A-Level — mathsAdvanced2 min read

The Modulus Function

Graph and solve equations involving the modulus function at A-Level. Handle |f(x)| and f(|x|).

Tutor AI Team2026-02-25T01:54:11.662Z

The modulus (absolute value) function $|x|$ returns the non-negative value of $x$ . It creates V-shaped graphs and requires careful handling when solving equations and inequalities.

Core Concepts

Definition

$|x| = x$ if $x \geq 0$ , $|x| = -x$ if $x < 0$ .

Graphs

$y = |f(x)|$ : reflect the negative parts of $f(x)$ in the x-axis.
$y = f(|x|)$ : keep the graph for $x \geq 0$ , reflect it in the y-axis.

Solving $|f(x)| = g(x)$

Method 1: Sketch both graphs, find intersections. Method 2: Solve $f(x) = g(x)$ and $f(x) = -g(x)$ , then check validity.

Solving $|f(x)| = |g(x)|$

Square both sides: $[f(x)]^2 = [g(x)]^2$ .

Worked Example: Example 1

Problem

Solve $|2x - 3| = 5$ .

$2x - 3 = 5$ → $x = 4$ . $2x - 3 = -5$ → $x = -1$ .

Solution

Worked Example: Example 2

Problem

Solve $|x + 1| = 2x$ .

$x + 1 = 2x$ → $x = 1$ (check: $|2| = 2$ ✓). $x + 1 = -2x$ → $x = -\frac{1}{3}$ (check: $|\frac{2}{3}| = -\frac{2}{3}$ ✗ — rejected).

Solution: $x = 1$ only.

Solution

Worked Example: Example 3

Problem

Solve $|3x - 1| < 5$ .

$-5 < 3x - 1 < 5$ → $-4 < 3x < 6$ → $-\frac{4}{3} < x < 2$ .

Solution

Practice Problems

1. Sketch $y = |x^2 - 4|$ .
1. Solve $|4x + 1| = |2x - 3|$ .
1. Solve $|x - 2| > 3$ .

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Key Takeaways

✓
$|f(x)|$ : reflect negative parts above x-axis.
✓
$f(|x|)$ : reflect right side to create symmetry about y-axis.
✓
Solve by considering both cases and checking validity.
✓
For $|f(x)| < a$ : $-a < f(x) < a$ .

Tags:

#a-level#maths#pure#functions#modulus#absolute-value

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Core Concepts

Definition

Graphs

Solving ∣f(x)∣=g(x)|f(x)| = g(x)∣f(x)∣=g(x)

Solving ∣f(x)∣=∣g(x)∣|f(x)| = |g(x)|∣f(x)∣=∣g(x)∣

Worked Example: Example 1

Worked Example: Example 2

Worked Example: Example 3

Practice Problems

Key Takeaways

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Solving $|f(x)| = g(x)$

Solving $|f(x)| = |g(x)|$