Complement of a Set and Its Properties

By De Morgan’s law, A' ∪ B' = (A ∩ B)'. An element belongs to A ∩ B when it is divisible by both 3 and 7, so it must be divisible by lcm(3, 7) = 21. Between 1 and 42, the multiples of 21 are 21 and 42; hence |A ∩ B| = 2. Therefore, |A' ∪ B'| = |U| − |A ∩ B| = 42 − 2 = 40. Thus option B is correct.

The complement is taken in R. The interval (-∞, -1) excludes -1, so -1 belongs to the complement and the left endpoint is included. The interval [3, 8) includes 3 but excludes 8, so 3 is omitted from the complement while 8 is included. All real numbers from -1 through values less than 3, together with 8 and larger numbers, give A' = [-1, 3) ∪ [8, ∞). Hence option A is correct.

A' ∩ B consists of elements that are divisible by 7 but not divisible by 5. From 1 to 70, B contains 70/7 = 10 numbers. The numbers divisible by both 5 and 7 are multiples of lcm(5, 7) = 35, namely 35 and 70, so there are 2 such numbers. Therefore, |A' ∩ B| = 10 − 2 = 8, making option C correct.

Factor the quadratic: x² − 4x − 12 = (x − 6)(x + 2). Because the parabola opens upward, the expression is negative between its roots, so A = (-2, 6). The complement in R contains the endpoints and the two outside intervals. Therefore, A' = (-∞, -2] ∪ [6, ∞), which is option A.

First calculate the difference B − A. The elements 6 and 12 occur in both sets, so they are removed from B, leaving B − A = {9, 15, 18}. The complement is taken with respect to U = {1, ..., 18}; therefore remove 9, 15, and 18 from U. The result is {1,2,3,4,5,6,7,8,10,11,12,13,14,16,17}, so option A is correct.

The symmetric difference A Δ B consists of elements belonging to exactly one of A or B. In logical terms, these elements satisfy exclusive OR. Taking its complement selects all elements for which exclusive OR is false: an element is either in both A and B or in neither set. Thus (A Δ B)' = (A ∩ B) ∪ (A' ∩ B'), represented by option A.

A' contains the odd numbers, while B' contains the non-prime numbers. Therefore, A' ∩ B' consists of odd composite numbers and 1, because 1 is not prime. From 1 to 30, these elements are {1, 9, 15, 21, 25, 27}. The number 1 must not be excluded merely because it is neither prime nor composite; it is non-prime. Hence the cardinality is 6, so option A is correct.

Because \(A\subseteq B\), every element of \(A\) is also an element of \(B\). Now take any element \(x\in B'\). It is not in \(B\); therefore it cannot be in the smaller set \(A\), so \(x\in A'\). Hence every element of \(B'\) belongs to \(A'\), giving \(B'\subseteq A'\). Complementation reverses inclusion.

The identity \(A'\cap B'=A'\) says that intersecting \(A'\) with \(B'\) removes nothing from \(A'\). This can happen only when every element of \(A'\) is already in \(B'\). Therefore \(A'\subseteq B'\). Taking complements would equivalently give \(B\subseteq A\), not necessarily \(A\subseteq B\).

An element belongs to both \(A\) and \(B\) exactly when it is divisible by both 12 and 21. Such numbers are multiples of \(\operatorname{lcm}(12,21)=84\). Within \(\{1,2,\ldots,84\}\), the only common multiple is 84, so \(|A\cap B|=1\). Hence \(|(A\cap B)'|=84-1=83\).

The complement B′ contains every element of U that is not in B. To form A ∩ B′, retain only those members of A that are absent from B. The set A is {1, 4, 9, 16}; among these, 9 is also in B, so it must be removed. The remaining elements 1, 4, and 16 are in A but not in B, giving A ∩ B′ = {1, 4, 16}. Thus option A is correct; option B is the common part A ∩ B.

By De Morgan’s law, A′ ∪ B′ = (A ∩ B)′. The intersection of A = (−2, 4] and B = [1, 7) is [1, 4], because both intervals contain every real number from 1 through 4, including both endpoints. Taking the complement in ℝ excludes [1, 4], giving (−∞, 1) ∪ (4, ∞).

There are \(\lfloor32/4\rfloor=8\) multiples of 4 in \(U\), so \(|A'|=32-8=24\). Every multiple of 16 is also a multiple of 4, hence \(B\subseteq A\), which means \(A'\cap B=\varnothing\). The multiples of 16 in the universe are 16 and 32, so \(|B|=2\). Therefore \(|A'\cup B|=24+2=26\).

