representation on number line

A closed circle includes (7), and the left side contains numbers less than (7). In exams, add equality when you see a closed circle.

The bracket (() means (-2) is out and (]) means (4) is in. In exams, check left and right brackets separately.

The first part is less than (-1), and the second part is (3) and greater. In exams, read (\cup) as or.

The common part is greater than (4) and up to (6). In exams, conditions taken together mean intersection.

The governing concept is the intersection of two inequalities: a value must satisfy both conditions at the same time. The inequality x ≥ −5 includes −5, so the left endpoint is marked with a closed bracket. The inequality x < 2 excludes 2, so the right endpoint is marked with an open bracket. Combining the common values gives −5 ≤ x < 2, which is written in interval notation as [−5, 2). Therefore, option A is correct. Option B reverses the inclusion signs, while options C and D ignore one of the two restrictions and extend too far along the number line.

([3,\infty)) extends endlessly to the right, so it has no final right endpoint. In exams, do not treat (\infty) as an actual point.

((-\infty,0)) excludes (0) and contains all real numbers less than it. In exams, pay attention to the open bracket.

Only separate integer points are filled, so (x\in\mathbb{Z}). In exams, distinguish a shaded line segment from separate points.

Solving gives (-2x\le -4), and dividing by a negative gives (x\ge 2). In exams, reversing the sign is essential.

Subtracting (1) from all parts gives (3<x\le 7). In exams, change both boundaries together in a compound inequality.

(-2) is not included, so the integer just before it is (-3). In exams, do not choose an open endpoint.

(-4) is not included, so the next integer is (-3). In exams, take the integer after an open left endpoint.

The direct answer is option B: only a closed point at 2. The interval \([2,2]\) starts at 2 and ends at 2, and both square brackets say that the endpoint is included. There is no space between the endpoints because they are the same number. Therefore the set contains exactly one real number: 2. On the number line, represent it by one filled or closed dot at 2. Option A, no point, is wrong because 2 is included. Option B is correct because it shows exactly the single member of the interval. Option C, a ray to the right, represents something like \([2,\infty)\), containing numbers greater than 2, not just 2. Option D, a ray to the left, represents a set containing numbers less than 2, also not correct. The notation \([a,a]\) always reduces to the singleton set \(\{a\}\). Exam cue: equal closed endpoints mean one filled point; equal open endpoints would mean an empty set.

No number can be greater than (3) and less than (3) at the same time. In exams, an open interval with equal endpoints is empty.

In (x\neq 2), all real numbers except (2) are included. In exams, show the excluded point with an open circle.

For any real (x), (x^2) is never negative. In exams, represent an always true inequality by the whole line.

The square of a real number is never less than (0). In exams, write an impossible condition as the empty set.

The governing concept is the distance interpretation of absolute value. The statement |x| < 3 means that the distance of x from zero is less than 3. Therefore, x must lie strictly between −3 and 3, which can be written as −3 < x < 3. Since the inequality is strict, neither endpoint is included, so both ends use round brackets. The interval is consequently (−3, 3), making option A correct. Option B includes the endpoints and would represent |x| ≤ 3. Option C describes values whose distance from zero is greater than 3, and option D includes only the right-hand portion, so neither matches the given condition.

The governing concept is the distance meaning of absolute value together with a union of two intervals. The inequality |x| ≥ 4 says that x is at least 4 units from zero. Hence x ≤ −4 or x ≥ 4. The word “or” requires a union: (−∞, −4] ∪ [4, ∞). The endpoints −4 and 4 are included because the sign is ≥, so square brackets are required at both finite endpoints. Therefore, option C is correct. Option A is the central interval and represents distance at most 4; option B excludes the endpoints and represents distance less than 4; option D incorrectly omits the separate right-side condition and includes values between −4 and 4.

The direct answer is option B: \(-1,0,1,2\). The condition \(x\in\mathbb Z\) means x must be an integer, so we list only whole numbers, including negative whole numbers, zero, and positive whole numbers. The inequality is \(-2<x\le2\). Because the left sign is strict, -2 is excluded. Because the right sign includes equality, 2 is included. The integers strictly greater than -2 and up to 2 are therefore \(-1,0,1,2\). Option A incorrectly includes -2. Option B lists exactly all allowed integers. Option C incorrectly omits 2, even though \(x=2\) satisfies the inequality. Option D incorrectly omits -1 and 0, both of which satisfy it. On a number line for integers, mark separate points rather than shading every real number. Memory trick: strict left boundary removes -2; inclusive right boundary keeps 2.

The two parts take all values less than and greater than (1), but not (1). In exams, combine both parts for union.

The governing concept is intersection, which means the set of values satisfying both conditions simultaneously. From x ≥ 0, the possible values are 0 and all numbers to its right. From x ≤ 0, the possible values are 0 and all numbers to its left. The only common value in these two sets is exactly 0. Both inequalities include 0 because their signs are ≥ and ≤, so the point is included rather than excluded. Therefore, the intersection is the singleton set {0}, and option C is correct. The entire line would satisfy neither restriction, the empty set overlooks the common endpoint, and the line excluding 0 contains values that violate one of the inequalities.

In ((1,\infty)), (1) is not included but greater values are included. In exams, exclude the open left endpoint.

Both boundaries are equal and included, so only the point (-6) is obtained. In exams, show it as one closed point.

The intersection of two sets contains only the numbers that belong to both sets. The interval \((-\infty,5]\) contains every number less than or equal to 5. The interval \((2,\infty)\) contains every number greater than 2, but it does not contain 2 itself. We therefore keep only the overlap of these conditions.

A number in the intersection must satisfy \(x>2\) and \(x\le5\). Thus all numbers between 2 and 5 are included, 2 is excluded, and 5 is included. In interval notation this is \((2,5]\), represented by an open circle at 2 and a closed circle at 5 with shading between them. Hence option A is correct.