Solve the inequality |x - 3| ≤ 2. [2 marks]

SEAB Original-style practice: Solve the inequality |x - 2| < 3x.

Since the left side is non-negative, we need 3x > 0, that is x > 0. On x > 0 both sides are non-negative, so squaring is valid. (x - 2)^2 0, so (2x - 1)(x + 1) > 0. This holds for x 12. Combined with x > 0, the solution is x > 12. Markers reward establishing x > 0, valid squaring, factorising the quadratic, and intersecting with the domain restriction.

SingaporeMathsSyllabus dot point

How does the modulus function behave, and how do we solve equations and sketch graphs that involve it?

Define the modulus function, sketch graphs involving the modulus of a function, and solve equations and inequalities involving the modulus

A focused answer to the H2 Mathematics outcome on the modulus function. The definition, sketching the modulus of a linear and a curved function, and solving modulus equations and inequalities by cases and by squaring.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to define the modulus (absolute value) function, sketch graphs involving $|\mathrm{f}(x)|$ , and solve equations and inequalities containing a modulus, either by considering cases or by squaring both sides when both are non-negative.

The answer

Definition

The modulus of $x$ is its size ignoring sign:

|x| = \begin{cases} x & x \geq 0 \\ -x & x < 0 \end{cases}

Equivalently $|x| = \sqrt{x^2}$ , and $|x| \geq 0$ always. Geometrically $|a - b|$ is the distance between $a$ and $b$ on the number line.

Sketching the modulus of a function

To sketch $y = |\mathrm{f}(x)|$ , first sketch $y = \mathrm{f}(x)$ , then reflect any part below the $x$ -axis up to above it. Parts already at or above the axis are unchanged. The result is never negative and has a sharp corner wherever the original graph crossed the axis.

This is different from $y = \mathrm{f}(|x|)$ , where you keep the right half ( $x \geq 0$ ) and reflect it in the $y$ -axis to produce an even (symmetric) graph.

Solving modulus equations

Two reliable methods:

Cases. Split at the point where the inside changes sign, solve each linear or quadratic equation, then check each solution lies in the correct case.
Squaring. If both sides are non-negative, $|A| = B$ is equivalent to $A^2 = B^2$ . Always check the right side is non-negative first.

Solving modulus inequalities

The key facts:

|x| < a \iff -a < x < a, \qquad |x| > a \iff x < -a \text{ or } x > a

For more complex inequalities, square both sides only when both are non-negative, or split into cases. Always sketch to confirm the solution set.

Worked example

Solve the inequality $|x + 1| \geq |2x - 4|$ .

Step 1: Both sides are non-negative, so square

Both sides are moduli, hence non-negative, so the inequality is equivalent to $(x + 1)^2 \geq (2x - 4)^2$ .

Step 2: Expand

x^2 + 2x + 1 \geq 4x^2 - 16x + 16

Step 3: Rearrange to one side

0 \geq 3x^2 - 18x + 15 \implies 0 \geq x^2 - 6x + 5 \implies x^2 - 6x + 5 \leq 0

Step 4: Factorise and solve

$(x - 1)(x - 5) \leq 0$ , which holds for $1 \leq x \leq 5$ .

Step 5: State the solution

The solution set is $1 \leq x \leq 5$ . A quick sketch of the two V-shaped graphs confirms the first graph is on or above the second exactly on this interval.

Examples in context

Example 1. Tolerance in measurement. A component is acceptable if its length $L$ is within $0.2$ mm of $50$ mm, written $|L - 50| \leq 0.2$ . This unfolds to $49.8 \leq L \leq 50.2$ , the allowed range. The modulus captures "within a distance" cleanly.

Example 2. A graph with two corners. The graph of $y = |x^2 - 4|$ is the parabola $y = x^2 - 4$ with the dipped middle section (between $x = -2$ and $x = 2$ , where it was negative) reflected upward, producing a W-like shape with corners at $x = \pm 2$ on the $x$ -axis.

Try this

Q1. Sketch $y = |3 - x|$ , marking the corner and intercepts. [2 marks]

Cue. A V with corner at $(3, 0)$ , $y$ -intercept $3$ , slopes of $-1$ then $+1$ .

Q2. Solve $|2x + 1| = 7$ . [2 marks]

Cue. $2x + 1 = 7$ gives $x = 3$ ; $2x + 1 = -7$ gives $x = -4$ .

Q3. Solve the inequality $|x - 3| \leq 2$ . [2 marks]

Cue. $-2 \leq x - 3 \leq 2$ , so $1 \leq x \leq 5$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksSolve the equation

|2x - 3| = x + 1

Show worked answer →

Split into cases on the sign of $2x - 3$ .

Case 1 ( $2x - 3 \geq 0$ , that is $x \geq 1.5$ ): $2x - 3 = x + 1$ , so $x = 4$ . Check $x = 4 \geq 1.5$ , valid.

Case 2 ( $2x - 3 < 0$ , that is $x < 1.5$ ): $-(2x - 3) = x + 1$ , so $-2x + 3 = x + 1$ , giving $3x = 2$ , $x = \tfrac{2}{3}$ . Check $\tfrac{2}{3} < 1.5$ , valid.

Both must satisfy $x + 1 \geq 0$ (the right side is a modulus output and is non-negative); both do. Solutions: $x = 4$ and $x = \tfrac{2}{3}$ .

Markers reward splitting on the correct boundary, solving each case, and checking each solution against its case condition and the non-negativity of the right side.

Original4 marksSolve the inequality

|x - 2| < 3x

Show worked answer →

Since the left side is non-negative, we need $3x > 0$ , that is $x > 0$ . On $x > 0$ both sides are non-negative, so squaring is valid.

$(x - 2)^2 < 9x^2$ , so $x^2 - 4x + 4 < 9x^2$ , giving $0 < 8x^2 + 4x - 4$ , that is $2x^2 + x - 1 > 0$ , so $(2x - 1)(x + 1) > 0$ .

This holds for $x < -1$ or $x > \tfrac{1}{2}$ . Combined with $x > 0$ , the solution is $x > \tfrac{1}{2}$ .

Markers reward establishing $x > 0$ , valid squaring, factorising the quadratic, and intersecting with the domain restriction.