For \(n=1,2,3,4,5,6\), the distinct values of \(2^n\) are \(2,4,8,16,32,64\). Thus \(|A|=6\). The universal set \(U=\{1,2,\ldots,64\}\) contains 64 elements. Since \(A'\) contains all elements of \(U\) not in \(A\), \(|A'|=|U|-|A|=64-6=58\). Neither 0 nor 1 belongs to \(A\).

Every multiple of 6 is already a multiple of both 2 and 3, so C is contained in A ∩ B and does not add any new element to the union. In U there are 18 multiples of 2 and 12 multiples of 3; 6 numbers are multiples of both. Therefore |A ∪ B| = 18 + 12 − 6 = 24. The complement has 36 − 24 = 12 elements, so option C is correct.

The interval A contains every real number from −3 inclusive up to, but not including, 2. The interval B contains numbers greater than 2 up to and including 6. Thus their union leaves exactly the point 2 uncovered between the intervals. Numbers less than −3 and greater than 6 are also outside the union. Hence the complement in R is (−∞, −3) ∪ {2} ∪ (6, ∞), which is option A.

The set A contains all odd numbers from 1 through 15, so A' within U is {2, 4, 6, 8, 10, 12, 14, 16}. To obtain A' ∩ B', remove from this list the elements that occur in B. The only such element is 8. Therefore the result is {2, 4, 6, 10, 12, 14, 16}; however, because 2 is in B, it must also be removed. Thus A' ∩ B' = {4, 6, 10, 12, 14, 16}, option A.

The first part of A includes −5 but excludes −1, while the second part excludes 2 but includes 7. Therefore numbers outside A are less than −5, from −1 through 2 including both endpoints, and greater than 7. In interval notation the complement is (−∞, −5) ∪ [−1, 2] ∪ (7, ∞). Hence option A is correct.

There are floor(77/7) = 11 multiples of 7 and floor(77/11) = 7 multiples of 11. The common multiples are multiples of lcm(7, 11) = 77, so only 77 is common and |A ∩ B| = 1. Hence |A ∪ B| = 11 + 7 − 1 = 17. By De Morgan’s law, A' ∩ B' = (A ∪ B)', so its size is 77 − 17 = 60. Option A is correct.

The even numbers from 1 through 33 are 2, 4, ..., 32, so there are floor(33/2) = 16 of them. An even number that is also divisible by 3 is a multiple of 6. The multiples of 6 up to 33 are 6, 12, 18, 24, and 30, giving 5 numbers. Therefore the number of even elements not divisible by 3, and hence in A', is 16 − 5 = 11. Option B is correct.

The equality \(A'\cup B=B\) means that adding every element of \(A'\) to \(B\) does not enlarge \(B\). Therefore, every element of \(A'\) must already be an element of \(B\), so \(A'\subseteq B\). The reverse inclusion, equality of the sets, and disjointness are not forced by the given condition.

Every multiple of 18 is also a multiple of 6, so \(B\subseteq A\). Hence \(A\cup B=A\). Applying De Morgan’s law gives \(A'\cap B'=(A\cup B)'=A'\). Notice that although \(A'\subseteq B'\), the intersection of the two complements is the smaller complement, namely \(A'\).

By De Morgan’s law, \(A'\cup B'=(A\cap B)'\). A number belonging to both \(A\) and \(B\) must be divisible by \(\operatorname{lcm}(6,9)=18\). The multiples of 18 from 1 through 54 are 18, 36, and 54, so \(|A\cap B|=3\). Therefore, \(|A'\cup B'|=54-3=51\).

The universal set is U = {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}. The condition x² < 10 means |x| < √10. Since 3 < √10 < 4 and x must be an integer, the possible values are −3 through 3, so A = {−3, −2, −1, 0, 1, 2, 3}. The complement contains every element of U that is not in A. Therefore, A′ = U \ A = {−5, −4, 4, 5}.

Among the integers from 1 to 30, 15 are divisible by 2 and 10 are divisible by 3. The numbers divisible by both 2 and 3 are the multiples of 6, and there are 5 of them: 6, 12, 18, 24, and 30. By inclusion–exclusion, n(A ∪ B) = 15 + 10 − 5 = 20. The complement contains the remaining elements of U, so n((A ∪ B)′) = 30 − 20 = 10. Hence option A is correct.