Sets and their representations

Correct answer: A. {7, 14, 21, 28}

Explanation: The positive natural-number multiples of 7 are 7, 14, 21, 28, 35, and so on. The condition x < 30 retains only 7, 14, 21, and 28. The number 35 is excluded because it is greater than 30; 0 is not included under the natural-number convention used here, and 1 is not a multiple of 7. Thus S = {7, 14, 21, 28}.

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

Explanation: The positive perfect squares are obtained by squaring natural numbers: 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, and 7² = 49. The next square, 8² = 64, is not less than 50. Under the convention ℕ = {1,2,3,...}, zero is not included. Hence T = {1, 4, 9, 16, 25, 36, 49}.

Correct answer: B. 5

Explanation: Because x belongs to the integers, we list only whole-number values, not decimals or fractions. The inequality -2 ≤ x includes -2, while x < 3 excludes 3. Thus the elements are V = {-2, -1, 0, 1, 2}. Counting them gives five elements, so the cardinality of the set is n(V) = 5. The endpoint symbols are essential: the left endpoint is included because of ≤, and the right endpoint is excluded because of <.

Correct answer: B. {-2, -1, 0, 1, 2}

Explanation: The condition |x| < 3 means that the distance of x from zero is less than 3. Therefore, x must lie strictly between −3 and 3. Since x is restricted to integers, the possible values are −2, −1, 0, 1, and 2. The endpoints −3 and 3 are excluded because the inequality is strict, so the roster form is {-2, −1, 0, 1, 2}.

Correct answer: C. {1, 2, 3, 4, 5}

Explanation: Here x is a natural number satisfying both inequalities 1 ≤ x and x ≤ 5. Because both inequalities are non-strict, the endpoints 1 and 5 are included. Listing every natural number from 1 through 5 gives A₁ = {1, 2, 3, 4, 5}. Zero is not included because the lower bound begins at 1.

Correct answer: D. ∅

Explanation: The variable x is required to be an integer and must satisfy both 0 < x and x < 1. There is no integer strictly between 0 and 1; numbers such as 0.5 are real numbers, not integers. The endpoints are also excluded by the strict inequalities. Therefore, the set contains no element and is the empty set, ∅.

Correct answer: C. 4

Explanation: The positive factors of 15 are the positive integers that divide 15 without a remainder. They are 1, 3, 5, and 15, so B₂ = {1, 3, 5, 15}. Counting these distinct members gives four elements. Negative factors such as -1 and -3 are excluded because the question specifically requires positive factors.

Correct answer: B. {6, 12, 18}

Explanation: A natural number divisible by both 2 and 3 must be divisible by their least common multiple, lcm(2, 3) = 6. The positive multiples of 6 not exceeding 18 are 6, 12, and 18. Therefore C₂ = {6, 12, 18}. Numbers divisible by only one of 2 or 3 do not satisfy the word both.

Correct answer: B. {3}

Explanation: The positive natural-number factors of 9 are 1, 3, and 9. Among them, 3 is prime because it has exactly two positive divisors, 1 and 3. The number 1 is not prime, and 9 is composite because it has more than two positive divisors. Therefore, the set C₃ contains only 3, so C₃ = {3}.

Correct answer: C. It is an infinite set.

Explanation: The natural-number multiples of 5 are 5, 10, 15, 20, 25, and so on. This sequence continues indefinitely because multiplying 5 by every positive natural number produces another member. There is no greatest natural-number multiple of 5, so D₁ has infinitely many elements and is an infinite set.

Correct answer: A. {−4, −2, 0, 2, 4}

Explanation: The condition −5 < x < 5 restricts x to the integers −4, −3, −2, −1, 0, 1, 2, 3, and 4. From these integers, the even numbers are −4, −2, 0, 2, and 4. Both endpoints −5 and 5 are excluded because the inequalities are strict. Therefore, the roster form is {−4, −2, 0, 2, 4}. Zero is included because zero is an even integer.

Correct answer: B. {23, 29}

Explanation: The natural numbers strictly between 20 and 30 are 21, 22, 23, 24, 25, 26, 27, 28, and 29. Among them, 23 and 29 are prime: each has exactly two positive divisors, 1 and itself. The other candidates are composite because 21 = 3 × 7, 22 = 2 × 11, 24 is even, 25 = 5 × 5, 26 = 2 × 13, 27 = 3 × 9, and 28 is even. Hence D₃ = {23, 29}.

Correct answer: B. {1, 3, 5, 7, 9, 11}

Explanation: Assuming the standard school convention ℕ = {1, 2, 3, …}, the odd natural numbers not exceeding 11 are 1, 3, 5, 7, 9, and 11. The symbol ≤ means “less than or equal to,” so the endpoint 11 must be included. The set contains only odd numbers, so the even numbers are excluded. Therefore, the roster form is {1, 3, 5, 7, 9, 11}, which is option B.

