What is wrong partial-fraction form for a repeated factor?

A factor (x + a)^2 needs both (A)/(x + a) and (B)/((x + a)^2), not a single term.

CCEA CCEA 2021-style practice: Express (3x + 5)/((x + 1)(x + 3)) in partial fractions.

Write (3x + 5)/((x + 1)(x + 3)) = (A)/(x + 1) + (B)/(x + 3). Multiplying through by (x + 1)(x + 3): 3x + 5 = A(x + 3) + B(x + 1). Let x = -1: 3(-1) + 5 = A(2), so 2 = 2A and A = 1. Let x = -3: 3(-3) + 5 = B(-2), so -4 = -2B and B = 2. Therefore (3x + 5)/((x + 1)(x + 3)) = (1)/(x + 1) + (2)/(x + 3). Markers reward the correct form, the cover-up or substitution method, both constants, and the final expression.

CCEA CCEA 2019-style practice: The functions are f(x) = 2x - 1 and g(x) = x^2. Find fg(x) and gf(x), and solve fg(x) = 17.

fg(x) means apply g first, then f: fg(x) = f(x^2) = 2x^2 - 1. gf(x) means apply f first, then g: gf(x) = g(2x - 1) = (2x - 1)^2. Solve fg(x) = 17: 2x^2 - 1 = 17, so 2x^2 = 18, x^2 = 9 and x = 3. Markers reward the correct order for each composite, the two expressions, and both solutions of the equation.

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How do we describe, combine and invert functions, and split a rational expression into partial fractions?

Functions, domain and range, composite and inverse functions, the modulus function and modulus equations, and expressing a rational function as partial fractions.

A CCEA A-Level Mathematics answer on functions with their domain and range, composite and inverse functions, the modulus function and solving modulus equations, and expressing a rational function as partial fractions.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

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

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What this dot point is asking

CCEA wants you to work with functions and their domain and range, form composite and inverse functions, handle the modulus function and solve modulus equations, and express a rational function in partial fractions. These tools are essential for the calculus that follows, where partial fractions make many integrals tractable.

The answer

Functions, domain and range

Composite functions

Inverse functions

The modulus function

The modulus $|x|$ is the non-negative size of $x$ , so $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$ . The graph of $y = |f(x)|$ reflects any part of $y = f(x)$ below the $x$ -axis up above it. To solve $|f(x)| = k$ , solve both $f(x) = k$ and $f(x) = -k$ .

Partial fractions

Before decomposing, check the fraction is proper (the numerator has lower degree than the denominator); if not, divide first to separate a polynomial part. Then factorise the denominator fully, because each factor produces its own partial fraction.

Worked example: solving a modulus equation

Examples in context

Example 1. Undoing a conversion. A temperature formula $F = \frac{9}{5}C + 32$ is a function; its inverse $C = \frac{5}{9}(F - 32)$ converts the other way. Finding an inverse is exactly reversing a real-world conversion.

Example 2. Preparing an integral. The fraction $\frac{1}{(x-1)(x+2)}$ is hard to integrate as it stands, but as partial fractions $\frac{1/3}{x-1} - \frac{1/3}{x+2}$ each piece integrates to a logarithm. Partial fractions are the standard preparation step for integrating rational functions.

Try this

Q1. For $f(x) = 3x + 2$ , find $f^{-1}(x)$ . [2 marks]

Cue. $y = 3x + 2 \Rightarrow x = \frac{y - 2}{3}$ , so $f^{-1}(x) = \frac{x - 2}{3}$ .

Q2. Solve $|x + 1| = 4$ . [2 marks]

Cue. $x + 1 = 4$ or $x + 1 = -4$ , so $x = 3$ or $x = -5$ .

Q3. For $f(x) = x + 1$ and $g(x) = 2x$ , find $fg(x)$ . [2 marks]

Cue. $fg(x) = f(2x) = 2x + 1$ .

Exam-style practice questions

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

CCEA 20216 marksExpress

\frac{3x + 5}{(x + 1)(x + 3)}

in partial fractions.

Show worked answer →

Write $\frac{3x + 5}{(x + 1)(x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 3}$ .

Multiplying through by $(x + 1)(x + 3)$ : $3x + 5 = A(x + 3) + B(x + 1)$ .

Let $x = -1$ : $3(-1) + 5 = A(2)$ , so $2 = 2A$ and $A = 1$ .

Let $x = -3$ : $3(-3) + 5 = B(-2)$ , so $-4 = -2B$ and $B = 2$ .

Therefore $\frac{3x + 5}{(x + 1)(x + 3)} = \frac{1}{x + 1} + \frac{2}{x + 3}$ .

Markers reward the correct form, the cover-up or substitution method, both constants, and the final expression.

CCEA 20195 marksThe functions are

f(x) = 2x - 1

and

g(x) = x^2

. Find

fg(x)

and

gf(x)

, and solve

fg(x) = 17

Show worked answer →

$fg(x)$ means apply $g$ first, then $f$ : $fg(x) = f(x^2) = 2x^2 - 1$ .

$gf(x)$ means apply $f$ first, then $g$ : $gf(x) = g(2x - 1) = (2x - 1)^2$ .

Solve $fg(x) = 17$ : $2x^2 - 1 = 17$ , so $2x^2 = 18$ , $x^2 = 9$ and $x = \pm 3$ .

Markers reward the correct order for each composite, the two expressions, and both solutions of the equation.

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Sources & how we know this

CCEA GCE Mathematics specification — CCEA (2018)