Sets and their representations

The condition 1<x<2 is strict at both ends. Therefore, x may be any real number between 1 and 2, but it cannot equal 1 or 2. Strict inequalities are represented with round brackets, so the interval notation is (1,2). Square brackets would incorrectly include the corresponding endpoint; hence A, C, and D do not match the given set-builder description.

The inequality 1 ≤ x ≤ 5 includes both endpoints because equality is allowed at 1 and at 5. In interval notation, an included endpoint is written with a square bracket. Therefore the correct interval is [1, 5]. Option A excludes both endpoints, option C excludes 5, and option D excludes 1, so none of them represents the given double inequality exactly.

The condition x > 7 includes all real numbers greater than 7 but excludes 7 itself. Exclusion of a finite endpoint is shown by a round parenthesis, so the interval is (7, ∞). Infinity is never an actual endpoint and is always written with a round parenthesis. The other choices either include 7 or describe numbers less than 7.

The endpoint −1 is included, so a square bracket is used at the left side: [−1. The endpoint 2 is excluded, so a round bracket is used at the right side: 2). Therefore, the required interval is [−1, 2). This is called a left-closed, right-open or half-open interval. The corresponding inequality is −1 ≤ x < 2.

The inequality −4 < x means that x is greater than −4, but x cannot equal −4; therefore, −4 is excluded and receives a round bracket. The inequality x ≤ 1 means that x may equal 1; therefore, 1 is included and receives a square bracket. Hence A is written in interval notation as (−4, 1].

Both inequalities are strict: 2 < x excludes the endpoint 2, and x < 7 excludes the endpoint 7. An excluded endpoint is represented by a round bracket in interval notation. Since neither endpoint is included, the set is the open interval (2, 7). Equivalently, it contains every real number strictly between 2 and 7, but not 2 or 7 themselves.

The square bracket at 3 shows that 3 is included, so the inequality must contain 3 ≤ x. The round bracket at 8 shows that 8 is excluded, so the inequality must contain x < 8. Combining these conditions gives {x : x ∈ ℝ, 3 ≤ x < 8}. Thus option B is the correct set-builder form of [3, 8).

The inequality x ≤ 3 includes every real number less than 3 as well as 3 itself. The interval extends without bound to the left, represented by −∞, and ends at 3. Since 3 is included because of the symbol ≤, a square bracket is used at 3. Infinity is never included as an endpoint, so a parenthesis is used at −∞. Therefore, the interval is (−∞,3].

The condition x > −2 represents all real numbers strictly greater than −2. The number −2 itself is not allowed because the inequality is strict, so the left endpoint must use a parenthesis. The solution continues indefinitely to the right, which is shown by ∞; infinity is not an included real endpoint, so it also uses a parenthesis. Hence the interval is (−2,∞).

The condition 0 ≤ x means that x may equal 0, so the left endpoint is included and receives a square bracket. The condition x < 4 means that x must be strictly less than 4, so 4 is excluded and receives a parenthesis. Combining these endpoint rules gives the half-open interval [0,4). Both boundaries must be checked separately in a mixed inequality.

The closed interval [2,2] represents all real numbers x satisfying 2 ≤ x ≤ 2. A number cannot be both less than and greater than 2 under this condition; the only possible value is exactly x = 2. Since the endpoint is included at both sides, the interval contains one element only. Therefore, [2,2] is the singleton set {2}, not the empty set.

The set of all real numbers extends without bound in both the negative and positive directions. In interval notation it is written as (-∞, ∞), which represents every real number x satisfying -∞ < x < ∞. Infinity is not an actual real number or endpoint, so it is always written with round brackets, never square brackets. Therefore, option A is correct.

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.

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 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}.

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 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.