Complement of a Set and Its Properties

First find the intersection: A∩B contains the elements common to both sets, so A∩B={2,4}. The prime symbol denotes complement relative to the stated universal set U. Therefore (A∩B)'=U\(A∩B)={1,2,3,4,5,6,7}\{2,4}={1,3,5,6,7}. Option B is only the intersection, not its complement; the other options omit elements that must remain in the complement.

The union contains every element appearing in A or B: A∪B={2,3,4,6,8,9,10,12}. It has 8 elements. Since the universal set U has 12 elements, its complement has 12−8=4 elements, namely {1,5,7,11}. The same result follows from inclusion–exclusion: n(A∪B)=6+4−2=8 because 6 and 12 are counted in both sets.

A' consists of students who are not in A, so they do not study Mathematics. Similarly, B' consists of students who do not study Physics. Their intersection A'∩B' contains students satisfying both conditions simultaneously: they study neither Mathematics nor Physics. By De Morgan's law, A'∩B'=(A∪B)', which confirms that these are students outside the group studying at least one of the two subjects.

The complement of A is defined relative to the universal set U: A'=U\A, or equivalently A'={x∈U : x∉A}. Since A is empty, none of the elements of U are removed. Consequently every element of U belongs to A', so A'=U={1,2,3,4,5}. The set {∅} is not the same as ∅, and 0 is not even an element of the given universal set.

The complement \(A'\) contains precisely those elements of the universal set \(U\) that are not in \(A\). Therefore, to recover \(A\), remove \(b,d,f\) from \(U\): \(A=U\setminus A'=\{a,b,c,d,e,f\}\setminus\{b,d,f\}=\{a,c,e\}\). Hence option A is correct. Option B is the complement itself, while option C incorrectly includes every element of \(U\).

The complement of \(A\) is formed by taking all elements of the universal set that are not in \(A\). Here \(A\) contains the perfect squares \(1,4,9\) from \(U\). Therefore, \(A'=U\setminus A=\{2,3,5,6,7,8\}\), which is precisely the set of non-perfect-square numbers in \(U\). Hence option B is correct; option A describes \(A\), not its complement.

All complements are taken relative to \(U\). Thus \(A'=U\setminus A=\{5,6,7,8\}\) and \(B'=U\setminus B=\{1,2,7,8\}\). Their union contains every element appearing in either complement: \(A'\cup B'=\{1,2,5,6,7,8\}\). This also agrees with De Morgan’s law, \(A'\cup B'=(A\cap B)'\), since \(A\cap B=\{3,4\}\). Therefore option A is correct.

The governing idea is that the complement reverses the stated condition within U. Numbers in U divisible by neither 2 nor 5 are 1, 3, 7, and 9, so A = {1, 3, 7, 9}. Consequently A′ contains every element divisible by 2 or by 5: the even numbers 2, 4, 6, 8, 10 together with 5. Thus A′ = {2, 4, 5, 6, 8, 10}, making option A correct; option C wrongly omits multiples of 2 other than 10.

The universal set consists of the natural numbers from 1 through 15. Set A contains all non-prime numbers, including 1, because 1 is neither prime nor composite. Therefore, the complement A′ contains exactly the prime numbers in U: 2, 3, 5, 7, 11, and 13. Thus option A is correct.

The multiples of 4 in U are 4, 8, and 12, so A = {4,8,12}. The complement contains all members of U that are not in A. Removing these three elements from U gives A′ = {1,2,3,5,6,7,9,10,11}. Notice that the complement includes even numbers such as 2, 6, and 10 because they are not divisible by 4.

Because the universal set is all integers, A contains every integer from −2 through 3, including both endpoints: {−2,−1,0,1,2,3}. Its complement therefore contains all integers outside this closed interval. In set-builder form, A′ = {x ∈ Z : x < −2 or x > 3}; the endpoints are excluded from the complement.

First form the union: A ∪ B = {a, c, g} ∪ {b, c, h} = {a, b, c, g, h}. The complement is taken relative to the stated universal set U, so we select the elements of U that are absent from this union. The remaining elements are d and e. Therefore, (A ∪ B)' = {d, e}, making option A correct.

The complement A' contains all elements of the universal set U that are not in A. If A' is equal to the entire universal set U, then every element of U must be outside A. Therefore, A contains no elements and must be the empty set: A = ∅. If A were U, its complement would be empty instead. Hence, option A is correct.

The complement of A relative to U is A' = {5, 6, 7, 8, 9}, because these are the elements of U not belonging to A. Now intersect A' with B = {4, 5, 6}. The common elements are 5 and 6; 4 is excluded because it belongs to A. Hence, A' ∩ B = {5, 6}, so option A is correct.

The universal set U contains all 26 lowercase English letters. The set V contains the five vowels a, e, i, o, and u. Its complement V' therefore contains every lowercase letter that is not a vowel. Since V is a subset of U, |V'| = |U| − |V| = 26 − 5 = 21. Hence, option B is correct.

The common elements of A and B are 2 and 4, so A∩B={2,4}. The complement is taken relative to the stated universal set U. Therefore, remove 2 and 4 from U: (A∩B)'=U−{2,4}={1,3,5,6,7,8}. Hence option A is correct. Option B is the intersection itself, while options C and D omit or include elements incorrectly. This also agrees with De Morgan’s law: (A∩B)'=A'∪B'.

The universal set is U={1,2,...,20}, which has 20 elements. The perfect squares in this range are 1, 4, 9, and 16, so A has four elements. The complement A' contains every element of U that is not a perfect square. Therefore, |A'|=|U|−|A|=20−4=16. The complement must be taken relative to U, not to all natural numbers. Hence option C is correct.

Because U contains only nonnegative integers, the condition x^2<25 is satisfied by x=0,1,2,3,4. The value x=5 is excluded because 5^2=25, and the inequality is strict. Thus A={0,1,2,3,4}. The relative complement contains the elements of U outside A, namely A'=U−A={5,6,7,8,9}. Therefore, option A is correct. Checking the boundary value 5 prevents confusing <25 with ≤25.

The universal set U contains the integers −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, and 5, so |U| = 11. The condition |x| ≤ 2 gives A = {−2, −1, 0, 1, 2}, which has 5 elements. The complement relative to U therefore has |A′| = |U| − |A| = 11 − 5 = 6 elements. Hence option C is correct.

Within the universal set U, the numbers not divisible by 2 are the odd numbers, so A = {1,3,5,7,9}. The complement A′ contains every element of U that is not in A. Therefore, A′ consists of the even members of U, namely {2,4,6,8}; 10 is excluded because it is not in U.