Sets and their representations

The governing concept is translating interval bounds for integers into a roster. The lower inequality −2≤x is inclusive, so −2 must be listed. The upper inequality x<3 is strict, so 3 must not be listed. The integers between these bounds are −2, −1, 0, 1, and 2. Option B shifts the lower endpoint and includes 3, C omits 0, and D incorrectly includes 3. Hence A is correct.

By definition, a triangle is a polygon with exactly three sides. Therefore, an object cannot simultaneously be a triangle and have four sides. No element satisfies the defining condition of N, so N contains no members. A set with no members is called an empty set, written as ∅; hence option A is correct.

A divisor of 24 is a natural number that divides 24 without leaving a remainder. Checking factor pairs gives 1 × 24, 2 × 12, 3 × 8, and 4 × 6. Therefore, the positive natural divisors are 1, 2, 3, 4, 6, 8, 12, and 24. The number 5 is not a divisor, and 16 does not divide 24 exactly.

The governing concept is substitution in a bounded set-builder rule. Because 2≤n≤6 includes both endpoints, n takes the integer values 2, 3, 4, 5, and 6. Substituting into x=3n gives 6, 9, 12, 15, and 18. The value 3 would require n=1, which is outside the range, while 21 would require n=7. Thus the complete roster is option A.

A two-digit number has a tens digit and a units digit, and the condition requires these two digits to be identical. The possible repeated digits are 1 through 9, giving 11, 22, 33, 44, 55, 66, 77, 88, and 99. The number 1 is excluded because it has only one digit, while numbers such as 12 have unequal digits. Therefore, option A is the correct roster form.

To find the members of U, solve the defining equation. We factor x² − 5x + 6 as (x − 2)(x − 3). Hence, the product is zero when x = 2 or x = 3. Both values are integers, so both belong to the set. There are no other roots of this quadratic equation. Thus the roster form is U = {2, 3}, making option A correct.

The question specifies the ordering in which Sunday is followed by Monday. Therefore, the only day immediately preceding Monday in this weekly order is Sunday. A set records the qualifying day once, so V = {Sunday}. The explicit convention removes the ambiguity that could arise from different calendar conventions.

A common divisor must divide both 36 and 48 without leaving a remainder. Their greatest common divisor is 12, and the positive divisors of 12 are 1, 2, 3, 4, 6, and 12. Each of these divides both original numbers, whereas 8, 9, and 24 fail for at least one of them.

The governing concept is classification of natural numbers into prime, composite, or neither. From 1 through 10, the primes are 2, 3, 5, and 7; the composites are 4, 6, 8, 9, and 10. The number 1 has only one positive divisor, so it is neither prime nor composite. Under the usual school convention ℕ = {1, 2, 3, …}, X = {1}; therefore option A is correct. Option B incorrectly includes 0, while C and D do not satisfy the condition.

A set records distinct elements, so repeated occurrences are written only once. The word “level” contains the letters l, e, v, e, l. After removing repetitions, the distinct letters are l, e, and v. The order of elements in a set does not matter, so {l, e, v} and {e, l, v} represent the same set. Therefore, option A is correct.

First solve the inequality: 2x + 1 < 12 gives 2x < 11 and therefore x < 11/2, or x < 5.5. The natural numbers satisfying this condition are 1, 2, 3, 4, and 5. The value 6 is too large, and 0 is not included under the usual school convention ℕ = {1, 2, 3, …}. Thus option A is the correct roster form.

The inequality |x| < 3 means that x lies within a distance of less than 3 from zero. Equivalently, −3 < x < 3. The integers strictly between −3 and 3 are −2, −1, 0, 1, and 2. The endpoints −3 and 3 are excluded because the inequality is strict. Therefore, the roster form is given in option A.

The listed numbers are precisely the positive multiples of 3. If n is a natural number beginning with 1, the rule x = 3n produces 3, 6, 9, 12, and so on. Option B starts with 4, option C gives odd numbers, and option D also includes zero and negative multiples.

The positive even numbers less than 15 are 2, 4, 6, 8, 10, 12, and 14. We must then apply the second condition: the number must be composite. Although 2 is even and positive, it is prime, not composite, so it must be removed. Every remaining listed number has factors other than 1 and itself. Hence C₁ = {4, 6, 8, 10, 12, 14}, so option A is correct.

A number with tens digit 2 lies in the interval from 20 through 29. Every one of these numbers is a natural number and is less than 30, so all of them satisfy the defining conditions. Therefore, the roster form contains every integer from 20 to 29 exactly once. The restriction is about the tens digit, not the units digit, so odd and even numbers are both included.

Because x is a natural number, x is non-negative; the inequality x² ≤ 100 gives x ≤ 10. The natural numbers from 1 through 10 that are odd are 1, 3, 5, 7, and 9. The number 11 is excluded because its square is greater than 100, and even numbers do not satisfy the oddness condition. Hence option A is correct.

Starting with the positive natural multiple 18, the successive multiples are 18, 36, 54, 72, 90, and 108. The condition x < 100 keeps the first five values and excludes 108. Under the usual school convention used here, natural numbers begin with 1, so zero is not included. Therefore, the correct finite roster is option A.

To find the members, solve the defining equation x³ = x. Moving all terms to one side gives x³ − x = 0, which factors as x(x² − 1) = x(x − 1)(x + 1) = 0. Hence x is −1, 0, or 1. All three values are integers and satisfy the original equation, so the roster form is option A.

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.

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}.