Complement of a Set and Its Properties

Correct answer: A. (−1, 4]

Explanation: 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.

Correct answer: A. 40

Explanation: 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.

Correct answer: A. A ⊆ B

Explanation: 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.

Correct answer: A. B ⊆ A

Explanation: 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.

Correct answer: B. 24

Explanation: 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 \(|A'\cap B'|=36-12=24\).

Correct answer: A. \((-1,1)\)

Explanation: 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 \(A\).

Correct answer: C. 42

Explanation: 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.

Correct answer: A. \((-,-2]\cup(6,)\)

Explanation: 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.

Correct answer: A. \(A\cap B=\varnothing\)

Explanation: 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.

Correct answer: D. 10

Explanation: 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 \(20-10=10\).

Correct answer: A. A ∪ B'

Explanation: 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.

Correct answer: A. (-∞, -2] ∪ [8, ∞)

Explanation: 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.

Correct answer: C. \(\varnothing\)

Explanation: 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.

Correct answer: A. ∅

Explanation: 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.

Correct answer: C. 22

Explanation: 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.

Correct answer: A. (1, 9]

Explanation: 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.

Correct answer: C. 25

Explanation: 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.

Correct answer: A. (-∞, 0) ∪ [4, 7] ∪ (10, ∞)

Explanation: 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, ∞).

Correct answer: A. 4

Explanation: 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.

Correct answer: B. 14

Explanation: 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.

Correct answer: B. 3

Explanation: 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.

Correct answer: A. (−4,3]

Explanation: 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].

Correct answer: A. \(A \subseteq B'\)

Explanation: 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.

Correct answer: C. 58

Explanation: 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 \(|(A\cap B)'|=60-2=58\).

Correct answer: A. 1 and prime numbers

Explanation: 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.