Complement of a Set and Its Properties

The universal set contains the integers from 0 through 15. Among the options, 2, 3, and 13 are prime, so they belong to A. The number 9 is composite because \(9=3\times3\), and it is in U but not in A. Therefore, 9 belongs to the complement \(A'=U\setminus A\), making C correct.

For any two subsets of the same universal set, taking complements does not change which elements belong to exactly one of the sets. Algebraically, \(A'\Delta B'=(A'\cap B)\cup(B'\cap A)=A\Delta B\). Here \(A\Delta B=\{1,2,5,6\}\), since these elements occur in exactly one of A and B. Therefore, A is correct.

Set A contains all real numbers less than 2 together with all real numbers at least 8. Its complement must contain numbers satisfying neither condition: they are not less than 2 and are less than 8. Thus \(2\le x<8\), represented by \([2,8)\). The endpoint 2 is included because it is excluded from A, while 8 is excluded because it belongs to A.

The inequality \(x^2<16\) is equivalent to \(|x|<4\), which means \(-4<x<4\). Therefore, \(A=(-4,4)\). Since the universal set is all real numbers, the complement consists of the two outside intervals, including the boundary points where equality holds: \(x\le-4\) or \(x\ge4\). Hence \(A'=(-\infty,-4]\cup[4,\infty)\), option A.

The sets A and A' partition the universal set: every element belongs to exactly one of them. Consequently, \(B=(A\cap B)\cup(A'\cap B)\), and similarly \(C=(A\cap C)\cup(A'\cap C)\). The two given equalities make the corresponding parts equal, so their unions are equal. Therefore, \(B=C\), and option A is correct.

The universal set contains the integers from 1 through 25. Since A is the set of odd numbers, its complement A' within U is the set of even numbers. The perfect squares in U are {1, 4, 9, 16, 25}. Among them, the even squares are 4 and 16. Therefore, A' ∩ B = {4, 16}, and its cardinality is 2.

The complement of a set consists of all elements of the universal set that are not in that set. Here A contains the natural numbers up to 100 that are not divisible by 2. Therefore, A' contains precisely those elements of U that are divisible by 2, namely the even natural numbers from 1 to 100.

The condition (A ∪ B)' = ∅ means that A ∪ B = U, because the only set whose complement is empty is the universal set itself. The condition A ∩ B = ∅ says that A and B are disjoint. Thus A and B partition U, so every element outside A must belong to B. Hence B = A'.

The relevant governing concept is the double-complement law. For any subset A of a universal set U, A' contains exactly the elements of U outside A. Taking the complement again removes those outside elements and restores the original members of A, so (A')' = A. Although A is the set of even numbers from 2 through 18, the identity does not require listing them. Therefore option A is correct; A' is only the first complement, while U and ∅ are not generally the result.

The complement A' consists of elements in U that are not in A. The multiples of 3 from 1 to 20 are 3, 6, 9, 12, 15 and 18, so |A| = 6. Since U has 20 elements, the complement-count rule gives |A'| = |U| − |A| = 20 − 6 = 14. Hence option B is correct. The other numerical choices result from miscounting the multiples or subtracting incorrectly.

A complement must be determined relative to the stated universal set, not relative to all integers. The set U contains the integers −3 through 6. Within U, the condition x ≥ 2 gives A = {2, 3, 4, 5, 6}. Therefore the elements left outside A are −3, −2, −1, 0 and 1. Thus A' = {−3, −2, −1, 0, 1}, which is option A. Option B is A itself, while C and D omit or include boundary elements incorrectly.

Use inclusion–exclusion because numbers divisible by both 2 and 3 are counted in both groups. There are 30/2 = 15 multiples of 2 and 30/3 = 10 multiples of 3. Their common multiples are multiples of 6, numbering 30/6 = 5. Hence |A| = 15 + 10 − 5 = 20, and |A'| = 30 − 20 = 10. Therefore option C is correct; adding 15 and 10 without subtracting the overlap would give the wrong count.

The set difference B − A means the elements that are in B but not in A. By definition, “not in A” is represented by the complement A' relative to U. Intersecting B with A' keeps exactly those members of B outside A, so B − A = B ∩ A'. For these sets, the actual elements are {6, 10, 12, 14}, which confirms the identity. The union or the other intersections select different regions.

The interval A begins at -2 and includes -2, while B extends through 7 and includes 7. Together, the intervals overlap and form A ∪ B = [-2, 7]. Since the universal set is R, the complement contains real numbers strictly less than -2 or strictly greater than 7. Thus (A ∪ B)' = (-∞, -2) ∪ (7, ∞).

Apply De Morgan’s law: A' ∩ B' = (A ∪ B)'. The intervals A = (−3, 4] and B = [1, 6) overlap, so their union is (−3, 6). The endpoint −3 is excluded because A excludes it, and 6 is excluded because B excludes it. The complement in R therefore includes both endpoints: (−∞, −3] ∪ [6, ∞). Hence option A is correct; option B incorrectly excludes the boundary points.

The universal set has |U| = 100 elements. The perfect squares in this range are 1², 2², 3², ..., 10², ending at 100, so A contains exactly 10 elements. The complement contains every member of U that is not a perfect square. Therefore |A'| = |U| − |A| = 100 − 10 = 90. Option B is correct. Counting 0 or a square above 100 would be inappropriate because neither belongs to the stated universal set.

First find the intersection of A and B. The only common element is 4, so A ∩ B = {4} and |A ∩ B| = 1. The complement is taken with respect to U, which contains 15 elements. Therefore, |(A ∩ B)′| = |U| − |A ∩ B| = 15 − 1 = 14. Hence, option B is correct.

A′ consists of all elements of the universal set U that do not belong to A. Assuming A ⊆ U, the elements of U are divided into two disjoint parts: A and A′. Hence, |U| = |A| + |A′|, so n = r + |A′|. Rearranging gives |A′| = n − r. Thus option B is correct.

For every element of the universal set U, either it belongs to A or it does not belong to A. The elements not in A form A′, so combining A and A′ gives the whole universal set: A ∪ A′ = U. No element can simultaneously belong to A and its complement, so A ∩ A′ = ∅. Therefore, option A is correct.

A ⊆ B means that every element of A is also an element of B. Consider any element x in B′. Since x is not in B, it cannot be in A either; otherwise A ⊆ B would force x to be in B. Therefore x belongs to A′, proving B′ ⊆ A′. Taking complements reverses the direction of inclusion, so option A is correct.