The Modulus Function

This lesson covers the modulus function as required by the AQA A-Level Mathematics specification (7357). The modulus (or absolute value) of a number is its distance from zero on the number line, regardless of sign. The modulus function appears in equations, inequalities, and graph transformations, and is a topic that many students find challenging.

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Definition of the Modulus

The modulus of a real number x, written |x|, is defined as:

|x| = { x    if x ≥ 0
       { −x   if x < 0

In words: |x| is always non-negative. It gives the magnitude of x without regard to sign.

Examples:

• |5| = 5
• |−3| = 3
• |0| = 0
• |−7.2| = 7.2

Properties of the Modulus

Property	Statement
Non-negativity
	x
	xy
	x/y
	x
Triangle inequality

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The Graph of y = |x|

The graph of y = |x| is a V-shape with vertex at the origin. It consists of:

• y = x for x ≥ 0 (the right-hand branch)
• y = −x for x < 0 (the left-hand branch, which is the reflection of y = x in the y-axis)

The graph always lies on or above the x-axis.

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Graphs of y = |f(x)| and y = f(|x|)

y = |f(x)|

To sketch y = |f(x)| from y = f(x):

1. Draw the graph of y = f(x).
2. Any part of the graph below the x-axis (where f(x) < 0) is reflected in the x-axis.
3. Parts on or above the x-axis remain unchanged.

Example: Sketch y = |x² − 4|.

Start with y = x² − 4 (a U-shaped parabola crossing the x-axis at x = −2 and x = 2, with minimum at (0, −4)).

For |x² − 4|: the portion between x = −2 and x = 2 (where x² − 4 < 0) is reflected upward. The minimum value becomes +4 at x = 0, and the curve touches the x-axis at x = ±2.

y = f(|x|)

To sketch y = f(|x|) from y = f(x):

1. Draw the part of y = f(x) for x ≥ 0 only.
2. Reflect this part in the y-axis to get the left-hand side.

The resulting graph is always symmetrical about the y-axis.

Example: Sketch y = (|x| − 1)(|x| − 3).

For x ≥ 0: y = (x − 1)(x − 3), which is a U-shaped parabola with roots at x = 1 and x = 3, minimum at (2, −1).

The graph of y = f(|x|) reflects this in the y-axis, giving a graph symmetrical about x = 0 with roots at x = ±1 and x = ±3.

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Solving Modulus Equations

Type 1: |f(x)| = k (where k > 0)

|f(x)| = k  ⟹  f(x) = k  or  f(x) = −k

Example: Solve |2x − 5| = 3.

2x − 5 = 3   →   2x = 8   →   x = 4
2x − 5 = −3  →   2x = 2   →   x = 1

Solutions: x = 1 and x = 4.

Check: |2(4) − 5| = |3| = 3 ✓ and |2(1) − 5| = |−3| = 3 ✓.

Type 2: |f(x)| = |g(x)|

|f(x)| = |g(x)|  ⟹  f(x) = g(x)  or  f(x) = −g(x)

Example: Solve |3x − 1| = |x + 5|.

Case 1: 3x − 1 = x + 5 → 2x = 6 → x = 3. Case 2: 3x − 1 = −(x + 5) → 3x − 1 = −x − 5 → 4x = −4 → x = −1.

Solutions: x = 3 and x = −1.

Type 3: |f(x)| = g(x)

If the right-hand side is a function (not just a constant), solve f(x) = g(x) and f(x) = −g(x), then check that g(x) ≥ 0 for each solution (since |f(x)| ≥ 0, we need g(x) ≥ 0).

Example: Solve |x − 4| = 2x − 1.

Case 1: x − 4 = 2x − 1 → −3 = x → x = −3. Check: |−3 − 4| = 7 and 2(−3) − 1 = −7. Since −7 ≠ 7, reject x = −3.

