|A'\cap B'|=36-12=24\).

Factor the inequality as \(x^2-1=(x-1)(x+1)\ge0\). The product is non-negative outside the roots, so \(A=(-\infty,-1]\cup[1,\infty)\). Since the universal set is all real numbers, the complement consists of the real numbers strictly between the roots. Thus \(A'=(-1,1)\), and the endpoints are excluded because they belong to \(A\).

An element belongs to both \(A\) and \(B\) exactly when it is divisible by both 3 and 5. Therefore, it must be a multiple of \(\operatorname{lcm}(3,5)=15\). Within \(\{1,\ldots,45\}\), these elements are 15, 30, and 45, so \(|A\cap B|=3\). The complement has \(45-3=42\) elements, making option C correct.

The interval \(A=(-2,6]\) contains every real number greater than \(-2\) and less than or equal to 6. Its complement in \(\mathbb{R}\) therefore contains the numbers not satisfying that condition: all numbers \(x\le-2\), together with all numbers \(x>6\). Hence \(A'=(-\infty,-2]\cup(6,\infty)\). The endpoint rules reverse in the complement.

There are \(72/4=18\) multiples of 4 and \(72/9=8\) multiples of 9. Numbers counted in both sets are multiples of \(\operatorname{lcm}(4,9)=36\); these are 36 and 72, so there are 2. Thus \(|A\cup B|=18+8-2=24\). The complement within a 72-element universal set has \(72-24=48\) elements.

Solve the inequality by factoring: \(x^2-4x=x(x-4)\le0\). Therefore, the real solution interval is \([0,4]\). Since the universal set contains only integers from -8 through 8, \(A=\{0,1,2,3,4\}\). Removing these five integers from \(U\) leaves \(A'=\{-8,-7,-6,-5,-4,-3,-2,-1,5,6,7,8\}\), which is option A.

The statement \(A\subseteq B'\) says that every element of \(A\) lies outside \(B\). Consequently, no element can belong to both sets, so \(A\cap B=\varnothing\). The union need not be empty: both sets may contain many elements while remaining disjoint. Also, the condition does not imply that \(B\subseteq A\) or that the two complements are equal.

The set \(B\) contains the 20 odd numbers from 1 through 40. The odd prime numbers in this range are 3, 5, 7, 11, 13, 17, 19, 23, 29, and 31, giving 10 elements. Notice that 1 is odd but not prime, while 2 is prime but not odd. Therefore, the odd numbers that are not prime, namely \(A'\cap B\), number \(20-10=10\).

Apply De Morgan’s law to the complement of an intersection: (A' ∩ B)' = (A')' ∪ B'. The complement of A' is A, so the result becomes A ∪ B'. Thus, option A is correct. The important rule is that complementation changes an intersection into a union and complements each individual set.

Solve the absolute-value inequality: |x − 3| < 5 means −5 < x − 3 < 5. Adding 3 throughout gives −2 < x < 8, so A = (−2, 8). Since the universal set is all real numbers, the complement contains the endpoints and all values outside the interval: A' = (−∞, −2] ∪ [8, ∞). Therefore, option A is correct.

By De Morgan’s law, the intersection of the complements equals the complement of the union: \(A'\cap B'=(A\cup B)'\). The question states that \(A\cup B=U\), where \(U\) is the universal set. The complement of the universal set contains no elements, so \(U'=\varnothing\). Hence \(A'\cap B'=\varnothing\), and option C is correct. Option D is a different expression: by De Morgan’s law, \(A'\cup B'=(A\cap B)'\), so it cannot be selected.

Use inclusion–exclusion. There are 45 multiples of 2, 30 of 3, and 18 of 5. Pairwise overlaps contain 15 multiples of 6, 9 of 10, and 6 of 15; the triple overlap contains 3 multiples of 30. Thus |A ∪ B ∪ C| = 45 + 30 + 18 − 15 − 9 − 6 + 3 = 66. Therefore, the complement has 90 − 66 = 24 elements, so option B is correct.

By De Morgan’s law, (A ∪ B)' = A' ∩ B'. The given sets A' = {b, e, i} and B' = {a, d, j} have no common element. Their intersection is therefore empty: A' ∩ B' = ∅. Hence (A ∪ B)' = ∅, so option A is correct. The other listed sets either combine elements or select only one complement.

A' ∩ B' contains elements that are neither perfect squares nor multiples of 5. Among the choices, 16 is a perfect square, 20 is a multiple of 5, and 25 is both a perfect square and a multiple of 5. The number 22 is neither a perfect square nor divisible by 5, so it belongs to both A' and B'. Therefore, option C is correct.

The difference A − B consists of elements that belong to A but do not belong to B; equivalently, A − B = A ∩ B'. If this set is empty, there is no element of A outside B. Therefore every element of A must be in B, which is exactly the statement A ⊆ B. The other conclusions do not necessarily follow.