Sets and their representations

Every member of P is a positive multiple of 3, so it can be written as 3n, where n is natural. The first term 3 corresponds to n = 1, and the last term 30 corresponds to n = 10 because 30 = 3 × 10. Thus n ranges from 1 through 10, giving P = {3n : n ∈ N, 1 ≤ n ≤ 10}. Option D stops at 27 and omits 30.

We inspect the natural numbers strictly between 10 and 25 and retain only primes. The prime numbers in that interval are 11, 13, 17, 19, and 23. Numbers such as 15 and 21 are composite, so they cannot be included. The endpoints 10 and 25 are excluded by the strict inequalities; 25 is also composite. Hence option A is the correct roster form.

A divisor of 48 is a natural number that divides 48 without leaving a remainder. The relevant positive divisors are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48. The condition x < 10 removes 12 and every larger divisor. Therefore the required roster form is T = {1, 2, 3, 4, 6, 8}. Option A includes every divisor below 10, while the other options either omit a valid divisor or include 12.

To determine the members of U, solve the defining equation x² − 1 = 0. Using the difference of squares, x² − 1 = (x − 1)(x + 1). Hence either x − 1 = 0, giving x = 1, or x + 1 = 0, giving x = −1. Both values belong to the integers, so both must be included. Therefore U = {-1, 1}, making option A correct. Zero does not satisfy the equation because 0² − 1 = −1.

The positive divisors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, and 100. The phrase two-digit means that the number must be at least 10 and at most 99. Filtering the divisor list with this digit condition leaves 10, 20, 25, and 50. The number 100 is a divisor but has three digits, while 1, 2, 4, and 5 have one digit. Thus V = {10, 20, 25, 50}, so option A is correct.

The integers satisfying −3 ≤ x ≤ 3 are −3, −2, −1, 0, 1, 2, and 3. The square of an integer is even exactly when the integer itself is even, because an even number has an even factor and an odd number has an odd square. Among the listed integers, the even values are −2, 0, and 2. Therefore W = {-2, 0, 2}, so option A is correct.

The prime numbers less than 10 are 2, 3, 5, and 7, because each has exactly two positive factors: 1 and itself. Thus the description “x is a prime number less than 10” produces precisely X = {2, 3, 5, 7}. Option B is incomplete because it excludes 2, the only even prime number. Factors of 10 and composite numbers less than 10 produce different sets, so options C and D are incorrect.

Rewrite the equation as x² + x − 12 = 0. Factoring gives (x + 4)(x − 3) = 0, so the algebraic solutions are x = −4 and x = 3. However, the set is restricted to natural numbers. Under the usual school convention, −4 is not a natural number, whereas 3 is natural. Consequently only 3 belongs to Y, and the required set is Y = {3}. Therefore option A is correct.

First solve x² ≤ 9. This gives −3 ≤ x ≤ 3, so the possible integers are −3, −2, −1, 0, 1, 2, and 3. Next solve x + 1 > 0, which gives x > −1. Among the possible integers, this retains 0, 1, 2, and 3; −1 is excluded because the inequality is strict. The intersection is therefore Z = {0, 1, 2, 3}, making option A correct.

The positive multiples of 6 are 6, 12, 18, 24, 30, 36, 42, and so on. The condition x ≤ 36 stops the list at 36, so 42 and all later multiples are excluded. The roster therefore contains exactly 6, 12, 18, 24, 30, and 36. Zero is not listed because the question is using the positive natural multiples intended by the options; 1 is not a multiple of 6. Hence option A is correct.

List the positive divisors of 18 first: 1, 2, 3, 6, 9, and 18. The set requires divisors that are also even. From this list, 2, 6, and 18 are even, whereas 1, 3, and 9 are odd. Therefore the elements satisfying both conditions are C = {2, 6, 18}. Number 4 is not a divisor of 18, so option C is wrong, and option D omits the valid divisor 2. Hence option A is correct.

The elements 1, 8, 27, and 64 are the cubes of 1, 2, 3, and 4 respectively: 1 = 1³, 8 = 2³, 27 = 3³, and 64 = 4³. Therefore, the set can be described as D = {n³ : n ∈ ℕ and 1 ≤ n ≤ 4}. The other options generate squares, powers of 2, or multiples of 4, so they do not produce exactly the given set.

Solve the defining inequality: 2x − 1 ≤ 9 gives 2x ≤ 10 and hence x ≤ 5. Since x belongs to the natural numbers, and natural numbers here are taken as 1, 2, 3, …, the possible values are 1, 2, 3, 4, and 5. Thus the roster form is E = {1, 2, 3, 4, 5}. The equality sign includes 5.

The strict inequality −4 < x < 4 permits the integers −3, −2, −1, 0, 1, 2, and 3; the endpoints −4 and 4 are excluded. The additional condition x ≠ 0 removes 0 from this list. Therefore, F = {−3, −2, −1, 1, 2, 3}. Option C incorrectly retains zero, while option B incorrectly includes the endpoints.

To find G, first list the positive divisors of 42: 1, 2, 3, 6, 7, 14, 21, and 42. From this list, identify the numbers that are prime. The numbers 2, 3, and 7 are prime because each has exactly two positive factors, 1 and itself. The number 1 is neither prime nor composite, while 6, 14, 21, and 42 are composite. Therefore, the set of prime divisors of 42 is G = {2, 3, 7}, so option A is correct.

To determine B, test positive natural numbers against x² < 20. We get 1² = 1, 2² = 4, 3² = 9, and 4² = 16, all below 20. However, 5² = 25, so 5 and every larger positive natural number are excluded. Hence B = {1, 2, 3, 4}, exactly the same set as A. Therefore A = B, making option A correct; B is neither empty nor infinite.

The digits of 2026, in order, are 2, 0, 2, and 6. However, a set records whether an element is present, not how many times it occurs. The digit 2 appears twice but is listed only once in the set. Therefore the distinct-digit set is A = {0, 2, 6}. Option B is correct; options A and D incorrectly preserve repetition, while C omits the digit 0.

A prime number is greater than 1 and has exactly two positive divisors: 1 and itself. Checking the natural numbers below 10 gives 2, 3, 5, and 7 as primes. The number 1 is excluded because it has only one positive divisor, and 4, 6, and 8 are composite. Therefore the correct roster, or listing, form is P = {2, 3, 5, 7}.

The governing concept is converting set-builder notation into roster form. The symbol ℤ restricts x to integers, and the inclusive inequality −2 ≤ x ≤ 2 includes every integer from −2 through 2. Listing them gives −2, −1, 0, 1, and 2, so the set has five elements and option A is correct. Option B lists only the endpoints, option C omits both endpoints, and option D wrongly treats the set as empty.

The interval [2, 5] is a closed interval because square brackets are used at both ends. This means that every real number from 2 through 5 is included, including both endpoints 2 and 5. Among the choices, 2 is included, whereas 1, 6, and 5.5 lie outside the interval. Thus option A is correct.