Sets and their representations

A common multiple of 9 and 12 must be a multiple of their least common multiple. Since 9 = 3² and 12 = 2² × 3, LCM(9, 12) = 2² × 3² = 36. The positive multiples of 36 not exceeding 100 are 36 and 72; the next one, 108, is too large. Therefore V = {36, 72}, so option A is correct.

To convert the set-builder form into roster form, first solve the condition given for x. Factor the quadratic as x² − 4x + 3 = (x − 1)(x − 3) = 0. Therefore, x = 1 or x = 3. Both values are integers, so both belong to A. Hence, the elements listed in roster form are A = {1, 3}. The order of elements is not important in a set.

Examine the elements using their positions. For n = 1, 2, 3, 4, 5, the expression n² − 1 gives 0, 3, 8, 15, and 24 respectively. Thus every listed element is generated by x = n² − 1, and the restriction 1 ≤ n ≤ 5 produces exactly the five members of B. The other expressions generate different numbers, so option A is correct.

The strict inequalities require numbers greater than 10 and less than 30, so 10 and 30 are excluded. Checking the integers in that interval, the prime numbers are 11, 13, 17, 19, 23, and 29. Numbers such as 12, 15, and 21 are composite, so they cannot be included.

For natural numbers, test the positive integers against x² < 50. The values 1 through 7 work because 7² = 49, which is less than 50. The next natural number, 8, does not work because 8² = 64, which is greater than 50. Hence the roster is {1, 2, 3, 4, 5, 6, 7}.

Every member of C is an even natural number, and the smallest and largest members are 2 and 12. Therefore, the property and the inclusive bounds must both be stated: x is even and 2 ≤ x ≤ 12. The other choices include odd, prime, or additional natural numbers not belonging to C.

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.

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.

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.

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.

A number that is both a multiple of 10 and divisible by 25 must be a common multiple of 10 and 25. Their least common multiple is 50, so the positive common multiples are 50, 100, 150, and so on. The condition x < 100 excludes 100 and all larger values. Therefore, the only member is 50, making option A correct.

The inequality x² < 10 implies −√10 < x < √10. Since √10 is slightly greater than 3, the possible integer values are −3, −2, −1, 0, 1, 2, and 3. Applying the additional condition that x must be even leaves only −2, 0, and 2. Hence the roster form is {−2, 0, 2}; values such as ±4 do not satisfy the inequality.

Test natural numbers less than 10 against both prime conditions. For x = 3, 10 − x = 7, and both are prime. For x = 5, 10 − x = 5, so both are prime. For x = 7, 10 − x = 3, and both are prime. The other natural numbers below 10 fail at least one condition, so T₁ = {3, 5, 7}.

List the integers from 1 through 20 and reject every number divisible by 2 or by 3. The numbers divisible by 2 are all even numbers, while the multiples of 3 include 3, 6, 9, 12, 15, and 18. The numbers that belong to neither group are 1, 5, 7, 11, 13, 17, and 19, so option A is correct.

Start with x² + x = 20 and bring all terms to one side: x² + x − 20 = 0. Factoring gives (x + 5)(x − 4) = 0, so the algebraic solutions are x = −5 and x = 4. However, the set specifies x ∈ ℕ, and −5 is not a natural number. Only 4 satisfies both the equation and the domain condition, so Z₁ = {4}. Therefore, option A is correct.

Factor the quadratic expression: x² − 7x + 10 = (x − 2)(x − 5). Setting each factor equal to zero gives x = 2 or x = 5. Both values are natural numbers and satisfy the original equation: 4 − 14 + 10 = 0 and 25 − 35 + 10 = 0. Therefore, the roster form is A = {2, 5}.

Because x must be an integer, list the integers lying strictly greater than −3 and less than or equal to 2. They are −2, −1, 0, 1, and 2. The lower endpoint −3 is excluded because the inequality is strict, while 2 is included because the upper inequality allows equality. Thus C has five elements, so option B is correct.

The inequality x² < 10 is equivalent to |x| < √10. Since √10 is approximately 3.16, the integers satisfying this condition range from −3 to 3. Checking them confirms that (−3)² = 9 < 10, while (−4)² = 16 > 10; the positive side is symmetric. Zero must also be included because 0² = 0 < 10. Hence E = {-3, -2, -1, 0, 1, 2, 3}, making option B correct.

The governing concept is converting a bounded parameterized set into roster form. The integers satisfying −2 ≤ n ≤ 2 are −2, −1, 0, 1, and 2. Substitution into x = 2n + 1 gives −3, −1, 1, 3, and 5 respectively. Since a set lists each distinct value once, F = {-3, -1, 1, 3, 5}. Therefore option A is correct. Option B starts with the value obtained from n = −3, which is outside the stated range.

The elements 4, 9, 16, and 25 are the squares of 2, 3, 4, and 5 respectively: 4 = 2², 9 = 3², 16 = 4², and 25 = 5². Therefore the rule is x = n² with n restricted to the natural numbers from 2 through 5. Starting at n = 1 would incorrectly include 1, so option B is not correct. Hence option A gives the exact set-builder form.

The elements of K are the squares 0², 1², 2², 3², and 4². Therefore, allowing n to be an integer from 0 through 4 produces exactly {0, 1, 4, 9, 16}. Option A also includes n = 5 and therefore includes 25, so it is not equal to K. The other options produce consecutive or even numbers, not precisely the required squares.

First list the multiples of 3 not exceeding 20: 3, 6, 9, 12, 15, and 18. The phrase “not divisible by 2” removes the even numbers 6, 12, and 18. The remaining odd multiples of 3 are 3, 9, and 15. The number 21 is excluded because it is greater than 20. Hence L = {3, 9, 15}.