Case 2: −(x − 4) = 2x − 1 → −x + 4 = 2x − 1 → 5 = 3x → x = 5/3. Check: |5/3 − 4| = |−7/3| = 7/3 and 2(5/3) − 1 = 10/3 − 3/3 = 7/3. ✓

Solution: x = 5/3.

Exam Tip: When solving |f(x)| = g(x), always check your solutions. Spurious solutions often arise from the case that produces negative values of g(x).

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Solving Modulus Inequalities

Type 1: |x| < a (where a > 0)

|x| < a  ⟺  −a < x < a

Type 2: |x| > a (where a > 0)

|x| > a  ⟺  x < −a  or  x > a

General Approach

Example: Solve |2x − 3| < 5.

−5 < 2x − 3 < 5
−2 < 2x < 8
−1 < x < 4

Solution: −1 < x < 4.

Example: Solve |x + 1| ≥ 4.

x + 1 ≥ 4    →   x ≥ 3
or
x + 1 ≤ −4   →   x ≤ −5

Solution: x ≤ −5 or x ≥ 3.

Graphical Approach to Modulus Inequalities

For more complex modulus inequalities, sketching both sides of the inequality and identifying the intersection points is often the clearest method.

Example: Solve |x² − 4| < 3x.

Sketch y = |x² − 4| and y = 3x on the same axes. The solution is the set of x-values where the V-shaped curve lies below the line.

Solve the equations:

• x² − 4 = 3x → x² − 3x − 4 = 0 → (x − 4)(x + 1) = 0 → x = 4 or x = −1.
• −(x² − 4) = 3x → −x² + 4 = 3x → x² + 3x − 4 = 0 → (x + 4)(x − 1) = 0 → x = −4 or x = 1.

From the graph, the curve |x² − 4| is below 3x when 1 < x < 4. (Rejecting negative x-values because 3x must be positive for the inequality to make sense.)

Solution: 1 < x < 4.

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Summary

• The modulus |x| gives the magnitude of x: |x| = x if x ≥ 0 and |x| = −x if x < 0.
• y = |f(x)|: reflect parts below the x-axis upward.
• y = f(|x|): keep x ≥ 0 part and reflect it in the y-axis.
• |f(x)| = k: solve f(x) = k and f(x) = −k.
• |f(x)| = |g(x)|: solve f(x) = g(x) and f(x) = −g(x).
• |x| < a means −a < x < a; |x| > a means x < −a or x > a.
• Always check solutions when solving modulus equations.
• The graphical approach is often the clearest method for modulus inequalities.

Exam Tip: Drawing a sketch is almost always the best strategy for modulus questions. It prevents sign errors and helps you identify all solutions. When solving |f(x)| = g(x), you must check that g(x) ≥ 0 at each solution. Always state your final answer clearly in set or interval notation.

A-Level Deep Dive: The Modulus Function

Spec mapping

AQA 7357 specification, Pure section B (Algebra and functions), Year 2 content covers the modulus function, including the notation $|x|$∣x∣, and use relations such as $|a + b| \leq |a| + |b|$∣a+b∣≤∣a∣+∣b∣ and $|ax + b| = c$∣ax+b∣=c in problem-solving contexts (refer to the official specification document for exact wording). The modulus is examined on Paper 2 alongside calculus, sequences and the broader Year 2 algebra strand. It interlocks with graph transformations (Year 1 — sketching $|f(x)|$∣f(x)∣ and $f(|x|)$f(∣x∣)), with functions (composite and inverse, where modulus often gives a non-invertible map without domain restriction), with inequalities (modulus inequalities such as $|2x + 1| < |x - 3|$∣2x+1∣<∣x−3∣), and with trigonometric / exponential equations when their solution sets cross sign boundaries. The AQA formula booklet does not list modulus identities — $|x|^2 = x^2$∣x∣2=x2, the case-split definition, and the triangle inequality must be memorised.

