Sets and their representations

For an absolute-value inequality, |x − 2| ≤ 3 means that x is at most 3 units away from 2. Thus, −3 ≤ x − 2 ≤ 3. Adding 2 throughout gives −1 ≤ x ≤ 5. Since x must be an integer, the complete list is −1, 0, 1, 2, 3, 4, and 5. Hence the roster form is option A.

For option A, x² ≤ 4 is equivalent to -2 ≤ x ≤ 2. Restricting x to the integers gives exactly -2, -1, 0, 1, and 2, which are all the elements of S. The other options either restrict x to natural numbers or use strict inequalities that exclude one or both boundary values.

Let the tens digit range from 1 through 9. For a digit sum of 9, the corresponding units digits are 8, 7, 6, 5, 4, 3, 2, 1, and 0. This gives 18, 27, 36, 45, 54, 63, 72, 81, and 90. The number 9 is not two-digit, so it is excluded, while 90 is included because 9 + 0 = 9.

The primes less than 20 are 2, 3, 5, 7, 11, 13, 17, and 19. Test the additional condition that x + 2 must also be prime: 3 gives 5, 5 gives 7, 11 gives 13, and 17 gives 19. The other candidates fail this condition, so Y = {3, 5, 11, 17}, making option A correct.

The correct answer is A. A factor of 72 divides 72 exactly, while a multiple of 6 is obtained by multiplying 6 by a natural number. First list the positive factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72. Now keep only the numbers divisible by 6: 6, 12, 18, 24, 36 and 72. Thus B₁ = {6, 12, 18, 24, 36, 72}. Option A contains every number satisfying both conditions. Option B wrongly omits 18, although 72 ÷ 18 = 4. Option C includes 54, but 54 does not divide 72. Option D omits 6 and 18. Memory cue: apply both filters—factor of 72 and multiple of 6.

The elements can be written as 1 = 1³, 8 = 2³, 27 = 3³, and 64 = 4³. Therefore every element has the form n³, where n is a natural number from 1 through 4. Option A gives squares, option C gives powers of 2, and option D gives an arithmetic sequence, so none of those represents the given set.

Since x/2 is an integer, x must be divisible by 2; therefore x must be an even integer. The strict inequality −5 < x < 5 allows the integers −4 through 4, but among them the even integers are −4, −2, 0, 2, and 4. Zero is included because 0/2 = 0, which is an integer. Thus option A is correct.

Factor the quadratic: x² − 5x + 6 = (x − 2)(x − 3). This product is negative strictly between its roots, so 2 < x < 3. There is no natural number strictly between 2 and 3. The roots themselves do not qualify because the inequality is strict and gives zero at x = 2 or x = 3. Therefore, J₁ is the empty set, option A.

Factor the quadratic: x² − 5x + 6 = (x − 2)(x − 3). Since the parabola opens upward, the expression is less than or equal to zero between the roots, including the endpoints. Thus 2 ≤ x ≤ 3. The natural numbers in this interval are only 2 and 3, so the roster form is K₁ = {2, 3}.

A natural number that is both a perfect square and a perfect cube must be a perfect sixth power, because the least common multiple of the exponents 2 and 3 is 6. The sixth powers below 100 are 1⁶ = 1 and 2⁶ = 64; 3⁶ = 729 is already greater than 100. Therefore, the required set in roster form is {1, 64}, so option A is correct. Option B lists ordinary cubes, while option C lists squares without applying both conditions.

Because x is an integer, we test the integers near zero. The inequality x² < 4 is satisfied by x = -1, 0, and 1, since their squares are 1, 0, and 1. The values x = -2 and x = 2 are excluded because their squares equal 4, not a number less than 4. Thus J = {-1, 0, 1}, which has 3 elements, so option B is correct.

A two-digit number is at least 10, and the condition x < 15 restricts the possible numbers to 10, 11, 12, 13, and 14. Among these, only 11 and 13 are prime: 10, 12, and 14 are composite because they have factors other than 1 and themselves. Therefore L = {11, 13}, so option C is correct.

Solving x + 3 = 3 gives x = 0. However, the question restricts x to the natural numbers ℕ = {1, 2, 3, ...}, which does not contain 0. Therefore no permitted natural number satisfies the equation, so the set has no elements and is the empty set, written as ∅. Option C is correct.

The positive factors of 18 are 1, 2, 3, 6, 9, and 18. We then apply the second condition, namely that the factor must be even. Among these factors, 2, 6, and 18 are even, while 1, 3, and 9 are odd. Therefore the roster form is W = {2, 6, 18}; numbers such as 4 and 8 are excluded because they are not factors of 18.

A number that is a multiple of both 4 and 6 must be a multiple of their least common multiple. Since lcm(4, 6) = 12, the common multiples below 30 are 12 and 24. The number 36 is already beyond the bound. Thus the required set is X = {12, 24}. Option D lists multiples of 6 but incorrectly includes numbers that are not multiples of 4.

B₁ contains every real number strictly between 1 and 4, including numbers such as 1.5, 2.25, and 3.9. It is an infinite set, so its members cannot be completely listed in an ordinary finite roster. The interval notation (1, 4) and the given set-builder notation describe it correctly, but {2, 3} contains only two members and omits infinitely many others.

The listed members 2, 3, 5, 7, and 11 are exactly the prime natural numbers less than 12. Option A states both necessary conditions: x must be prime and x must be less than 12. Option B also includes non-primes such as 1 and 9, while C includes every smaller natural number and D wrongly excludes 2.

Using ℕ = {1, 2, 3, …}, consider the natural numbers less than 20, so 20 itself is excluded. Multiples of 2 below 20 are 2, 4, 6, 8, 10, 12, 14, 16, and 18. Multiples of 5 below 20 are 5, 10, and 15. Taking the union gives {2, 4, 5, 6, 8, 10, 12, 14, 15, 16, 18}; 10 is listed only once. Thus A is correct.

The listed elements are consecutive cubes: 1 = 1³, 8 = 2³, 27 = 3³, and 64 = 4³. Therefore, the elements can be described as x³, where x is a natural number from 1 through 4 inclusive. Option A produces squares, option C produces natural numbers, and option D omits 1 and 64. Hence B is correct.

For integer x, the inequality x² ≤ 1 means that the absolute value of x is at most 1, or −1 ≤ x ≤ 1. The integers in this interval are −1, 0, and 1. Directly checking confirms that (−1)² = 1, 0² = 0, and 1² = 1, all of which satisfy the inequality. The integers −2 and 2 have square 4 and are excluded. Hence G₂ = {−1, 0, 1}, so A is correct.

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.

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 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}.

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.