Correct answer: B. It is a finite set with 12 elements

Explanation: The set consists of the names of the twelve months in a year: January through December. Its membership is clearly defined, and no additional month names are possible in the ordinary calendar-year context. Because it has exactly 12 members, it is finite, not infinite, empty, or a singleton. Therefore, option B is correct.

Correct answer: A. {−3, −1, 1, 3}

Explanation: The inclusive interval −4 ≤ x ≤ 4 contains the integers −4, −3, −2, −1, 0, 1, 2, 3, and 4. Selecting only the odd integers removes −4, −2, 0, 2, and 4, leaving −3, −1, 1, and 3. The endpoints −4 and 4 are included in the interval, but they are even and therefore do not belong to the set. Hence the roster form is {−3, −1, 1, 3}, option A.

Correct answer: B. {1}

Explanation: Solve the equation x² − 1 = 0 by factoring: (x − 1)(x + 1) = 0. Thus the algebraic solutions are x = 1 and x = −1. However, the set-builder condition restricts x to the natural numbers. Under the usual school convention, −1 is not a natural number, whereas 1 is. Therefore only 1 belongs to F₂, so F₂ = {1}. This is a singleton set, making option B correct.

Correct answer: A. {8, 12, 24}

Explanation: The positive natural-number factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. The additional condition x > 6 removes 1, 2, 3, 4, and 6, because 6 is not greater than 6. The remaining factors are 8, 12, and 24. Therefore G₁ = {8, 12, 24}. Option B incorrectly includes 6, while option D contains numbers that are not all factors of 24.

Correct answer: B. A = {0, 1}

Explanation: To find the roster form, solve the condition x² = x. Rearranging gives x² − x = 0, so x(x − 1) = 0. Therefore, x = 0 or x = 1. Both values are integers and satisfy the original equation: 0² = 0 and 1² = 1. Hence the set contains exactly these two distinct elements, so A = {0, 1}.

Correct answer: A. B = {1, 4, 9, 36}

Explanation: The positive factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36. We now retain only those factors that are perfect squares. Among them, 1 = 1², 4 = 2², 9 = 3² and 36 = 6². The number 16 is not a factor of 36, and 6 is not a perfect square. Therefore, B = {1, 4, 9, 36}, making option A correct.

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

Explanation: The inequality |x − 2| ≤ 1 means that x is at a distance of at most 1 from 2. Equivalently, −1 ≤ x − 2 ≤ 1. Adding 2 throughout gives 1 ≤ x ≤ 3. Since x must be an integer, the possible values are 1, 2 and 3. Thus the correct roster form is C = {1, 2, 3}. The endpoints are included because the inequality is non-strict.

Correct answer: A. B = {1, 4, 6, 9, 12, 18, 36}

Explanation: First list the positive divisors of 36: 1, 2, 3, 4, 6, 9, 12, 18 and 36. The prime divisors in this list are 2 and 3, so they must be removed. The number 1 is not prime because a prime number has exactly two positive divisors, whereas 1 has only one. All remaining divisors are composite or 1, giving option A.

Correct answer: A. C = {1, 2, 3, 4, 5, 6}

Explanation: Solve the inequality: 2x + 3 ≤ 15 gives 2x ≤ 12, and hence x ≤ 6. Under the convention used here, ℕ denotes the positive natural numbers 1, 2, 3, and so on. Therefore, the natural numbers satisfying x ≤ 6 are 1, 2, 3, 4, 5 and 6. Thus the roster form is option A. Option B incorrectly includes zero, while C omits 6.

Correct answer: C. 7

Explanation: Use the standard absolute-value inequality rule: |x − 2| ≤ 3 is equivalent to −3 ≤ x − 2 ≤ 3. Adding 2 to all parts gives −1 ≤ x ≤ 5. The integers in this closed interval are −1, 0, 1, 2, 3, 4 and 5. Counting them gives 7 elements, so option C is correct. The closed endpoints must be included because the original sign is ≤.

Correct answer: A. G is a finite set

Explanation: The English alphabet has exactly five commonly recognized vowels: a, e, i, o, and u. These elements are clearly specified, so the set is well-defined. Because its elements can be listed completely and their number is limited to five, G is a finite set. It is neither empty nor infinite.

Correct answer: A. Empty set

Explanation: For any number x, the equation x + 5 = x would require subtracting x from both sides, giving 5 = 0. This is impossible, so no natural number satisfies the stated condition. Therefore H contains no elements. A set containing no element is called the empty set, usually denoted by ∅.