representation on number line

The inequality 1 ≤ x ≤ 4 includes every real number from 1 through 4. Because equality is allowed at both ends, the numbers 1 and 4 are included in the solution set. Therefore, square brackets are used at both endpoints, giving the closed interval [1, 4]. Parentheses would incorrectly exclude an endpoint.

The inequality x > 3 means that x can be any real number strictly greater than 3. Since 3 itself is not allowed, a parenthesis is placed at 3. The values continue without bound toward positive infinity, and infinity is never an included number, so a parenthesis is also used there. Hence the interval is (3, ∞).

The inequality x < 5 describes all real numbers that are smaller than 5. The endpoint 5 is excluded because the inequality is strict, so a parenthesis is used at 5. The values extend indefinitely to the left, toward negative infinity. Infinity is never included, so a parenthesis is used there as well. Thus, the correct interval is (-∞, 5).

The interval (-1, 1) is an open interval, so it contains all real numbers strictly greater than -1 and strictly less than 1. The endpoints -1 and 1 are excluded because parentheses are used. The number 0 lies between these endpoints, so it belongs to the interval. The numbers 1, -1, and 2 do not satisfy the required strict bounds.

The interval [-3, 2) includes -3 because the left square bracket indicates inclusion. It contains every real number greater than -3 and less than 2. The right parenthesis excludes 2, so 2 is not part of the interval. The other choices, -3, 0, and 1.5, all lie within the included range and are therefore members of the interval.

The open interval (1, 4) contains all real numbers strictly greater than 1 and strictly less than 4. Its endpoints, 1 and 4, are excluded because round parentheses are used. The number 3 lies between the endpoints, so 3 belongs to the interval. The numbers 1, 4, and 5 do not satisfy both strict inequalities.

For an interval to be a subset of [2, 5], every number in that interval must lie between 2 and 5, including the relevant endpoints. Every number in [3, 4] is also in [2, 5], so [3, 4] is a subset. The other intervals contain numbers below 2 or above 5, so they cannot be subsets.

The statement asks whether every element of the first interval belongs to the second interval. Any x in (0, 1) satisfies 0 < x < 1, which automatically implies 0 ≤ x ≤ 1. Hence x belongs to [0, 1], and the subset statement is true. The endpoints 0 and 1 need not belong to the first interval because they are not elements of it.

The inequality x ≥ −3 includes −3 itself and every real number greater than −3. In interval notation, inclusion of the left endpoint is shown with a square bracket, so the left side is [−3. Infinity is never an actual endpoint, so it always uses a round parenthesis. Hence the solution set is [−3, ∞), answer B.

The open interval (1, 2) contains all real numbers strictly greater than 1 and strictly less than 2. Although the endpoints 1 and 2 are excluded, there are still infinitely many numbers between them. For example, 3/2, 4/3, 5/4, and many decimal values lie inside the interval, and more can always be found. Hence D is correct.

The inequality −3 < x excludes −3 because x must be strictly greater than −3. Therefore, the left endpoint uses a parenthesis. The inequality x ≤ 0 includes 0 because equality is permitted, so the right endpoint uses a square bracket. Reading the two endpoint conditions independently gives the interval (−3,0]. This is a half-closed, half-open interval.

Non-negative real numbers are all real numbers greater than or equal to zero, so they satisfy x ≥ 0. Because zero is included, the left endpoint must use a square bracket. The set continues without bound toward positive infinity, and infinity always uses a round bracket because it is not a real number. Thus the correct interval is [0, ∞), making option B correct.