Sets and their representations

Correct answer: A. P = {2, 3}

Explanation: Factor the quadratic expression: x² - 5x + 6 = (x - 2)(x - 3). For the product to be zero, either x - 2 = 0 or x - 3 = 0. Thus x = 2 or x = 3. Both values belong to the integers, so both are elements of P. Consequently, the roster form is P = {2, 3}, making option A correct.

Correct answer: A. Q = {4, 16, 36, 64}

Explanation: The natural-number perfect squares less than 100 are 1, 4, 9, 16, 25, 36, 49, 64, and 81. Selecting only the even squares leaves 4, 16, 36, and 64. The number 100 is excluded because the condition is strictly less than 100. Therefore Q = {4, 16, 36, 64}.

Correct answer: A. S = {1, 2, 4, 7, 8, 11, 13, 14}

Explanation: The positive integers less than 15 are 1 through 14. Since 15 = 3 × 5, a number is coprime to 15 precisely when it is divisible by neither 3 nor 5. Testing the integers gives 1, 2, 4, 7, 8, 11, 13, and 14. Their greatest common divisor with 15 is 1, while every omitted number shares a factor with 15. Hence option A is correct.

Correct answer: A. T = {1, 2, 3, 4, 6, 12}

Explanation: For x = 12/n to be a natural number, n must be a positive divisor of 12. The positive divisors of 12 are 1, 2, 3, 4, 6, and 12. Substitution gives x-values 12, 6, 4, 3, 2, and 1. Since sets do not depend on order, these values form T = {1, 2, 3, 4, 6, 12}. Zero is not obtained, so option A is correct.

Correct answer: A. V = {2, 4, 6}

Explanation: Because x/2 must be a natural number, x must be twice a positive natural number; therefore x is a positive even integer. The interval −5 < x < 7 contains the even integers −4, −2, 0, 2, 4, and 6, but only 2, 4, and 6 produce positive natural quotients. Thus V = {2, 4, 6}, so option A is correct.

Correct answer: A. W = {5, 15, 25, 35, 45}

Explanation: First list the multiples of 5 between 1 and 50: 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50. The phrase “but not by 10” removes 10, 20, 30, 40, and 50, because these are multiples of 10. The remaining values are 5, 15, 25, 35, and 45. Thus the roster form is option A.

Correct answer: A. X = {6, 8, 12, 24}

Explanation: The positive natural-number divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. We now apply x² > 24. For x = 1, 2, 3, and 4, the squares are not greater than 24; in particular, 4² = 16. For 6, 8, 12, and 24, the squares exceed 24. Therefore X = {6, 8, 12, 24}, so option A is correct.

Correct answer: A. A = {4, 12, 20}

Explanation: The positive multiples of 4 that do not exceed 25 are 4, 8, 12, 16, 20, and 24. Among these, 8, 16, and 24 are divisible by 8, so they must be excluded. The remaining numbers, 4, 12, and 20, are divisible by 4 but not by 8 and are all at most 25. Therefore option A is correct.

Correct answer: A. B = {x : x = n² + 1, n ∈ N, 1 ≤ n ≤ 5}

Explanation: For n = 1, 2, 3, 4, and 5, the expression n² + 1 gives 2, 5, 10, 17, and 26 respectively. Therefore, the rule generating every element is x = n² + 1. The restriction 1 ≤ n ≤ 5 is essential because it produces exactly the five listed elements; without this restriction, option D would generate infinitely many additional values.

Correct answer: A. D = {-3, 2}

Explanation: The governing concept is roster representation using the integer solutions of the defining equation. Factor the quadratic: x² + x − 6 = (x + 3)(x − 2). Therefore (x + 3)(x − 2) = 0 gives x = −3 or x = 2. Both values belong to the integers, so D = {−3, 2}. The other options arise from incorrect signs or from using roots that do not satisfy the equation. Hence A is correct.

Correct answer: A. 14

Explanation: Because 2 is positive, multiplying the compound inequality by 2 preserves both inequality signs: −6 ≤ x < 8. Since x must be an integer, the values are −6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6, and 7. Counting them gives 8 − (−6) = 14 integers, with the upper endpoint 8 excluded. Therefore option A is correct.

Correct answer: A. F = {1, 2, 6, 18}

Explanation: The set is formed by filtering the positive divisors of 18 according to a second condition. The divisors are 1, 2, 3, 6, 9, and 18. Adding 1 gives 2, 3, 4, 7, 10, and 19. The prime results are 2, 3, 7, and 19, corresponding to x = 1, 2, 6, and 18. Thus F = {1, 2, 6, 18}; option A is correct.

Correct answer: A. G = {1, 2, 6, 18}

Explanation: To convert the condition into roster form, list the positive divisors of 18: 1, 2, 3, 6, 9, and 18. Check x + 1 for each: 2, 3, 4, 7, 10, and 19. Only 2, 3, 7, and 19 are prime, so the corresponding x-values are 1, 2, 6, and 18. Therefore G = {1, 2, 6, 18}, making option A correct; the other options omit or include unsuitable divisors.

