Power Set and Subsets

The power set P(A) contains every subset of A, including A itself. A proper subset must be strictly smaller than the original set, so A is not a proper subset of itself. Every other subset of A has fewer than four elements and is proper. Thus exactly one element of P(A), namely A, is not a proper subset. The answer is option B.

An element of P(A) must be a subset of A, so all of its elements must be 1, 2, or 3. The sets {1, 2}, ∅, and {1, 2, 3} satisfy this requirement. However, {{1}} has the single element {1}, which is itself a set and is not an element of A. Therefore {{1}} is not a subset of A and is not an element of P(A).

The set A has exactly two elements: the empty set ∅ and the singleton set {1}. The fact that ∅ itself is an element does not change the count. For any finite set with n elements, its power set contains 2ⁿ subsets, because each element can either be selected or not selected. Therefore, n(P(A)) = 2² = 4.

The universal set is U = {−2, −1, 0, 1, 2}. The condition x² = 1 gives x = −1 or x = 1, so A = {−1, 1}. Therefore, the complement relative to U is A′ = {−2, 0, 2}, which has 3 elements. A set with n elements has 2ⁿ subsets in its power set. Hence n(P(A′)) = 2³ = 8, so option B is correct. The complement must always be taken with respect to the stated universal set.

For finite sets, n(P(A)) = 2ⁿ(A) and n(P(B)) = 2ⁿ(B). If these two powers are equal, their nonnegative integer exponents must be equal, so n(A) = n(B). This does not imply A = B, because different sets can have the same number of elements; for example, {1,2} and {3,4}.

P(A) = {∅, {1}, {2}, {1,2}} and P(B) = {∅, {3}, {4}, {3,4}}. The only common subset is ∅, because no nonempty subset can simultaneously contain only elements from A and only elements from B when A and B are disjoint. Therefore, their intersection is the set containing ∅, namely {∅}.

First find the power sets: P(A) = {∅, {1}, {2}, {1,2}} and P(B) = {∅, {2}, {3}, {2,3}}. Each power set has 4 elements. Their common elements are ∅ and {2}, so the intersection has 2 elements. Using the union formula, n(P(A) ∪ P(B)) = 4 + 4 − 2 = 6. Therefore, option B is correct. It is important to compare the subsets themselves, not merely the original elements of A and B, when removing duplicates.

The complement of A in U is A′ = {4, 5, 6}. Since A = {1, 2, 3} and A′ = {4, 5, 6} have no common element, the only subset common to both sets is the empty set ∅. Therefore, the only common member of their power sets is ∅ itself, so P(A′) ∩ P(A) = {∅}. It is {∅}, not ∅, because the power set contains the empty set as an element.

A ⊆ U means that every element of A is also an element of U. Now take any element X of P(A). By definition, X is a subset of A, so every element of X belongs to A and therefore also belongs to U. Hence X is a subset of U, which means X ∈ P(U). Since this holds for every X in P(A), we conclude that P(A) ⊆ P(U).

For every finite set X, n(P(X)) = 2ⁿ⁽ˣ⁾. Thus 64 = 2⁶ gives n(A′) = 6. Because A′ is the complement of A within U, A and A′ partition U, so n(A) + n(A′) = n(U). Therefore n(A) = 10 − 6 = 4. Option A is correct. The value 6 is the cardinality of the complement, not of A; the other values do not satisfy the power-set relation.

“At most 2” means the subset may contain 0, 1, or 2 elements. The numbers of such subsets are C(5,0)=1, C(5,1)=5, and C(5,2)=10. Adding these mutually exclusive cases gives 1 + 5 + 10 = 16. Therefore, 16 elements of P(A) have at most two elements, so option C is correct. The count must include the empty set because zero is allowed.

The elements a and b must be included in every desired subset, so they are fixed choices. The remaining three elements c, d, and e may each either be included or omitted independently. Hence there are 2³ = 8 possible choices for the remaining elements. Every such choice produces one subset containing both a and b, so the required number of elements of P(A) is 8. Option B is correct.

There are 2⁵ = 32 total subsets of A. It is easier to count the complement: subsets containing neither a nor b. Such subsets can use only c, d, and e, and each of these three elements may be chosen or omitted, giving 2³ = 8 subsets. Therefore the number containing at least one of a or b is 32 − 8 = 24. This includes subsets containing a only, b only, or both, so option B is correct.

The complement of A with respect to U is A′ = U − A = {2,4,6,8}. It has four elements, and every one of them is even. A set with n elements has 2ⁿ subsets, so P(A′) contains 2⁴ = 16 subsets. Since A′ itself contains only even numbers, every subset of A′ also contains only even numbers, including the empty set. Therefore, the correct answer is 16.

A has three elements, so its power set has 2³ = 8 elements. The power set includes both the empty set ∅ and the original set A. The expression P(A) − {∅, A} removes exactly these two distinct elements from the power set. Therefore, the remaining number of elements is 8 − 2 = 6. These remaining sets are precisely the non-empty proper subsets of A, so option C is correct.

The element 1 is forbidden, while 2 is compulsory. After fixing these conditions, the remaining elements 3, 4, and 5 may each be either included or excluded independently. Thus there are 2 choices for each of three unrestricted elements, giving 2³ = 8 valid subsets. The compulsory element 2 does not add a choice, and the forbidden element 1 has no choice. Therefore option B is correct.

The complement is taken relative to U. Thus A′ = U − A = {1, 3, 5, 7}. We need subsets of A′ that must contain 1 and 7. Those two elements are fixed as included. The remaining elements, 3 and 5, can independently be included or excluded, giving 2 choices for each. Therefore the number of valid subsets is 2² = 4. Option B is correct; 16 counts all subsets of A′ without the required-element condition.

An element of P(A) is itself a subset X of A. Since X ⊆ A, we have X ∩ A = X. For this intersection to be empty, X must be the empty set ∅. Thus the only member of P(A) disjoint from A is ∅, so the number is 1. Although P(A) has 2⁴ = 16 members, every non-empty subset shares at least one element with A. Therefore option B is correct.

The even elements of U form A = {2, 4, 6, 8, 10}. Therefore its complement is A′ = {1, 3, 5, 7, 9}, which has 5 elements. The elements of P(A′) are all subsets of A′. A subset with exactly two elements can be selected in C(5, 2) ways = 5!/(2!3!) = 10. Hence P(A′) contains 10 two-element subsets, so option B is correct.

The set A' ∩ B consists of elements that are in B but not in A. Set B is partitioned into two disjoint parts: A ∩ B and A' ∩ B. Therefore n(B) = n(A ∩ B) + n(A' ∩ B). Substituting the given values gives 53 = 22 + n(A' ∩ B), so n(A' ∩ B) = 53 − 22 = 31. Thus option A is correct.

Rewrite the condition as y = 1 - x and test the elements of A. For x = -1, y = 2, giving (-1,2); for x = 0, y = 1, giving (0,1); for x = 1, y = 0, giving (1,0); and for x = 2, y = -1, which is not in B. Thus exactly three ordered pairs belong to A × B and satisfy the equation. Therefore, option B is correct.