Power Set and Subsets

The elements of A are the empty set ∅, the singleton set {∅}, and the number 1. A set is a subset of A only when every one of its elements is an element of A. Option A contains ∅ and 1, both of which belong to A. Option B contains 0, option C contains {{∅}}, and option D contains 2; these are not elements of A.

A subset with three or more elements can have exactly 3, 4, or 5 elements. Using combinations, the counts are C(5,3) = 10, C(5,4) = 5, and C(5,5) = 1. Therefore the total is 10 + 5 + 1 = 16, so option B is correct. The value 10 counts only triples, 20 does not correspond to this sum, and 26 includes an incorrect extra category.

A set records membership, not the number of times an item appears. The distinct letters in LEVEL are L, E, and V; repeated occurrences of L and E do not create new elements. Thus A = {L, E, V}, which is exactly B. Therefore A = B and option A is correct.

Solving x² ≤ 4 for integer x gives -2 ≤ x ≤ 2, so A = {-2, -1, 0, 1, 2}. A subset is proper only when it is strictly smaller than the original set. Option A contains every element of A and is therefore equal to A itself; it is a subset but not a proper subset. The other options omit at least one element.

Substitute the allowed values n = 1, 2, 3, and 4 into x = 2n. The resulting values are x = 2, 4, 6, and 8, so B = {2, 4, 6, 8}. This is exactly the same collection of elements as A, hence A = B. Option B is false because A is not a proper subset of B; equal sets are not proper subsets. Option C lists consecutive natural numbers, and option D is false because the intersection is A itself, not the empty set.

The power set P(A) is the set of all subsets of A. Therefore it contains ∅, {1, 2}, {3}, {1}, {2}, {1, 3}, {2, 3}, and {1, 2, 3}. The number 2 by itself is an element of A, but it is not a subset of A. The singleton subset containing 2 would be {2}, which is different from 2. Hence 2 is not an element of B, so option D is correct.

The power set P(A) contains every subset of A as an element. For A = {a, b}, the subsets are the empty set ∅, the singleton sets {a} and {b}, and the set A itself, {a, b}. Thus P(A) = {∅, {a}, {b}, {a, b}}, which is option A. Notice the difference between a and {a}: the first is an element, whereas the second is a singleton subset.

The element 1 is compulsory, so it has no choice: it must be included. The element 4 is forbidden, so it also has no choice: it must be excluded. Only 2 and 3 are free, and each can independently be included or excluded. Therefore the number of valid subsets is 2 × 2 = 2² = 4. They are {1}, {1, 2}, {1, 3}, and {1, 2, 3}. Thus option B is correct.

A is a subset of B, so every subset of A is automatically a subset of B. Conversely, any subset of B that is also a subset of A must use only the elements 1, 2, and 3; it cannot contain 4 or 5. Thus the required subsets are precisely all subsets of A. Since A has three elements, its power set has 2³ = 8 subsets, including the empty set and A itself. Therefore option C is correct.

For natural numbers, x² < 50 gives x < √50, so the possible values are 1 through 7 (and possibly 0 if the convention includes 0). In particular, 1, 2, 3, 4, 5, 6, and 7 belong to A, but 8 does not because 8² = 64, which is greater than 50. Therefore {1, 5, 8} is not a subset of A. The other listed sets contain only elements of A, and the empty set is a subset of every set.

The prime divisors of 24 are 2 and 3, because 24 = 2^3 × 3. The prime divisors of 36 are also 2 and 3, because 36 = 2^2 × 3^2. Therefore, the prime divisors common to both numbers are exactly 2 and 3, so A = {2, 3}. Number 1 is not prime, and 6 or 12 are composite, not prime.

The positive multiples of 25 that are less than 100 are 25, 50, and 75. Thus A has three elements. A set with n elements has 2^n subsets, because each element can either be included or excluded from a subset. Therefore, A has 2^3 = 8 subsets. The strict phrase “less than 100” excludes 100 itself.

If a set has n elements, its power set has 2^n subsets in total. Exactly one of these subsets is the set itself, so the number of proper subsets is 2^n − 1. Given 2^n − 1 = 15, we obtain 2^n = 16. Therefore, the power set has 16 elements. The value 15 counts only proper subsets, not all subsets.

The set A has exactly three elements: {1}, {2}, and {1,2}. Therefore, the set {1,2} itself is an element of A, so option A is correct. The numbers 1 and 2 are elements of the inner sets, but they are not direct elements of A. Option D is another set that is not listed as an element of A. This question tests the important distinction between an element and a set containing elements.

The positive divisors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40. The condition x ≤ 10 allows only 1, 2, 4, 5, 8, and 10. Hence A = {1,2,4,5,8,10}, which is exactly option A. Option B contains divisors greater than 10, while option C omits valid divisors and option D includes numbers that violate the upper-bound condition.

The elements 2 and 3 must be included in every required subset, so they are fixed. The remaining elements, 1 and 4, may each either be included or omitted independently. Thus there are 2 choices for 1 and 2 choices for 4, giving 2 × 2 = 4 subsets. Therefore, option B is correct.

A set with n elements has 2^n total subsets because each element has two choices: included or excluded. For n = 5, A has 2^5 = 32 subsets. Exactly one of them is A itself. Therefore, the number of subsets not equal to A is 32 − 1 = 31, so option C is correct.

For every real number x, x² is non-negative, so x² + 1 is at least 1 and can never equal zero. Consequently, the defining condition has no real solution, and A contains no elements. Therefore A is the empty set ∅. The values i and −i solve the equation only in the complex number system, not in R.

The elements of A are the set {1,2}, the number 3, and the number 4. A subset may contain only elements that are directly in A. Options A, B, and C use 3, 4, and/or the whole element {1,2}, so they are subsets. Option D contains 1 and 2 separately, but neither is a direct element of A; therefore it is not a subset.

Taking N to mean the positive natural numbers here, the condition x ≤ 3 gives A = {1,2,3}. Each element of A generates one singleton subset: {1}, {2}, and {3}. All three singleton subsets belong to P(A), the power set of A. Therefore P(A) has exactly 3 elements that are singleton sets, so option C is correct.