Complement of a Set and Its Properties

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.

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.

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 set A contains all real numbers satisfying x ≤ 1 or x > 9. To find its complement, both conditions must fail simultaneously: x > 1 and x ≤ 9. Their intersection is the interval (1, 9]. The number 1 is excluded because it is already in A, while 9 is included because A contains only numbers strictly greater than 9. Hence option A is correct.

Use De Morgan’s law: A' ∪ B' = (A ∩ B)'. A ∩ B consists of numbers divisible by both 2 and 3, or equivalently by their least common multiple, 6. Between 1 and 30 these are 6, 12, 18, 24, and 30, so there are 5 such numbers. Therefore |A' ∪ B'| = 30 − 5 = 25, making option C correct.

The complement A′ contains all real numbers that are not in A. Since 0 is included in [0, 4), it is excluded from the complement; 4 is excluded from [0, 4), so 4 is included in A′. Similarly, 7 is excluded from (7, 10], while 10 is included in A. Therefore, A′ = (-∞, 0) ∪ [4, 7] ∪ (10, ∞).

Within U, the set B consists of the multiples of 6: 6, 12, 18, 24, 30, 36, 42, and 48, so |B| = 8. Numbers belonging to both A and B must be divisible by lcm(4, 6) = 12. These are 12, 24, 36, and 48, so |A ∩ B| = 4. Hence |A′ ∩ B| = |B| − |A ∩ B| = 8 − 4 = 4.

A contains the numbers in U that are not divisible by 7. Therefore, its complement A′ contains exactly the multiples of 7 in U. These are 7, 14, 21, …, 98. Their count is floor(100/7) = 14, because 7 × 14 = 98 while 7 × 15 = 105 exceeds 100. Hence |A′| = 14.

A′ contains the odd numbers, while B′ contains the non-prime numbers. We need numbers from 1 to 20 that are both odd and non-prime. They are 1, 9, and 15. Note that 1 is not prime, and 3, 5, 7, 11, 13, 17, and 19 are excluded because they are prime. Thus A′ ∩ B′ = {1,9,15}, whose cardinality is 3.

The set A contains every real number less than or equal to −4, and every real number greater than 3. Therefore, the numbers not in A lie between −4 and 3. Because −4 is already included in A, it must be excluded from the complement. Because 3 is not included in A, it must be included in the complement. Hence A′ = (−4,3].

For any sets \(X\) and \(Y\), the equality \(X\cup Y=Y\) holds exactly when every element of \(X\) is already an element of \(Y\). Thus \(X\subseteq Y\). Substituting \(X=A\) and \(Y=B'\) gives \(A\subseteq B'\). The other options either reverse the inclusion or assert equalities that do not follow from the given condition.

An element belongs to both \(A\) and \(B\) exactly when it is divisible by both 6 and 10. Such numbers are multiples of \(\operatorname{lcm}(6,10)=30\). Within \(1\) to \(60\), these are 30 and 60, so \(|A\cap B|=2\). Therefore \(|(A\cap B)'|=60-2=58\).

The complement \(A'\) contains all members of the universal set that are not composite. Among natural numbers, every number greater than 1 is either prime or composite, while 1 is neither prime nor composite. Therefore, within \(1\) to \(35\), \(A'\) consists of 1 together with all prime numbers. Option A is complete.