Sets and their representations

The governing concept is converting a bounded parameterized set into roster form. The integers satisfying −2 ≤ n ≤ 2 are −2, −1, 0, 1, and 2. Substitution into x = 2n + 1 gives −3, −1, 1, 3, and 5 respectively. Since a set lists each distinct value once, F = {-3, -1, 1, 3, 5}. Therefore option A is correct. Option B starts with the value obtained from n = −3, which is outside the stated range.

The elements 4, 9, 16, and 25 are the squares of 2, 3, 4, and 5 respectively: 4 = 2², 9 = 3², 16 = 4², and 25 = 5². Therefore the rule is x = n² with n restricted to the natural numbers from 2 through 5. Starting at n = 1 would incorrectly include 1, so option B is not correct. Hence option A gives the exact set-builder form.

The governing concept is interpreting strict inequalities while selecting members satisfying a divisibility condition. Multiples of 6 near the interval are 12, 18, 24, and 30. Because 12 < x < 30 is strict on both sides, 12 and 30 are excluded. The remaining multiples are 18 and 24, so H = {18, 24}. Hence option B is correct; the other choices include at least one boundary value or a number below the interval.

For every real number x, x² is non-negative, so x² ≥ 0. Consequently, x² + 1 ≥ 1, which means that x² + 1 can never equal zero when x is real. The equation would have complex solutions x = i and x = −i, but those are not real numbers. Hence the set of real solutions is empty: I = ∅.

The word STATISTICS contains the letters S, T, A, T, I, S, T, I, C, S. In a set, each element is written only once because repetition does not create a new element. Thus the distinct letters are S, T, A, I, and C. Their order in roster form is not important, so option A correctly represents J.

The elements of K are the squares 0², 1², 2², 3², and 4². Therefore, allowing n to be an integer from 0 through 4 produces exactly {0, 1, 4, 9, 16}. Option A also includes n = 5 and therefore includes 25, so it is not equal to K. The other options produce consecutive or even numbers, not precisely the required squares.

First list the multiples of 3 not exceeding 20: 3, 6, 9, 12, 15, and 18. The phrase “not divisible by 2” removes the even numbers 6, 12, and 18. The remaining odd multiples of 3 are 3, 9, and 15. The number 21 is excluded because it is greater than 20. Hence L = {3, 9, 15}.

For an absolute-value inequality, |x − 2| ≤ 3 means that x is at most 3 units away from 2. Thus, −3 ≤ x − 2 ≤ 3. Adding 2 throughout gives −1 ≤ x ≤ 5. Since x must be an integer, the complete list is −1, 0, 1, 2, 3, 4, and 5. Hence the roster form is option A.

First list the positive natural-number divisors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48. Now apply the condition x > 12. The divisors that are strictly greater than 12 are 16, 24, and 48. The number 12 is excluded because the inequality is strict. Therefore, P = {16, 24, 48}, which is option A.

Factor the quadratic expression: x² − 4x + 3 = (x − 1)(x − 3). By the zero-product property, either x − 1 = 0 or x − 3 = 0, giving x = 1 or x = 3. Both values are integers and satisfy the original equation. Therefore, the roster form of the solution set is Q = {1, 3}.

The governing concept is recognizing that √x ∈ N exactly when x is a perfect square of a natural number. The squares around the stated interval are 9, 16, 25, 36, 49, and 64. Applying the strict condition 10 < x < 50 removes 9 and 64, leaving 16, 25, 36, and 49. Therefore option A is correct. Option B wrongly retains 9, C omits 49, and D includes 11, which is not a perfect square.

For option A, x² ≤ 4 is equivalent to -2 ≤ x ≤ 2. Restricting x to the integers gives exactly -2, -1, 0, 1, and 2, which are all the elements of S. The other options either restrict x to natural numbers or use strict inequalities that exclude one or both boundary values.

Let the tens digit range from 1 through 9. For a digit sum of 9, the corresponding units digits are 8, 7, 6, 5, 4, 3, 2, 1, and 0. This gives 18, 27, 36, 45, 54, 63, 72, 81, and 90. The number 9 is not two-digit, so it is excluded, while 90 is included because 9 + 0 = 9.

Solve the defining equation: 3x + 2 = 11 gives 3x = 9 after subtracting 2 from both sides. Dividing by 3 gives x = 3. Since 3 belongs to the set of integers, it satisfies the stated domain restriction, and the solution set contains exactly this one element. Therefore U = {3}, so option B is correct.

The governing concept is listing positive multiples subject to an upper bound. The positive multiples of 4 begin 4, 8, 12, 16, 20, 24, 28, and so on. The condition x < 25 keeps 4 through 24 and excludes 28 and larger values. Because N⁺ contains positive natural numbers, zero is not included. Thus V = {4, 8, 12, 16, 20, 24}, making option A correct; B includes zero and C violates the bound.

An infinite set has endlessly many elements and no final element. The even natural numbers continue as 2, 4, 6, 8, 10, and so on, without an upper bound, so option C is infinite. Option A is bounded, option B contains only integers from -5 to 5, and option D has exactly twelve months; all three are finite.

To solve x² = 9, take both square roots of 9, giving x = 3 or x = -3. Both values are integers and both satisfy the equation: 3² = 9 and (-3)² = 9. Therefore the set must contain both solutions, not just one of them and not 9 itself. Hence the roster form is {-3, 3}, option C.

The governing concept is substituting every allowed natural-number value into a set-builder rule. Since 1 ≤ n ≤ 5, the possible values are 1, 2, 3, 4, and 5. Using x = 3n − 1 gives 2, 5, 8, 11, and 14 respectively. Therefore the roster form is X = {2, 5, 8, 11, 14}, so option A is correct. Option B reflects 3n − 2, while C lists 3n and D begins with an invalid n = 0.

The primes less than 20 are 2, 3, 5, 7, 11, 13, 17, and 19. Test the additional condition that x + 2 must also be prime: 3 gives 5, 5 gives 7, 11 gives 13, and 17 gives 19. The other candidates fail this condition, so Y = {3, 5, 11, 17}, making option A correct.

The governing concept is listing integers in a half-open interval. The lower condition −2 ≤ x includes −2, while the upper condition x < 4 excludes 4. The consecutive integers from −2 through 3 are −2, −1, 0, 1, 2, and 3. Hence option C is correct. Option A incorrectly includes 4, B omits the allowed endpoint −2, and D omits 0 even though zero is an integer in the interval.