Complement of a Set and Its Properties

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.

One of De Morgan’s laws states that the complement of an intersection equals the union of the complements: (A ∩ B)' = A' ∪ B'. The other related law is (A ∪ B)' = A' ∩ B'. Thus, complementation reverses the operation between intersection and union. Option B fails to change the operation, option C uses the wrong operation, and option D contradicts the double-complement law.

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, ∞).

A set and its complement are disjoint by definition, so A ∩ A′ is always the empty set. Also, A ∪ A′ = U and (A′)′ = A are standard complement identities. Since the universal set U is non-empty, the empty set cannot equal U. Therefore, A ∩ A′ = U is always false.

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.

The universal set U has 18 elements. The given complement A′ contains 7 elements. Since A and A′ partition U into disjoint parts, |A| + |A′| = |U|. Therefore, |A| = 18 − 7 = 11. The actual elements of A are the members of U that are not listed in A′, but only the cardinality is required.

Factor the inequality as x² − 9 = (x−3)(x+3). The product is positive when x < −3 or x > 3, so A = (−∞,−3) ∪ (3,∞). The points −3 and 3 make the expression zero and are not in A. Consequently, the complement in R contains every number from −3 through 3, including both endpoints: A′ = [−3,3].

The difference A − B contains the elements that are in A but not in B. Since A = {1, 2, 3, 4, 5, 6} and the common elements with B are {4, 5, 6}, we get A − B = {1, 2, 3}. The complement is taken with respect to U = {1, 2, …, 12}, so every element of U except 1, 2, and 3 remains. Thus (A − B)′ = {4, 5, 6, 7, 8, 9, 10, 11, 12}.

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.

The symmetric difference A △ B consists of elements belonging to exactly one of A and B. Its complement therefore consists of elements that belong to both sets or to neither set. The ‘both’ part is A ∩ B, and the ‘neither’ part is A′ ∩ B′. Taking their union gives (A △ B)′ = (A ∩ B) ∪ (A′ ∩ B′).

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 \(B'\), relative to \(U\), contains the elements not listed in \(B\); here these are the odd numbers from 1 to 15. From \(A=\{1,4,9,16\}\), the elements that are not in \(B\) are 1 and 9. Hence \(A\cap B'=\{1,9\}\), so option A is correct.

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.

By De Morgan’s law, \(A'\cup B'=(A\cap B)'\). The common part of \(A=[-1,5)\) and \(B=(2,8]\) is \((2,5)\); both endpoints are excluded because 2 is not in \(B\) and 5 is not in \(A\). The real-number complement of \((2,5)\) is \((-infty,2]\cup[5,\infty)\).

The complement is defined by \(A'=U\setminus A\), so it contains the elements of \(U\) that are outside \(A\). If this complement is empty, no element of \(U\) lies outside \(A\). Since \(A\) is a subset of \(U\), it must contain every element of \(U\), and therefore \(A=U\).

The multiples of 3 in \(U\) are \(3,6,9,12,15,18,21,24,27\), so \(|A|=9\) and \(|A'|=27-9=18\). The multiples of 9 are \(9,18,27\), so \(|B|=3\). Because every multiple of 9 is also a multiple of 3, \(B\subseteq A\), which means \(A'\cap B=\varnothing\). Therefore, \(|A'\cup B|=|A'|+|B|=18+3=21\). Thus, option D is correct.