Complement of a Set and Its Properties

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 \(A\subseteq B\).

Apply the distributive law \((X\cup Z)\cap(Y\cup Z)=Z\cup(X\cap Y)\), with \(X=A\), \(Y=A'\), and \(Z=B'\). The expression becomes \(B'\cup(A\cap A')\). A set and its complement are disjoint, so \(A\cap A'=\varnothing\). Therefore the result is \(B'\cup\varnothing=B'\).

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, ∞).

The complement of a set is taken with respect to the universal set. Every element of the universal set belongs to \(U\), so there is no element left outside it; consequently, the complement of the universal set is empty: \(U'=\varnothing\). Thus the given statement identifies \(U\) as the universal set. The other options are not generally implied.

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 complement of A ∪ B contains the elements of U that are divisible by neither 3 nor 5. The largest element of U is 45, but it is divisible by both 3 and 5. The next number, 44, is not divisible by 3 and is not divisible by 5. Therefore 44 belongs to the complement and is its greatest element. Option A is correct.

If the union of A and B is empty, neither set can contain any element; otherwise that element would belong to their union. Thus A = ∅ and B = ∅. Their complements relative to the universal set U are both U. Consequently A' ∩ B' = U ∩ U = U. This also follows directly from De Morgan’s law: A' ∩ B' = (A ∪ B)' = ∅' = U. Therefore option A is correct.

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.

By De Morgan’s law, A' ∩ B' = (A ∪ B)'. The given condition therefore says that the complement of A ∪ B is the entire universal set U. The only set whose complement is U is the empty set, because ∅' = U. Taking complements on both sides also gives A ∪ B = ∅. Thus option A is the only correct conclusion.

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.