Sets and their representations

First list the positive divisors of 36: 1, 2, 3, 4, 6, 9, 12, 18 and 36. The prime divisors in this list are 2 and 3, so they must be removed. The number 1 is not prime because a prime number has exactly two positive divisors, whereas 1 has only one. All remaining divisors are composite or 1, giving option A.

A set is well-defined when membership can be decided objectively and unambiguously. Whether a student finds mathematics “very easy” depends on personal feeling and has no fixed numerical or observable boundary. Different people may judge the same student differently. The other options use definite, checkable conditions, so A is not well-defined.

Solve the inequality: 2x + 3 ≤ 15 gives 2x ≤ 12, and hence x ≤ 6. Under the convention used here, ℕ denotes the positive natural numbers 1, 2, 3, and so on. Therefore, the natural numbers satisfying x ≤ 6 are 1, 2, 3, 4, 5 and 6. Thus the roster form is option A. Option B incorrectly includes zero, while C omits 6.

Use the standard absolute-value inequality rule: |x − 2| ≤ 3 is equivalent to −3 ≤ x − 2 ≤ 3. Adding 2 to all parts gives −1 ≤ x ≤ 5. The integers in this closed interval are −1, 0, 1, 2, 3, 4 and 5. Counting them gives 7 elements, so option C is correct. The closed endpoints must be included because the original sign is ≤.

Because x is a positive natural number and x² < 50, we have x < √50, which is approximately 7.07. Thus the possible natural numbers are 1, 2, 3, 4, 5, 6 and 7. Among these, the numbers divisible by 3 are only 3 and 6. Therefore, the roster form of F is {3, 6}, making option A correct. Zero is excluded under the stated positive-natural-number convention, and 9 fails the square bound.

The English alphabet has exactly five commonly recognized vowels: a, e, i, o, and u. These elements are clearly specified, so the set is well-defined. Because its elements can be listed completely and their number is limited to five, G is a finite set. It is neither empty nor infinite.

For any number x, the equation x + 5 = x would require subtracting x from both sides, giving 5 = 0. This is impossible, so no natural number satisfies the stated condition. Therefore H contains no elements. A set containing no element is called the empty set, usually denoted by ∅.

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}.

The natural numbers strictly between 10 and 20 that are prime are 11, 13, 17, and 19. Now test the second condition: for x = 11, 2x + 1 = 23, which is prime; for 13, it is 27, composite; for 17, it is 35, composite; and for 19, it is 39, composite. Hence only x = 11 satisfies both conditions, so J = {11}.

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.

Let the tens digit be a and the units digit be b. Since the number is two-digit, a cannot be zero, and the condition is a + b = 3. The possible digit pairs are (1,2), (2,1), and (3,0), giving the numbers 12, 21, and 30. The number 3 is not included because it has only one digit. Hence M = {12, 21, 30}.

Use the roster-form method by first listing all positive multiples of 6 not exceeding 30: 6, 12, 18, 24, and 30. The phrase “not a multiple of 12” removes 12 and 24, because both are divisible by 12. The remaining numbers, 6, 18, and 30, satisfy every condition in the definition. Thus option A is correct; B forgets the exclusion, while C lists only excluded values.

Factor the quadratic expression: x² - 5x + 6 = (x - 2)(x - 3). For the product to be zero, either x - 2 = 0 or x - 3 = 0. Thus x = 2 or x = 3. Both values belong to the integers, so both are elements of P. Consequently, the roster form is P = {2, 3}, making option A correct.

The natural-number perfect squares less than 100 are 1, 4, 9, 16, 25, 36, 49, 64, and 81. Selecting only the even squares leaves 4, 16, 36, and 64. The number 100 is excluded because the condition is strictly less than 100. Therefore Q = {4, 16, 36, 64}.

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.

The positive integers less than 15 are 1 through 14. Since 15 = 3 × 5, a number is coprime to 15 precisely when it is divisible by neither 3 nor 5. Testing the integers gives 1, 2, 4, 7, 8, 11, 13, and 14. Their greatest common divisor with 15 is 1, while every omitted number shares a factor with 15. Hence option A is correct.

For x = 12/n to be a natural number, n must be a positive divisor of 12. The positive divisors of 12 are 1, 2, 3, 4, 6, and 12. Substitution gives x-values 12, 6, 4, 3, 2, and 1. Since sets do not depend on order, these values form T = {1, 2, 3, 4, 6, 12}. Zero is not obtained, so option A is correct.

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 x/2 must be a natural number, x must be twice a positive natural number; therefore x is a positive even integer. The interval −5 < x < 7 contains the even integers −4, −2, 0, 2, 4, and 6, but only 2, 4, and 6 produce positive natural quotients. Thus V = {2, 4, 6}, so option A is correct.