Complement of a Set and Its Properties

Factor the quadratic: x² + 2x − 8 = (x + 4)(x − 2). Since the parabola opens upward, the inequality (x + 4)(x − 2) < 0 holds strictly between the roots, so A = (−4, 2). Because the universal set is R, the complement contains all real numbers outside this open interval, including the endpoints −4 and 2. Hence A′ = (−∞, −4] ∪ [2, ∞).

The multiples of 4 in U = {1, ..., 64} are 4, 8, 12, ..., 64, so their number is 64/4 = 16. Among them, the members of A are powers of 2: 4 = 2², 8 = 2³, 16 = 2⁴, 32 = 2⁵, and 64 = 2⁶, giving five numbers. These five are excluded from A′. Hence the required count is 16 − 5 = 11, so option A is correct.

Relative to U, A contains every even number, so A′ is the odd-number set {1, 3, 5, 7, 9, 11, 13, 15, 17}. To obtain A′ ∩ B′, remove from A′ every member that occurs in B. The odd members of B are 1, 3, 5, and 13. Removing them leaves {7, 9, 11, 15, 17}. Therefore option A is correct; option B is simply A′, and option C is A ∩ B.

By De Morgan’s law, A′ ∪ B′ = (A ∩ B)′. A number belongs to A ∩ B when it is divisible by both 10 and 15, hence by their least common multiple, lcm(10, 15) = 30. The multiples of 30 from 1 to 100 are 30, 60, and 90, so n(A ∩ B) = 3. Therefore n(A′ ∪ B′) = 100 − 3 = 97.

The intersection must contain numbers common to both intervals. A extends from −7 inclusive to −1 exclusive, while B extends from −3 exclusive to 5 inclusive. Therefore A ∩ B = (−3, −1); neither endpoint is included because −3 is excluded from B and −1 is excluded from A. The complement of this open interval in R includes both endpoints, giving (−∞, −3] ∪ [−1, ∞).

A ∩ B′ consists of numbers that are odd and not prime. The odd numbers from 1 to 24 are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, and 23. Removing the odd primes 3, 5, 7, 11, 13, 17, 19, and 23 leaves {1, 9, 15, 21}. Note that 1 is not prime, so it remains.

A set and its complement are disjoint, so A′ ∩ A = ∅. The complement is taken relative to the universal set U. The complement of the empty set is the entire universal set, because every element of U is outside ∅. Therefore (A′ ∩ A)′ = (∅)′ = U, making option A correct.

List the numbers from 1 to 50 whose last digit is 2 or 7: 2, 7, 12, 17, 22, 27, 32, 37, 42, and 47. Thus n(A) = 10. The universal set U has 50 elements, and the complement-count formula is n(A′) = n(U) − n(A). Therefore n(A′) = 50 − 10 = 40. Option A is correct. Option B counts A itself, while options C and D result from incorrect counting of the qualifying last digits.

The universal set contains every integer from −9 to 9. Set A consists of integers whose absolute value is greater than 4, namely −9 through −5 and 5 through 9. Its complement therefore contains the integers in U that do not satisfy |x| > 4. The opposite condition is |x| ≤ 4, which gives −4, −3, −2, −1, 0, 1, 2, 3, and 4. Hence option A is correct; option B describes A itself.

Complements are taken with respect to the universal set \(\mathbb{R}\). Set \(A\) contains every real number except 0, so \(A'=\{0\}\). Similarly, \(B\) contains every real number except 1, so \(B'=\{1\}\). Taking the union gives \(A'\cup B'=\{0\}\cup\{1\}=\{0,1\}\). Thus option A is the only correct answer.

Within the given universal set, \(B\) consists of all odd numbers. Therefore, \(B'\) consists of all even numbers from 1 through 35. The multiples of 5 in \(A\) are \(5,10,15,20,25,30,35\). Selecting only the even members of this list gives \(10,20,30\). Hence \(B'\cap A=\{10,20,30\}\), so option A is correct.

First form the union: \(A\cup B=\{1,2,3,4,6,8,9,12,15,16\}\). The complement relative to \(U\) contains every element of \(U\) that does not occur in this union. Checking the integers from 1 to 16 leaves \(5,7,10,11,13,14\). Therefore, \((A\cup B)'=\{5,7,10,11,13,14\}\), which is option A.

Since \(A=(-\infty,2]\), its complement in \(\mathbb{R}\) is \(A'=(2,\infty)\); the endpoint 2 is excluded because it belongs to A. Since \(B=[-1,\infty)\), its complement is \(B'=(-\infty,-1)\); the endpoint −1 is excluded because it belongs to B. No real number can be both greater than 2 and less than −1, so \(A'\cap B'=\varnothing\).

The set \(A\) contains all multiples of 3 in \(U\): \(\{3,6,9,12,15,18,21,24,27\}\). The set \(B\) contains the multiples of 9: \(\{9,18,27\}\). Thus \(B'\) excludes exactly these three numbers. Removing them from A leaves \(\{3,6,12,15,21,24\}\). Therefore, option A is correct.

The numbers from 1 to 49 that are divisible by 7 are 7, 14, 21, 28, 35, 42, and 49, giving seven numbers in total. Among them, only 49 is a perfect square because 49 = 7², so 49 belongs to A and is excluded from A′. The other six numbers are not perfect squares and therefore belong to the complement A′. Hence, the correct answer is 6, option D.

The set A contains all real numbers less than 3 and all real numbers greater than or equal to 8. Its complement must therefore contain the real numbers that are not less than 3 and are not at least 8. These conditions are \(x\ge3\) and \(x<8\), giving \(A'=[3,8)\). The endpoint 3 is included, while 8 is excluded, so option A is correct.

By De Morgan's law, \(A'\cap B'=(A\cup B)'\), so we count numbers from 1 to 60 divisible by neither 2 nor 3. There are 30 multiples of 2, 20 multiples of 3, and 10 multiples of both 2 and 3. Thus \(n(A\cup B)=30+20-10=40\). Therefore, \(n(A'\cap B')=60-40=20\), so option A is correct.

Within U, the condition \(x+1\ge4\) gives \(x\ge3\), so \(A=\{3,4,5,6,7,8\}\). Also, \(B=\{-3,-2,-1,0,1,2,3,4,5\}\). Therefore, \(A'\) contains \(-3,-2,-1,0,1,2\), while \(B'\) contains \(6,7,8\). Their union is \(\{-3,-2,-1,0,1,2,6,7,8\}\), option A.

The universal set U has 32 elements. Exactly 16 of them are even, so A contains 16 elements. The complement A' therefore contains the 16 odd numbers from 1 to 32. For any finite sets X and Y, n(X × Y) = n(X)n(Y). Hence n(A' × A') = 16 × 16 = 256. The order matters in a Cartesian product, and every first component can be paired with every second component.

De Morgan’s law states that the complement of a union is the intersection of the complements: (A ∪ B)' = A' ∩ B'. An element belongs to the left side precisely when it belongs to neither A nor B, which means it belongs to both A' and B'. Option A is incorrect because (A ∩ B)' = A' ∪ B', not A' ∩ B'. Also, the double-complement law gives (A')' = A, not A'.