Sets and their representations

We need all integers x whose square is 16. Solving x² = 16 gives x = √16 or x = -√16, so x = 4 or x = -4. Both values are integers and both satisfy the condition because 4² = 16 and (-4)² = 16. Therefore, the set must contain both values, written without repetition as I = {-4, 4}.

Because x is an integer and must satisfy −2 < x, the value −2 is excluded. The condition x ≤ 3 includes 3. The integers strictly greater than −2 and less than or equal to 3 are therefore −1, 0, 1, 2, and 3. Writing these elements in roster form gives K = {−1, 0, 1, 2, 3}, so option A is correct. The distractors mishandle one or both endpoints.

The governing idea is set-builder representation: describe every member by a rule and restrict the parameter to the required values. The listed numbers are consecutive cubes: 1 = 1³, 8 = 2³, 27 = 3³, 64 = 4³, and 125 = 5³. Therefore x = n³ with 1 ≤ n ≤ 5 gives exactly the five listed elements. Option B gives squares, option C gives multiples of 3, and option D gives infinitely many cubes. Hence A is correct.

We solve the defining equation x³ = x by bringing all terms to one side: x³ − x = 0. Factoring gives x(x² − 1) = 0, or x(x − 1)(x + 1) = 0. Therefore, x can be 0, 1, or −1. All three values are integers, so all satisfy the condition defining R. Hence the set is R = {−1, 0, 1}, which is option A.

A set records membership, not the number of times an element is written. Thus, repeated entries of 1 and 2 do not create new elements. Removing repetitions gives U = {0, 1, 2, 3}. Its cardinality is therefore 4, because there are four distinct elements. Zero is a valid element of a set and must be counted. Hence option A is the only correct statement.

Because n is a natural number satisfying 1 ≤ n ≤ 5, substitute n = 1, 2, 3, 4, and 5 into x = n² − 1. The resulting values are 1² − 1 = 0, 2² − 1 = 3, 3² − 1 = 8, 4² − 1 = 15, and 5² − 1 = 24. Therefore the roster form is A = {0, 3, 8, 15, 24}, making option A correct.

Using the usual school convention N = {1, 2, 3, ...}, every natural number is at least 1. Therefore, no natural number satisfies x < 1. A set whose defining condition is satisfied by no element has zero elements and is called the empty set, commonly denoted by ∅. Hence option A is correct.

A two-digit natural number has a nonzero tens digit. If both digits are equal, the repeated digit can be 1 through 9, producing 11, 22, 33, 44, 55, 66, 77, 88, and 99. The form 00 is not a two-digit natural number because its tens digit is zero. Option C contains unequal digits, while D contains one-digit numbers. Therefore A is correct.

Use the defining formula for each allowed value of n. When n = 1, 2, 3, 4, and 5, x = 3n − 1 gives 2, 5, 8, 11, and 14 respectively. The inequality includes both endpoints, so n = 5 must be used and 14 cannot be omitted. Option B lists multiples of 3, C follows 3n − 2, and D misses the final value. Hence A is correct.

A number divisible by both 3 and 5 must be divisible by their least common multiple, LCM(3, 5) = 15. The positive multiples of 15 that are at most 50 are 15, 30, and 45. Therefore, in roster form, M = {15, 30, 45}. The numbers 3 and 5 separately do not satisfy both conditions, and 60 is excluded because it is greater than 50.

The listed elements 5, 10, 15, 20, and so on are precisely the positive multiples of 5. Every positive multiple of 5 can be written as x = 5n, where n belongs to the natural numbers. The expression 5 + n gives consecutive numbers after 5, while 5ⁿ gives powers of 5, not all multiples. Hence option A is correct.

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, either x = 0 or x − 2 = 0, so x = 2. Both values are integers, as required. Hence the roster form is O = {0, 2}. Dividing by x at the beginning would incorrectly discard the valid solution x = 0.

