Sets and their representations

Since x² ≤ 4, taking square roots gives −2 ≤ x ≤ 2. The set is restricted to integers, so the possible values are −2, −1, 0, 1, and 2. These are five distinct elements. The endpoints are included because the inequality is less than or equal to 4. Hence the cardinality of T is 5, so option C is correct.

Solve the inequality: 2x + 1 < 12 gives 2x < 11 and therefore x < 11/2, or x < 5.5. Taking N as the positive natural numbers, the allowable values are 1, 2, 3, 4, and 5. The value 6 is too large, while 0 is excluded under this convention. Thus U = {1, 2, 3, 4, 5}, which is option A.

The listed numbers are consecutive cubes: 0 = 0³, 1 = 1³, 8 = 2³, 27 = 3³, and 64 = 4³. Therefore each element can be written as x = n³, where n is a whole number from 0 through 4. The square, linear, and exponential expressions in the other choices do not generate this complete list. Hence option A is correct.

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.

First list the integers strictly between −4 and 4: −3, −2, −1, 0, 1, 2, and 3. From this list, select the even integers. They are −2, 0, and 2. The endpoints −4 and 4 are excluded because the inequalities are strict. Also, zero is even because it is divisible by 2. Therefore option B is the correct roster form.

The square of an integer is odd if and only if the integer itself is odd. The natural numbers less than 8 are 1 through 7, and the odd members among them are 1, 3, 5, and 7. Therefore X = {1, 3, 5, 7}. The value 0 is not included under the positive-natural-number convention, and even numbers have even squares. Hence option A is correct.

A set is well-defined when it is possible to decide objectively and unambiguously whether any given object belongs to it. The odd natural numbers less than 10 are exactly 1, 3, 5, 7, and 9, so membership is definite. Words such as good, beautiful, and interesting depend on personal judgment, so they do not define well-defined sets.

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.

There are only twelve months, so only finitely many first letters can be collected. Because a set does not repeat identical elements, repeated initials are counted once, but this does not change the fact that the collection is finite. It is not empty and it is not a singleton because several different letters occur.

The condition requires integer values of x whose square is 9. Solving x² = 9 gives x = ±√9, so x = 3 or x = −3. Both values are integers and both satisfy the condition because 3² = 9 and (−3)² = 9. Therefore, the roster form must include both solutions: E = {−3, 3}. Zero and ±9 do not satisfy x² = 9.

The condition requires a natural number strictly greater than 5 and strictly less than 6. There is no integer, and therefore no natural number, between two consecutive integers 5 and 6. Since the endpoints are excluded by the strict inequalities, neither 5 nor 6 can be included. Hence F is the empty set.

A set is determined only by its distinct elements. The order in which elements are written does not affect the set, and repeating an element does not create a new member. After removing the repeated a from H and ignoring the order, H contains exactly a, b, and c, just like G. Therefore, G and H are equal, so option A is correct.

The set is defined as the set of vowels in the English alphabet. In the standard school convention, the vowels are a, e, i, o and u. A roster form lists every member explicitly inside braces, so I = {a, e, i, o, u}. The letter y is not included in this convention, and omitting u would make the list incomplete.

For n ∈ ℕ, taking the usual school convention ℕ = {1, 2, 3, ...}, the condition n < 5 gives n = 1, 2, 3, 4. Substituting these values in x = 2n + 1 gives 3, 5, 7, and 9. Therefore, the roster form is J = {3, 5, 7, 9}. The value n = 5 is excluded because the inequality is strict.

Whole numbers begin with 0 and continue as 1, 2, 3, and so on. The condition x < 4 allows 0, 1, 2, and 3, but it does not allow 4 because 4 is not less than 4. Negative integers are not whole numbers. Hence K = {0, 1, 2, 3}, which is option A.

The elements of L are consecutive perfect squares: 1 = 1², 4 = 2², 9 = 3², 16 = 4², and 25 = 5². Thus, allowing n to take the natural-number values from 1 through 5 gives L = {x : x = n², n ∈ ℕ, 1 ≤ n ≤ 5}. The other options produce even numbers, cubes, or many unwanted natural numbers.