Sets and their representations

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, ∅.

B₁ contains every real number strictly between 1 and 4, including numbers such as 1.5, 2.25, and 3.9. It is an infinite set, so its members cannot be completely listed in an ordinary finite roster. The interval notation (1, 4) and the given set-builder notation describe it correctly, but {2, 3} contains only two members and omits infinitely many others.

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.

The listed members 2, 3, 5, 7, and 11 are exactly the prime natural numbers less than 12. Option A states both necessary conditions: x must be prime and x must be less than 12. Option B also includes non-primes such as 1 and 9, while C includes every smaller natural number and D wrongly excludes 2.

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.

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}.

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.

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.

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}.

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.

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.

Using ℕ = {1, 2, 3, …}, consider the natural numbers less than 20, so 20 itself is excluded. Multiples of 2 below 20 are 2, 4, 6, 8, 10, 12, 14, 16, and 18. Multiples of 5 below 20 are 5, 10, and 15. Taking the union gives {2, 4, 5, 6, 8, 10, 12, 14, 15, 16, 18}; 10 is listed only once. Thus A is correct.

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.

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.

The listed elements are consecutive cubes: 1 = 1³, 8 = 2³, 27 = 3³, and 64 = 4³. Therefore, the elements can be described as x³, where x is a natural number from 1 through 4 inclusive. Option A produces squares, option C produces natural numbers, and option D omits 1 and 64. Hence B is correct.

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.

For integer x, the inequality x² ≤ 1 means that the absolute value of x is at most 1, or −1 ≤ x ≤ 1. The integers in this interval are −1, 0, and 1. Directly checking confirms that (−1)² = 1, 0² = 0, and 1² = 1, all of which satisfy the inequality. The integers −2 and 2 have square 4 and are excluded. Hence G₂ = {−1, 0, 1}, so A is correct.

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}.

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.

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.