Power Set and Subsets

The element 0 is required to be in every counted subset, so its inclusion is fixed and creates only one choice. The remaining three elements, 1, 2, and 3, may each either be included or excluded independently. Thus there are 2 choices for each of three elements, giving 2 × 2 × 2 = 2³ = 8 subsets. Equivalently, exactly half of the 2⁴ = 16 subsets of A contain any particular fixed element. Therefore option C is correct.

To form a subset that does not contain a, we may choose elements only from {b, c, d, e}. There are four remaining elements, and each element has two independent choices: it may be included or excluded. Therefore, the number of possible subsets is 2^4 = 16. This includes the empty set and every valid combination of the four allowed elements, so option B is correct.

Subset inclusion is transitive: from A ⊆ B and B ⊆ C, we obtain A ⊆ C. The additional condition A ≠ C rules out equality, so A must be a proper subset of C, written A ⊂ C. Nothing in the conditions forces B to equal either A or C, so options B and C are not always true.

A two-element subset is formed by choosing 2 elements from the 4 elements of A, and order does not matter. Hence the number is C(4,2) = 4!/(2!2!) = 6. For example, {p,q} and {q,p} are the same subset, so permutations must not be counted separately. Therefore option B is correct.

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.

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 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 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 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.

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.

Every element of C, namely 3 and 4, is present in A, and A also contains 2, so C is a proper subset of A. Every element of A is present in B, while B has the additional element 5, so A is a proper subset of B. Therefore, the complete proper-subset chain is C ⊂ A ⊂ B. The other options either reverse the inclusion, claim equality, or deny a relation that is clearly true.

Because x is an integer and 0 < x² < 10, x cannot be 0, and its absolute value must be less than √10, which is approximately 3.16. The possible integer values are therefore x = -3, -2, -1, 1, 2, and 3. Their squares are 9, 4, 1, 1, 4, and 9, all strictly between 0 and 10. Hence option A gives exactly A. Option B includes 0, option C omits ±3 and includes 0, and option D includes ±4, whose square is 16.

Any subset of A that contains B must include 1 and 3; these two elements are compulsory. The remaining elements of A are 2 and 4, and each may either be included or excluded independently. Thus there are 2 choices for 2 and 2 choices for 4, giving 2 × 2 = 2² = 4 possible subsets: {1,3}, {1,2,3}, {1,3,4}, and {1,2,3,4}. Therefore option B is correct.

If a finite set has n elements, then its total number of subsets is 2ⁿ. The question gives 2ⁿ = 64. Since 64 = 2⁶, it follows that n = 6. Therefore, A contains 6 elements, making option B correct. For comparison, a set with 5 elements has 32 subsets, one with 7 elements has 128 subsets, and one with 8 elements has 256 subsets. These checks confirm that no other option works.

To form a subset that does not contain 2, we must exclude 2 and may choose freely from the remaining five elements: 1, 3, 4, 5, and 6. Each of these five elements has two choices, included or excluded. Hence the number of valid subsets is 2⁵ = 32. The total number 2⁶ = 64 includes subsets both with and without 2, while 48 is not a power of two and 16 corresponds to only four freely chosen elements.

A subset with exactly 3 elements is formed by choosing 3 elements from the 4 elements of A. The number of such choices is C(4,3) = 4. They are {1,2,3}, {1,2,4}, {1,3,4}, and {2,3,4}. The order of elements does not matter in a set, so arrangements are not counted separately. Therefore, option B is correct.

The element 2 is compulsory, so it has only one choice: it must be included. The element 11 is forbidden, so it also has only one choice: it must be excluded. The remaining three elements, 3, 5, and 7, can each be included or excluded independently. Therefore the number of valid subsets is 2³ = 8, so option B is correct.

A has four distinct elements. Each element can independently be either included or excluded, so the total number of subsets is 2⁴ = 16. Exactly one of these subsets is empty, namely ∅. The phrase “at least one element” excludes that empty subset, so the number of non-empty subsets is 16 − 1 = 15. Hence option A is correct.