Sets and their representations

Negative real numbers are precisely the real numbers less than zero, so they satisfy x < 0. Since zero itself is neither negative nor less than zero, it must not be included. The interval therefore extends indefinitely to the left and ends at 0 with a round bracket: (-∞, 0). Infinity is also written with a round bracket because it is not a real endpoint. Hence option A is correct.

The condition x ∈ ℝ, x ≠ 0 means that x may be any real number except zero. The negative real numbers are represented by the open interval (-∞, 0), because zero is not included. The positive real numbers are represented by (0, ∞), again with zero excluded. Combining these two disjoint parts with union gives (-∞, 0) ∪ (0, ∞). Square brackets at zero would incorrectly include zero.

Under the standard school convention used here, natural numbers begin with 1. The one-digit natural numbers are 1 through 9, and the even ones among them are 2, 4, 6, and 8. Zero is not included because it is not being counted as a natural number in this convention, and 10 is not one-digit. Therefore the correct roster form is P = {2, 4, 6, 8}.

In interval notation, a square bracket means that the endpoint is included, while a round bracket means that the endpoint is excluded. To include both 2 and 5, the interval must have a square bracket at both ends. Therefore [2, 5] is correct. The intervals (2, 5), [2, 5), and (2, 5] each exclude at least one endpoint, so none of them contains both numbers.

The interval (3, 9] uses a round bracket at 3 and a square bracket at 9. The round bracket means that 3 is excluded, so 3 ∉ (3, 9]. The square bracket means that 9 is included, so 9 ∈ (3, 9]. Therefore statement B is correct. In general, an interval is interpreted by checking the bracket at each endpoint separately.

The inequalities are strict: −1 < x and x < 4. Therefore, x can be any real number greater than −1 and less than 4, but it cannot equal either endpoint. In interval notation, an excluded endpoint is shown with a round parenthesis, so the correct interval is (−1, 4). Square brackets would incorrectly include an endpoint.

In the interval [−2, 6), the square bracket at −2 means that −2 is included, while the round parenthesis at 6 means that 6 is excluded. Thus every real number x must satisfy −2 ≤ x < 6. Option B records both endpoint conditions correctly; the other options include or exclude at least one endpoint incorrectly.

The condition x ≥ 3 includes 3 because the inequality contains the equality sign. It also includes every real number greater than 3, continuing without an upper bound. Thus the interval begins with a square bracket at 3 and extends toward infinity: [3, ∞). Infinity is never an endpoint that can be reached, so a round bracket is always used beside ∞. Therefore option B is correct.

The strict inequality x < −2 means that −2 itself is excluded. Every real number smaller than −2 is included, and the set continues indefinitely toward negative infinity. Therefore its interval notation is (−∞, −2). A round bracket is required at −2 because equality is not allowed, and infinity also always uses a round bracket. Hence option A is correct.

The notation (2, 6) denotes an open interval. It contains every real number strictly greater than 2 and strictly less than 6, but it excludes both endpoints 2 and 6. The number 4 lies between these endpoints, so 4 belongs to A. The number 1 lies outside the interval. Consequently, the true statement is 4 ∈ A, which is option C.

The condition 0 ≤ x ≤ 1 describes all real numbers from 0 through 1, including both endpoints. In interval notation, a square bracket means that the endpoint is included, while a round bracket means that it is excluded. Since both inequalities contain equality, both 0 and 1 belong to A. Therefore, A = [0, 1].

The governing idea is that x² < 9 means |x| < 3, because x² measures the square of the distance of x from zero. Thus −3 < x < 3. The inequality is strict, so the endpoint values −3 and 3 are excluded; interval notation therefore uses parentheses. Option A, (−3, 3), is correct. Option B wrongly includes the endpoints, while C describes values outside the interval and D confuses x with x².

The relevant principle is |x| ≤ 4, since taking the nonnegative square root of x² ≤ 16 gives a distance from zero no greater than 4. Equivalently, −4 ≤ x ≤ 4. Equality is allowed, so both boundary points −4 and 4 belong to the set and square brackets are required. Therefore option B, [−4, 4], is correct. Option A excludes valid endpoints, C represents the outside region, and D lists squared values rather than possible x-values.

The symbol ℤ denotes the set of all integers, not all real numbers. We therefore list the integers satisfying -2 ≤ x ≤ 2, including both endpoints because the inequalities allow equality. These integers are -2, -1, 0, 1, and 2. Hence A = {-2, -1, 0, 1, 2}.

The symbol R denotes the set of all real numbers, so x may be any real number satisfying the inequality. The condition -2 ≤ x ≤ 2 includes both endpoint values -2 and 2. In interval notation, an included endpoint is represented by a square bracket. Therefore, the set is written as [-2, 2]. It is not a finite roster set because infinitely many real numbers lie between -2 and 2.

For a positive number 2, the inequality |x| < 2 means that the distance of x from zero is less than 2. Equivalently, -2 < x < 2. Both inequalities are strict, so neither -2 nor 2 is included; this requires round brackets at both ends. Therefore the correct interval notation is (-2, 2), not the outside region or a one-sided interval.

In interval notation, a round parenthesis means that the endpoint is excluded, while a square bracket means that the endpoint is included. Therefore, (-5, 2] contains all real numbers greater than -5 and less than or equal to 2. The correct inequality is -5 < x ≤ 2, so option A is correct.

The notation A = {∅, 1} means that A has exactly two elements: the empty set ∅ and the number 1. Therefore ∅ ∈ A is true. The set {1} is not an element of A, even though 1 itself is an element of A, so option B is false. No element belongs to ∅, making C false. Finally, {∅,1} equals A, so it is not a proper subset of A. Hence option A is correct.

The positive divisors of 18 are 1, 2, 3, 6, 9, and 18. The condition requires the divisor to be odd, so we retain only 1, 3, and 9. Therefore A = {1, 3, 9}. Option B contains the even divisors, option C contains every positive divisor without applying the oddness condition, and option D incorrectly includes 6, which is even. This is a roster-form representation of a set defined by a property.

A positive perfect square is the square of a positive integer. The positive integers whose squares are less than 25 are 1, 2, 3, and 4, giving the squares 1, 4, 9, and 16. Zero is not positive, and 25 is not included because the condition says less than 25, not less than or equal to 25. Therefore option A is correct.