Correct answer: A. H = {-3, -2, -1, 0, 1, 2, 3}

Explanation: For integer x, the inequality x² ≤ 9 is equivalent to |x| ≤ 3, or −3 ≤ x ≤ 3. Every integer in this closed interval satisfies the condition: −3, −2, −1, 0, 1, 2, and 3. Values outside the interval have square greater than 9. Hence option A is the complete roster form; B omits negative integers, C omits the endpoints, and D includes only two values.

Correct answer: A. L = {−3, −2, −1, 0, 1}

Explanation: The condition |x + 1| < 3 is equivalent to −3 < x + 1 < 3. Subtracting 1 throughout gives −4 < x < 2. Since x must be an integer, the possible values strictly between −4 and 2 are −3, −2, −1, 0, and 1. The endpoints −4 and 2 are excluded because the inequality is strict, so option A is correct.

Correct answer: A. V = {12, 6, 4, 3, 2, 1}

Explanation: For x = 24/n to be a natural number, n must be a positive divisor of 24. The even positive divisors of 24 are 2, 4, 6, 8, 12, and 24. Substitution gives x = 12, 6, 4, 3, 2, and 1, respectively. Therefore V = {12, 6, 4, 3, 2, 1}; the order of elements does not matter in a set.

Correct answer: A. A = {-2, -1, 0, 1, 2, 3}

Explanation: The variable x is restricted to integers. The inequality x² < 10 allows the integers -3 through 3, because 3² = 9 is less than 10 but 4² = 16 is not. The additional condition x > -3 excludes -3 itself, while -2, -1, 0, 1, 2, and 3 satisfy both conditions. Hence A = {-2, -1, 0, 1, 2, 3}.

Correct answer: A. B = {x : x is an even positive integer and 2 ≤ x ≤ 10}

Explanation: The listed elements are precisely the even positive integers from 2 through 10, inclusive. Option A states all three necessary facts: x is positive, x is even, and its value lies between 2 and 10. Option B includes infinitely many extra even numbers, while C and D include odd numbers. Therefore A is the exact set-builder description.

Correct answer: A. P = {1, 2, 3, 6}

Explanation: A member of P must be a natural-number divisor of both 18 and 24. The positive divisors of 18 are 1, 2, 3, 6, 9, and 18. The positive divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. The common elements in these two lists are 1, 2, 3, and 6. Therefore, P = {1, 2, 3, 6}.

Correct answer: A. S = {n² : n ∈ ℕ, 1 ≤ n ≤ 5}

Explanation: The listed elements are consecutive squares: 1 = 1², 4 = 2², 9 = 3², 16 = 4², and 25 = 5². Thus every element has the form n², where n is a natural number from 1 through 5. The correct set-builder description is S = {n² : n ∈ ℕ, 1 ≤ n ≤ 5}. The other options describe consecutive natural numbers, even numbers, or cubes.

Correct answer: A. C = {1, 3}

Explanation: The roster form lists every element that satisfies both conditions. The positive divisors of 12 are 1, 2, 3, 4, 6, and 12. Among these, only 1 and 3 are odd; 2, 4, 6, and 12 are even and must be excluded. Therefore, C = {1, 3}, so option A is correct. Option B lists all positive divisors without applying the oddness condition, while option D includes 5, which is not a divisor of 12.

Correct answer: A. {x ∈ N : x < 1}

Explanation: Under the stated convention, N = {1, 2, 3, ...}, so no natural number is less than 1. Therefore the set in option A has no elements and is empty. Option B contains 0, option C contains 1, and option D also contains the integer 0. Specifying the convention for N removes any ambiguity about whether zero is included.

Correct answer: A. D = {−1, 0, 1, 2, 3, 4, 5}

Explanation: The condition |x − 2| ≤ 3 means that x is at a distance of at most 3 from 2. Therefore, subtracting and adding 3 gives −1 ≤ x ≤ 5. Since x must be an integer, we list every integer in this closed interval: −1, 0, 1, 2, 3, 4, and 5. Hence the roster form is D = {−1, 0, 1, 2, 3, 4, 5}, so option A is correct.

Correct answer: A. T = {2, 3}

Explanation: To find the members of T, solve the condition x² − 5x + 6 = 0. Factoring gives (x − 2)(x − 3) = 0, so x = 2 or x = 3. Both values are natural numbers, so both satisfy the restriction x ∈ ℕ. Therefore, the set contains exactly these two elements: T = {2, 3}. Thus option A is correct; the other options either contain incorrect roots or include numbers that are not solutions.

Correct answer: A. L = {12, 21, 30}

Explanation: Let the tens digit be a and the units digit be b. For a two-digit number, a is at least 1, and a + b = 3. The possible digit pairs are (1,2), (2,1), and (3,0), producing 12, 21, and 30. The number 3 is not two-digit, and 11 has digit sum 2. Therefore option A is correct.