Sets and their representations

Correct answer: A. I₁ = {1, 4, 9, 16, 25, 36, 49}

Explanation: Perfect squares in the interval from 1 to 50 are obtained from 1² through 7²: 1, 4, 9, 16, 25, 36, and 49. The next square is 8² = 64, which is greater than 50 and must be excluded. Zero is below the stated interval, and the values in option D are mostly cubes rather than squares. Therefore, option A is correct.

Correct answer: A. J₁ = {1}

Explanation: The number 1 has exactly one positive divisor: 1 itself. Every natural number greater than 1 has at least two positive divisors, namely 1 and the number itself, so none of them qualifies. Under the convention in the question, natural numbers start at 1; zero is not considered because divisibility by zero is not defined in the usual positive-divisor sense. Hence J₁ is the singleton set {1}.

Correct answer: A. K₁ = {35, 42, 49, 56}

Explanation: The multiples of 7 around the required interval are 28, 35, 42, 49, 56, and 63. The strict inequalities 30 < x < 60 exclude 28 and 63, as well as the endpoints 30 and 60 if they were multiples. The remaining values are 35, 42, 49, and 56. Therefore, option A gives the complete roster form.

Correct answer: A. L₁ = {-4, -3, -2, -1, 0, 1}

Explanation: The condition −5 < x excludes −5 because the left inequality is strict. The condition x ≤ 1 includes 1 because the right inequality is inclusive. Therefore, the integers beginning at −4 and ending at 1 are −4, −3, −2, −1, 0, and 1. This complete list is option A; neither endpoint should be handled in the same way.

Correct answer: A. N₁ = {1, 2, 3, 4, 6, 12}

Explanation: Since x = 12/n must be a natural number and n is natural, n must be a positive divisor of 12. The possible divisors are 1, 2, 3, 4, 6, and 12. Their corresponding x-values are 12, 6, 4, 3, 2, and 1. A set is normally written in increasing order, giving {1, 2, 3, 4, 6, 12}. Option B contains the same members but is not the standard increasing roster form; option C incorrectly includes 5.

Correct answer: A. 8

Explanation: The word mathematics has the letters m, a, t, h, e, m, a, t, i, c, and s. In a set, repeated elements are counted only once. Thus the distinct-letter set is {m, a, t, h, e, i, c, s}, which contains 8 elements. Therefore option A is correct. The repeated letters m, a, and t do not increase the cardinality of the set.

Correct answer: A. P₁ = {3, 7}

Explanation: To find the members of P₁, solve the condition x² − 10x + 21 = 0. Factoring gives (x − 3)(x − 7) = 0, so x = 3 or x = 7. Both values are natural numbers, so both belong to the set. The roster form is therefore {3, 7}. Negative values are not roots of this equation, 1 and 21 are not solutions, and 0 is not a solution.

Correct answer: A. R₁ = {11, 13, 17, 19}

Explanation: A two-digit number is at least 10, and the condition x < 20 restricts the search to 10 through 19. Among these numbers, 11, 13, 17, and 19 have no positive divisors other than 1 and themselves, so they are prime. The other candidates are composite: 10, 12, 14, 15, 16, and 18. Therefore option A gives the correct roster form.

Correct answer: A. S₁ = {1, 3, 5, 15}

Explanation: The positive natural-number divisors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. The condition says that the divisor must not be even, so we retain only the odd divisors. Among the listed divisors, 1, 3, 5, and 15 are odd, while 2, 6, 10, and 30 are even. Therefore, the roster form of the set is S₁ = {1, 3, 5, 15}, so option A is correct.

Correct answer: A. U₁ = {1, 2, 4, 5, 10, 20}

Explanation: For 20/x to be a natural number, x must divide 20 exactly. The positive divisors of 20 are 1, 2, 4, 5, 10, and 20. Each of these values makes 20/x a natural number, whereas 3 does not divide 20 and values such as 40 give a non-natural fraction. Hence the roster form is U₁ = {1, 2, 4, 5, 10, 20}.

Correct answer: A. V₁ = {5, 14, 23}

Explanation: We examine every number from 1 through 30 and select those whose digits add to 5. The number 5 has digit sum 5, 14 has digit sum 1 + 4 = 5, and 23 has digit sum 2 + 3 = 5. The other listed alternatives contain numbers such as 15, 25, 30, or 32, whose digit sums are not 5 or whose values are outside the interval 1 to 30. Hence V₁ = {5, 14, 23}, making option A correct.

Correct answer: A. W₁ = {0, 2}

Explanation: Solve the defining equation x² = 2x by bringing all terms to one side: x² − 2x = 0. Factoring gives x(x − 2) = 0. By the zero-product property, x = 0 or x = 2. Both are integers and therefore satisfy the domain restriction. The set is consequently W₁ = {0, 2}. Dividing by x would incorrectly discard the valid solution x = 0.

Correct answer: A. X₁ = {7, 14, 21, 42} — X₁ = {7, 14, 21, 42}

Explanation: The positive divisors of 42 are 1, 2, 3, 6, 7, 14, 21, and 42. The condition x > 6 excludes 1, 2, 3, and 6 because they are not greater than 6. The remaining divisors are therefore 7, 14, 21, and 42. Hence the roster form is X₁ = {7, 14, 21, 42}, making option A correct. Notice that 6 is excluded because the inequality is strict, not x ≥ 6.

Correct answer: A. D = {1, 3, 9}

Explanation: The governing idea is to apply both restrictions in the definition: a member must be a positive factor of 36 and must be odd. The positive factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. Selecting only the odd factors leaves 1, 3, and 9. Thus D = {1, 3, 9}, so option A is correct. Option B omits the valid factor 1; C includes even 6; D includes 12, which is also even.

Correct answer: B. H = {18, 24}

Explanation: 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.

Correct answer: D. I = ∅

Explanation: 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 = ∅.

Correct answer: A. J = {S, T, A, I, C}

Explanation: 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.

Correct answer: A. P = {16, 24, 48}

Explanation: 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.

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

Explanation: 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}.

Correct answer: A. R = {16, 25, 36, 49}

Explanation: 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.

Correct answer: B. U = {3}

Explanation: 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.

Correct answer: A. V = {4, 8, 12, 16, 20, 24}

Explanation: 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.

Correct answer: C. {x : x ∈ N, x is even}

Explanation: 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.

Correct answer: C. W = {-3, 3}

Explanation: 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.

Correct answer: A. X = {2, 5, 8, 11, 14}

Explanation: 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.