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Worked example with full mark scheme

Question (8 marks):

(a) On the same axes, sketch the graphs of $y = |2x - 3|$y=∣2x−3∣ and $y = x + 1$y=x+1, labelling intercepts and the vertex of the modulus graph. (3)

(b) Hence solve $|2x - 3| = x + 1$∣2x−3∣=x+1. (3)

(c) State, with reasons, the values of $k$k for which $|2x - 3| = x + k$∣2x−3∣=x+k has exactly two real solutions. (2)

Solution with mark scheme:

(a) Step 1 — sketch $y = |2x - 3|$y=∣2x−3∣.

The vertex sits where the inner expression vanishes, $2x - 3 = 0 \implies x = \tfrac{3}{2}$2x−3=0⟹x=23​, giving the point $\bigl(\tfrac{3}{2}, 0\bigr)$(23​,0). The graph is a V with gradients $-2$−2 (for $x < \tfrac{3}{2}$x<23​) and $+2$+2 (for $x > \tfrac{3}{2}$x>23​), $y$y-intercept at $(0, 3)$(0,3).

M1 — V-shape with vertex on the $x$x-axis at $x = \tfrac{3}{2}$x=23​.

Step 2 — sketch $y = x + 1$y=x+1.

A straight line, gradient $1$1, $y$y-intercept $(0, 1)$(0,1), $x$x-intercept $(-1, 0)$(−1,0).

B1 — straight line correctly placed, with at least one intercept labelled.

A1 — both graphs on the same axes, vertex $\bigl(\tfrac{3}{2}, 0\bigr)$(23​,0) and the two intersection points clearly indicated (without needing exact coordinates yet).

(b) Step 1 — case split.

$|2x - 3| = 2x - 3$∣2x−3∣=2x−3 when $x \geq \tfrac{3}{2}$x≥23​, and $|2x - 3| = -(2x - 3) = 3 - 2x$∣2x−3∣=−(2x−3)=3−2x when $x < \tfrac{3}{2}$x<23​.

M1 — splitting into the correct two cases with the boundary stated.

Step 2 — solve each case.

Case 1 ($x \geq \tfrac{3}{2}$x≥23​): $2x - 3 = x + 1 \implies x = 4$2x−3=x+1⟹x=4. Check $4 \geq \tfrac{3}{2}$4≥23​: valid.

Case 2 ($x < \tfrac{3}{2}$x<23​): $3 - 2x = x + 1 \implies 2 = 3x \implies x = \tfrac{2}{3}$3−2x=x+1⟹2=3x⟹x=32​. Check $\tfrac{2}{3} < \tfrac{3}{2}$32​<23​: valid.

M1 — solving both linear equations correctly.

A1 — both solutions $x = \tfrac{2}{3}$x=32​ and $x = 4$x=4 stated, with the validity check carried out (or implied by domain remarks).

(c) Step 1 — geometric reasoning.

The line $y = x + k$y=x+k has fixed gradient $1$1, intersecting the V whose right arm has gradient $2$2 and left arm has gradient $-2$−2. The line is shallower than both arms in magnitude, so it cuts the V in exactly two points whenever it passes above the vertex $\bigl(\tfrac{3}{2}, 0\bigr)$(23​,0).

Two intersections occur precisely when $\tfrac{3}{2} + k > 0 \implies k > -\tfrac{3}{2}$23​+k>0⟹k>−23​.

M1 — geometric or algebraic reasoning about line-vs-V intersections.

A1 — answer $k > -\tfrac{3}{2}$k>−23​, with the boundary case $k = -\tfrac{3}{2}$k=−23​ (tangent at vertex, one solution) explicitly excluded.

Total: 8 marks (B1 M3 A4).

A subtlety frequently missed: when the line has the same gradient as one arm of the V, you get exactly one solution (parallel to the other arm) or no solution; and when it is steeper than both arms, intersections still come in pairs above the vertex but the algebra reverses. Sketch first, then case-split — never the other way round.