A number whose last digit is 2 appears in the sequence 2, 12, 22, 32, and so on, increasing by 10 each time. Restricting the numbers to the interval from 1 through 100 gives 2, 12, 22, 32, 42, 52, 62, 72, 82, and 92. The number 102 is not allowed because it exceeds 100. Therefore, option A is the complete roster form of P.

Perfect cubes in the relevant positive range are obtained from 1³, 2³, 3³, and so on. We have 1³ = 1 and 2³ = 8, both of which lie between 1 and 20. The next cube is 3³ = 27, which is outside the upper limit. Thus the members satisfying every condition are 1 and 8, so Q = {1, 8}. The numbers in option D are perfect squares, not perfect cubes.

The positive divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. We now retain only the odd divisors, meaning those not divisible by 2. These are 1, 3, and 9. Hence the roster form is R = {1, 3, 9}. The number 1 must be included because 1 divides every positive integer and 1 is odd.

First solve the inequality: 2x − 1 ≤ 11. Adding 1 to both sides gives 2x ≤ 12, and dividing by the positive number 2 gives x ≤ 6. Because x belongs to N, the admissible natural-number values are 1, 2, 3, 4, 5, and 6, using the convention N = {1, 2, 3, ...}. Therefore, S = {1, 2, 3, 4, 5, 6}.

The conditions translate to the strict double inequality 5 < x < 11, with x restricted to the integers. Since the inequalities are strict, neither endpoint 5 nor endpoint 11 belongs to the set. The integers strictly between them are 6, 7, 8, 9, and 10. Therefore, the roster form is T = {6, 7, 8, 9, 10}; this is a finite set of five integers.

A prime number is a natural number greater than 1 with exactly two positive divisors: 1 and itself. The prime numbers less than 10 are 2, 3, 5, and 7. The number 1 is not prime, and 9 is composite because it has divisors 1, 3, and 9. Therefore the correct roster form is option A.

Since x is restricted to integers, inspect the integers near zero. The values −1, 0, and 1 have squares 1, 0, and 1, all of which are less than 2. The next integers, −2 and 2, have square 4 and therefore fail the strict inequality. All integers farther from zero have even larger squares. Thus the complete roster form is {−1, 0, 1}, so option A is correct.

For x² = 9, the integer solutions are x = −3 and x = 3. For x² = 16, the integer solutions are x = −4 and x = 4. These four values are different, so E = {−4, −3, 3, 4}. Consequently, the cardinality of E is 4, making option C correct. Both positive and negative square roots must be included.

The roster form lists exactly the five English vowels: a, e, i, o, and u. A correct descriptive or set-builder form must include every one of these letters and no other letter. Option A describes precisely the vowels of the English alphabet. The other choices either describe consonants, include letters beyond the set, or include all alphabet letters.

Start with the inequality 3x + 1 < 16. Subtracting 1 from both sides gives 3x < 15, and dividing by the positive number 3 gives x < 5. Because x belongs to ℕ, using the usual school convention ℕ = {1, 2, 3, …}, the possible natural numbers less than 5 are 1, 2, 3, and 4. Therefore G = {1, 2, 3, 4}, making option A correct.

The inequality has two different boundary conditions. The sign −2 < x means that −2 is excluded, while x ≤ 3 means that 3 is included. The integers strictly greater than −2 and less than or equal to 3 are −1, 0, 1, 2, and 3. Consequently, H = {−1, 0, 1, 2, 3}, so option A is the only correct answer.

A factor of 20 is a natural number that divides 20 exactly, leaving no remainder. The factor pairs are 1 × 20, 2 × 10, and 4 × 5. Thus the complete list of positive natural factors is 1, 2, 4, 5, 10, and 20. Both 1 and 20 must be included, because 1 divides every natural number and every number divides itself. Hence option A is correct.

Two sets are equal when they contain exactly the same distinct elements; the order of listing and repetition do not matter. In option A, A contains 1, 2, and 3, while B also contains the same distinct elements because the repeated 2 is counted only once. Thus A = B. Every other option has at least one different or additional element.