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.

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.

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.

The governing concept is roster representation from a domain and an inequality. Whole numbers are 0, 1, 2, 3, 4, and so on. The strict condition x<4 includes 0, 1, 2, and 3 but excludes 4. Negative integers are not whole numbers, so they cannot be included. Consequently, the complete set is 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.

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.

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 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.

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.