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Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): The function $f$f is defined by $f(x) = |x^2 - 4|$f(x)=∣x2−4∣ for $x \in \mathbb{R}$x∈R.

(a) Sketch the graph of $y = f(x)$y=f(x), labelling all intercepts with the coordinate axes and the local maximum. (3)

(b) Solve $f(x) = 3$f(x)=3. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — recognising that $y = x^2 - 4$y=x2−4 is reflected in the $x$x-axis wherever it is negative (i.e. on $-2 < x < 2$−2<x<2).
• B1 (AO1.1b) — $x$x-intercepts at $(\pm 2, 0)$(±2,0) correctly marked, $y$y-intercept at $(0, 4)$(0,4) (formerly $-4$−4, now reflected) labelled.
• A1 (AO2.5) — local maximum at $(0, 4)$(0,4) identified, with the graph showing the characteristic shape: parabola for $|x| \geq 2$∣x∣≥2, inverted parabola for $|x| < 2$∣x∣<2.

(b)

• M1 (AO1.1b) — splitting into cases $x^2 - 4 = 3$x2−4=3 and $x^2 - 4 = -3$x2−4=−3.
• A1 (AO1.1b) — $x^2 = 7 \implies x = \pm\sqrt{7}$x2=7⟹x=±7​ from the first; $x^2 = 1 \implies x = \pm 1$x2=1⟹x=±1 from the second.
• A1 (AO3.2a) — all four solutions $x \in \{-\sqrt{7}, -1, 1, \sqrt{7}\}$x∈{−7​,−1,1,7​} presented and (ideally) cross-checked against the sketch.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. The AO3 mark is for bringing the graph and the algebra into agreement — a candidate who solves only one case algebraically but draws four intersections on the sketch should notice the inconsistency.

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Synoptic links

Connects to:

1. Graph transformations (Year 1, Pure section B): $y = |f(x)|$y=∣f(x)∣ reflects the negative parts of $f$f in the $x$x-axis; $y = f(|x|)$y=f(∣x∣) takes the right-hand half of $f$f and mirrors it in the $y$y-axis (discarding the original left half). These are different transformations and frequently confused. Mastery of the modulus function is, in practice, mastery of these two transformations.

2. Functions and their inverses (Year 2, Pure section B): the modulus is many-to-one, so $f(x) = |x|$f(x)=∣x∣ has no inverse over $\mathbb{R}$R. Restricting the domain to $x \geq 0$x≥0 gives $f^{-1}(x) = x$f−1(x)=x; restricting to $x \leq 0$x≤0 gives $f^{-1}(x) = -x$f−1(x)=−x. This is the cleanest example of domain restriction to enable inversion on the spec.

3. Inequalities (Year 1 / Year 2 algebra): modulus inequalities such as $|2x + 1| < |x - 3|$∣2x+1∣<∣x−3∣ are best solved either graphically (sketch both V-shapes, find intersections, read off intervals) or by squaring both sides (valid only when both sides are non-negative, which modulus guarantees). Squaring converts a modulus inequality to a polynomial inequality without sign ambiguity.

4. Trigonometric equations (Pure section E): equations like $|\sin x| = \tfrac{1}{2}$∣sinx∣=21​ are routinely set on Paper 2. Each branch of the modulus generates its own family of solutions; exhaustive listing across the requested interval is the marked skill.

5. Differentiation (Pure section G): $y = |x|$y=∣x∣ is not differentiable at $x = 0$x=0. The left-derivative is $-1$−1, the right-derivative is $+1$+1 — they do not agree. This is the simplest example of a corner / non-smooth point on the entire A-Level. It generalises: $y = |f(x)|$y=∣f(x)∣ has a corner wherever $f$f crosses zero with non-zero gradient.

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Mark-scheme literacy

Modulus questions on AQA 7357 distribute AO marks across all three categories more evenly than algebra-of-surds questions: