Sets and their representations

The governing concept is listing integers in a half-open interval. The lower condition −2 ≤ x includes −2, while the upper condition x < 4 excludes 4. The consecutive integers from −2 through 3 are −2, −1, 0, 1, 2, and 3. Hence option C is correct. Option A incorrectly includes 4, B omits the allowed endpoint −2, and D omits 0 even though zero is an integer in the interval.

A natural number has exactly two distinct positive factors precisely when it is prime: the number itself and 1. The prime numbers less than 15 are 2, 3, 5, 7, 11, and 13. The number 1 is not prime because it has only one positive factor, namely 1. Hence the required set is option B.

For an absolute-value equation |x + 1| = 3, there are two possible cases: x + 1 = 3 or x + 1 = −3. These give x = 2 and x = −4, respectively. Both values are integers and satisfy the original equation, so both must be included. Therefore, the roster form is L₁ = {−4, 2}.

The governing concept is finding the intersection of the divisors of 100 with the perfect squares. The positive divisors are 1, 2, 4, 5, 10, 20, 25, 50, and 100. Among them, 1 = 1², 4 = 2², 25 = 5², and 100 = 10² are squares. Thus M₁ = {1, 4, 25, 100}, so option A is correct. Option C lists all divisors rather than only squares, and 16 in D is square but does not divide 100.

A number divisible by both 4 and 6 must be divisible by their least common multiple. Since lcm(4, 6) = 12, the required numbers are multiples of 12. The positive multiples of 12 that are at most 50 are 12, 24, 36 and 48. Thus, the roster form is O₁ = {12, 24, 36, 48}. Numbers such as 4 or 6 satisfy only one of the two divisibility conditions.

The positive multiples of 5 less than 30 are 5, 10, 15, 20, and 25. The positive multiples of 7 less than 30 are 7, 14, 21, and 28. Since “or” means that either condition may hold, combine both lists and remove repetitions. The result is option A; 30 is excluded because the inequality is strict.

The governing concept is combining an inequality with an integer and parity condition. From x² ≤ 9, we get |x| ≤ 3, so the possible integers are −3, −2, −1, 0, 1, 2, and 3. Filtering these for odd values leaves −3, −1, 1, and 3. Therefore R₁ = {-3, -1, 1, 3}, making option A correct. B includes even integers, C omits valid endpoints, and D incorrectly includes zero.

The natural-number factors of 45 are 1, 3, 5, 9, 15 and 45. Adding 2 to these values gives 3, 5, 7, 11, 17 and 47, respectively. Every one of these results is prime. Hence every factor of 45 satisfies the second condition as well, so the complete roster form is S₁ = {1, 3, 5, 9, 15, 45}. Option C is incorrect because it omits valid factors.

Factor the quadratic equation: x² − 2x − 8 = (x − 4)(x + 2) = 0. Hence x = 4 or x = −2. Both values belong to the integers and satisfy the defining equation. A set has no order requirement, so {−2, 4} and {4, −2} would describe the same set; among the given choices, option A is correct.

The roster form lists every natural number from 10 through 40 that contains the digit 3 in at least one position. The numbers 13 and 23 contain 3 in the units place, while 30 through 39 contain it in the tens place. Number 40 does not contain 3, and 43 is outside the interval. Therefore option A is complete and correct.

To convert the set-builder form into roster form, substitute every permitted natural-number value of n into x = n/(n + 1). For n = 1, 2, 3, and 4, the values are respectively 1/2, 2/3, 3/4, and 4/5. Listing these distinct values gives A = {1/2, 2/3, 3/4, 4/5}. Option B reverses the fractions, while C and D do not use the stated formula correctly.

We need the natural numbers less than 10 that have exactly two positive divisors: 1 and the number itself. These are 2, 3, 5, and 7. The number 1 is not prime because it has only one positive divisor, and 9 is composite because it has divisors 1, 3, and 9. Hence A = {2, 3, 5, 7}.

In the usual elementary classification of the English alphabet, the vowels are a, e, i, o, and u. The letters b, c, and d are consonants, so option B is incorrect. The letter y is not included in the standard vowel list used in this question. Hence the complete roster form is option A.

Since x is an integer, we list the integers between the boundary values. The strict inequality -3 < x excludes -3, while x ≤ 2 includes 2. The integers satisfying both conditions are -2, -1, 0, 1, and 2. Therefore, the roster form is C = {-2, -1, 0, 1, 2}.

A mathematical set must be well-defined, meaning that membership can be decided objectively and consistently. The even natural numbers less than 10 are exactly 2, 4, 6, and 8, so there is no ambiguity. In contrast, good, beautiful, and interesting depend on personal opinion. Therefore option C describes a set.

Solving x² = 16 over the integers gives x = 4 or x = -4. However, the definition restricts x to the natural numbers, and -4 is not a natural number. The only permitted solution is x = 4, since 4² = 16. Consequently, the set contains exactly one element: D = {4}. Option A ignores the natural-number restriction, and options C and D omit the valid solution.

Using the usual school convention ℕ = {1, 2, 3, …}, the only natural number less than 2 is 1. Since 1 is odd, it satisfies both conditions and belongs to E. Therefore E contains exactly one element, namely 1, so E = {1}. Option C is correct; the set is not empty.

The element x must be a natural number strictly greater than 5 and strictly less than 6. Since 5 and 6 are consecutive natural numbers, there is no natural number between them. Both endpoints are excluded by the strict inequalities. Consequently, no element satisfies the condition, so F is the empty set and option D is correct.

The elements of G are precisely the positive even natural numbers that are less than 10: 2, 4, 6, and 8. Therefore, the condition must be “x is even and x < 10.” Option B also includes 10, while option C describes odd numbers and option D includes numbers that are not necessarily even and may include 1, 2, 3, and so on.

In a set, repetition of an element does not create a new element. The displayed members reduce to the distinct values 1, 2, and 3, so H can be written as {1, 2, 3}. Its cardinality, or number of distinct elements, is therefore 3. The repeated appearances of 2 and 3 must not be counted again.

A positive natural number divides 12 exactly when the remainder is zero. Checking the positive factors gives 1, 2, 3, 4, 6, and 12: 12 ÷ each of these is an integer. The number 1 and 12 itself must both be included. Zero cannot be a divisor, so option D is invalid, and option A lists all factors without omission.

The positive natural-number multiples of 10 that are less than or equal to 50 are 10, 20, 30, 40, and 50. The endpoint 50 must be included because the inequality is ≤, not <. Hence M = {10, 20, 30, 40, 50}, and counting its distinct members gives 5. Therefore, option B is correct. The other numerical choices result from omitting an endpoint or confusing the count with the common difference.

The condition x² = 9 means that x is a number whose square is 9. Since x must be an integer, we solve x² = 9 and obtain x = 3 or x = -3. Both values satisfy the condition because 3² = 9 and (-3)² = 9. Therefore, the roster form, which lists every element explicitly inside braces, is P = {-3, 3}. The order of elements in a set does not matter, so {-3, 3} and {3, -3} represent the same set.

For natural numbers, using the convention in this question, x starts at 1. The inequality x² ≤ 25 means x ≤ 5 because x is positive. Thus the possible natural numbers are 1, 2, 3, 4, and 5; each has a square no greater than 25. Negative integers are not natural numbers here, and 0 is excluded by the stated convention. Therefore Q = {1, 2, 3, 4, 5}.

The symbol ∅ denotes the empty set, which contains no elements. However, R = {∅} is a set whose only element is the empty set itself. Therefore R has exactly one element, so n(R) = 1. This is different from R = ∅, whose cardinality would be zero. Braces change the empty set into a member of a singleton set.