Sets and their representations

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.

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 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.

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.

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.

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 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.

The positive natural-number multiples of 7 are 7, 14, 21, 28, 35, and so on. The condition x < 30 retains only 7, 14, 21, and 28. The number 35 is excluded because it is greater than 30; 0 is not included under the natural-number convention used here, and 1 is not a multiple of 7. Thus S = {7, 14, 21, 28}.

The positive perfect squares are obtained by squaring natural numbers: 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, and 7² = 49. The next square, 8² = 64, is not less than 50. Under the convention ℕ = {1,2,3,...}, zero is not included. Hence T = {1, 4, 9, 16, 25, 36, 49}.

Because x belongs to the integers, we list only whole-number values, not decimals or fractions. The inequality -2 ≤ x includes -2, while x < 3 excludes 3. Thus the elements are V = {-2, -1, 0, 1, 2}. Counting them gives five elements, so the cardinality of the set is n(V) = 5. The endpoint symbols are essential: the left endpoint is included because of ≤, and the right endpoint is excluded because of <.

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.

The condition |x| < 3 means that the distance of x from zero is less than 3. Therefore, x must lie strictly between −3 and 3. Since x is restricted to integers, the possible values are −2, −1, 0, 1, and 2. The endpoints −3 and 3 are excluded because the inequality is strict, so the roster form is {-2, −1, 0, 1, 2}.

Here x is a natural number satisfying both inequalities 1 ≤ x and x ≤ 5. Because both inequalities are non-strict, the endpoints 1 and 5 are included. Listing every natural number from 1 through 5 gives A₁ = {1, 2, 3, 4, 5}. Zero is not included because the lower bound begins at 1.