Complement of a Set and Its Properties

The allowed values of \(n\) are 0, 1, 2, 3, 4, and 5. Hence, \(A=\{2^0,2^1,2^2,2^3,2^4,2^5\}=\{1,2,4,8,16,32\}\), which contains six distinct elements. Since the complement is taken relative to \(U\), every element of U not in A belongs to \(A'\). Therefore, \(|A'|=|U|-|A|=32-6=26\). Thus, option B is correct. The value 32 is included in A because the condition permits \(n=5\).

Every multiple of 4 is also a multiple of 2, so \(C\subseteq A\) and \(A\cup B\cup C=A\cup B\). In \(1\) through \(24\), \(|A|=12\), \(|B|=8\), and the common multiples of 2 and 3 are the multiples of 6, of which there are 4. Hence \(|A\cup B|=12+8-4=16\), and the complement has \(24-16=8\) elements.

The intervals A = (1,4] and B = [4,7) meet at 4, and 4 is included in both sets. Therefore, their union is (1,7): 1 is excluded and 7 is excluded. Since the universal set is ℝ, the complement contains every real number outside this open interval. Hence (A ∪ B)' = (-∞,1] ∪ [7,∞). The endpoints 1 and 7 belong to the complement because they do not belong to A ∪ B.

The complement (A ∪ B)' consists of elements of U that are not in A and not in B. Thus, its elements are numbers from 1 to 30 that are divisible by neither 2 nor 3. The number 1 is not divisible by 2 or 3, so it belongs to the complement. Because 1 is the smallest element of U, it is automatically the smallest element of (A ∪ B)'. Therefore, option A is correct.

De Morgan's law states that the complement of a union equals the intersection of the complements: (A ∪ B)' = A' ∩ B'. This identity is valid whether or not A and B overlap. The given condition A ∩ B = ∅ only tells us that A and B are disjoint; it does not change the De Morgan identity. Therefore, option A is the unique correct answer.

Relative to U, A contains every odd number from 1 through 13, so A' = {2,4,6,8,10,12,14}. The set B contains 2,3,5,7,11,13, so B' = {1,4,6,8,9,10,12,14}. The elements common to A' and B' are therefore {4,6,8,10,12,14}. This also follows directly from De Morgan's law: A' ∩ B' = (A ∪ B)'.

The set A contains the open interval (0,2), so its complement contains x ≤ 0 and x ≥ 2 around that interval. It also contains [5,9), so its complement contains x < 5 and x ≥ 9 in the corresponding region; combining the portions gives [2,5) between the two intervals. Therefore, over ℝ, A' = (-∞,0] ∪ [2,5) ∪ [9,∞). Endpoint symbols are determined by whether the endpoint was included in A.

Apply De Morgan's law to the left side: A' ∪ B' = (A ∩ B)'. The condition says that (A ∩ B)' is the entire universal set U. The only subset whose complement is U is the empty set, because no element can belong to A ∩ B. Hence A ∩ B = ∅. This means A and B are disjoint, although it does not imply that either set itself is empty.

By De Morgan's law, A' ∩ B' = (A ∪ B)'. In U, the multiples of 5 are counted by floor(55/5) = 11, and the multiples of 11 by floor(55/11) = 5. The common multiples are multiples of 55; only 55 lies in U, so inclusion–exclusion gives |A ∪ B| = 11 + 5 − 1 = 15. Therefore |A' ∩ B'| = 55 − 15 = 40, so option C is correct.

The universal set U contains the integers from 1 through 21. Its odd elements are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, and 21, so there are 11 odd numbers. Among these, 3, 9, 15, and 21 are divisible by 3 and therefore belong to A, not A′. Removing them leaves 11 − 4 = 7 odd elements in A′. Hence, option B is correct.

Factor the quadratic: x² − 2x − 8 = (x − 4)(x + 2). Since the parabola opens upward, the inequality (x − 4)(x + 2) ≤ 0 holds between and at the roots, so A = [−2,4]. The complement in ℝ contains numbers strictly outside this closed interval. Hence A' = (−∞,−2) ∪ (4,∞). The roots are excluded from the complement because they belong to A.

Let C be the set of students who play cricket and F the set who play football. The number playing at least one game is found by inclusion–exclusion: |C ∪ F| = |C| + |F| − |C ∩ F| = 46 + 38 − 20 = 64. Students playing neither game form the complement of C ∪ F in the class. Thus, 80 − 64 = 16 students play neither game, so option B is correct.

Since A ⊆ B, every element of A is also an element of B. Therefore, if an element is not in B, it cannot be in A. The elements outside B are consequently contained among the elements outside A, giving B′ ⊆ A′. Complementation reverses the direction of a subset relation; this is a fundamental property of complements.

By De Morgan’s law, A′ ∩ B′ = (A ∪ B)′. Among the numbers 1 to 36, 18 are divisible by 2, 12 are divisible by 3, and 6 are divisible by both because they are multiples of 6. Hence |A ∪ B| = 18 + 12 − 6 = 24. Its complement in U therefore has 36 − 24 = 12 elements, so option C is correct.

The first interval includes −3 but excludes 4, while the second includes 10 but excludes 7. Therefore, points not in A are all real numbers less than −3, the interval from 4 through 7 including both endpoints, and numbers greater than 10. Thus A′ = (−∞, −3) ∪ [4, 7] ∪ (10, ∞), which is option A.

A number belongs to both A and B precisely when it is divisible by the least common multiple of 6 and 14. Since lcm(6, 14) = 42, the common elements in U are 42 and 84 only. Thus |A ∩ B| = 2, and the complement contains the remaining 84 − 2 = 82 elements. Therefore, option C is correct.

There are floor(96/6) = 16 multiples of 6 and floor(96/8) = 12 multiples of 8. Numbers counted in both sets are multiples of lcm(6,8) = 24, giving floor(96/24) = 4. Hence |A ∪ B| = 16 + 12 − 4 = 24. The complement therefore has 96 − 24 = 72 elements, so option C is correct.

Factor the quadratic: x² − 9x + 20 = (x − 4)(x − 5). Since the parabola opens upward, the inequality is at most zero for 4 ≤ x ≤ 5. Because the universe contains integers, A = {4, 5}. Removing these two elements from U = {−10, …, 10} gives A′ = {−10, …, 3, 6, …, 10}, listed in option A.

The two intervals overlap, so their union extends continuously from −6 to 5. The left endpoint −6 is excluded because it is excluded from A, and the right endpoint 5 is excluded because it is excluded from B. Thus A ∪ B = (−6, 5). Taking the complement in ℝ gives (−∞, −6] ∪ [5, ∞), including both boundary points in the complement.

The complement operation reverses inclusion. Starting with A′ ⊆ B′ and taking complements of both sides gives (B′)′ ⊆ (A′)′, which simplifies to B ⊆ A. No equality or disjointness is forced by the given condition, so the other choices need not hold. Therefore, option A is the only conclusion that is always valid.