Roster Form and Inequality Conditions

This question applies set-builder notation with one strict and one inclusive inequality. The condition 4 < x excludes 4 because x must be greater than 4. The condition x ≤ 7 includes 7 because equality is permitted. The whole numbers satisfying both conditions are therefore 5, 6, and 7, so J = {5, 6, 7}; option A is correct. Option B incorrectly includes the excluded lower endpoint 4. Option C correctly excludes 4 but incorrectly omits the included upper endpoint 7. Option D makes both boundary errors by including 4 and omitting 7. The intersection of the two conditions must be used, rather than considering either inequality separately.

The governing concept is conversion from set-builder notation to roster form over the integers. The inequality -3 ≤ x includes the lower endpoint because equality is permitted. The inequality x < 1 excludes the upper endpoint because it is strict. Listing the integers from -3 through the greatest integer below 1 gives -3, -2, -1, and 0. Hence U = {-3, -2, -1, 0}, so option A is correct. Option B wrongly excludes -3 and includes 1. Option C also includes the forbidden endpoint 1. Option D omits 0, although 0 is an integer satisfying both inequalities. The endpoint symbols determine exactly which values are included.

The governing concept is converting a set-builder condition into roster form while respecting both the domain and the inequality. A negative integer must be less than 0, and x > −5 means that it must be strictly greater than −5. The only possible integers are therefore −4, −3, −2, and −1. Thus D = {−4, −3, −2, −1}, so option A is correct. The endpoint −5 is excluded because the inequality is strict. Zero is excluded because it is neither negative nor less than zero. Consequently, option B includes an invalid endpoint, while options C and D include zero. Listing values requires checking every stated condition, not just one of them.

The set-builder condition requires a number to be both even and prime, and it must be less than 20. The only even prime number is 2. Every even number greater than 2 is divisible by 2 and another positive integer, so it has more than two positive factors and is composite rather than prime. Since 2 is less than 20, it satisfies both conditions. Therefore the roster form is K = {2}, so option A is correct. Option B lists many even numbers, but most are composite. Option C lists odd primes and therefore fails the even condition. Option D is wrong because the set is not empty. This illustrates how a set-builder description is converted to roster form by identifying every element satisfying all stated conditions.

The set contains natural numbers satisfying both 15 ≤ x and x ≤ 17. Because both inequalities use the inclusive symbol ≤, the endpoints 15 and 17 must be included. The only natural number strictly between them is 16. Thus the complete roster is N = {15, 16, 17}, so option A is correct. Option B keeps only the middle value and omits both endpoints. Option C omits 17, while option D omits 15. The key concept is translating a bounded inclusive inequality into a list of all permitted integers. Since the interval is closed at both ends, neither boundary may be discarded. No other natural number lies between 15 and 17, so the answer is unique.