GCSE Maths · Higher · Chapter 24: Algebraic fractions and functions

Algebraic fractions, rearranging formulae, composite and inverse functions, and iteration

Grades 4–9About 85 minutesCalculator allowed92 marks

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NameClassDate

Before you start

Grades 4–5

You will need these in every question below.

A1
Factorise $x^{2} - 9$ .
[1 mark]
Answer

A2
Factorise $x^{2} + 9 x + 14$ .
[1 mark]
Answer
A3
Work out $\frac{3}{4} + \frac{1}{6}$ .
[1 mark]
Answer
A4
Make $x$ the subject of $y = 5 x - 2$ .
[2 marks]
Answer
A5
Solve $x^{2} - 2 x - 15 = 0$ .
[2 marks]
Answer

Examples, then your turn

Grades 5–6

Your teacher works through each example with you. Then try the one beside it on your own.

Example 1

Simplify fully $\frac{x ^{2} + 3 x - 10}{x ^{2} - 4}$ .

Hint Factorise the top and the bottom. Then cancel the bracket they share.

B1
Simplify fully $\frac{x ^{2} - 2 x - 8}{x ^{2} - 16}$ .
[3 marks]
Answer

Example 2

Write $\frac{3}{x + 1} + \frac{2}{x - 2}$ as a single fraction.

Hint Use

(x + 1) (x - 2)

as the common denominator.

B2
Write $\frac{4}{x - 3} - \frac{1}{x + 2}$ as a single fraction in its simplest form.
[3 marks]
Answer

Example 3

Make $y$ the subject of $a (y + 1) = 2 y - 5$ .

Hint Expand the bracket. Collect every term with

y

on one side, then take

y

out as a factor.

B3
Make $y$ the subject of $t (y - 2) = 3 y + 1$ .
[3 marks]
Answer

Example 4

$f (x) = 2 x + 3$ and $g (x) = x^{2}$ . Find $f g (2)$ and $f^{- 1} (x)$ .

Hint

f g (2)

means do

g

first. For the inverse, write

y = 2 x + 3

and make

x

the subject.

B4
$f (x) = 5 x - 1$ and $g (x) = x^{2} + 2$ . Find $g f (1)$ and $f^{- 1} (x)$ .
[3 marks]
Answer $g f (1) = \dots\dots$ $f^{- 1} (x) = \dots\dots$

Practice

Grades 5–7

Each question asks for something different. Show your working.

C1
Simplify fully $\frac{2 x ^{2} + 7 x + 3}{4 x ^{2} - 1}$ .
[3 marks]
Answer

C2
Simplify fully $\frac{x ^{2} - 25}{x + 2} \div \frac{x + 5}{2 x ^{2} + 4 x}$ .
[3 marks]
Answer
C3
Solve $\frac{3}{x - 1} + \frac{2}{x + 2} = 1$ . Give your answers to 2 decimal places.
[4 marks]
Answer
C4
Make $x$ the subject of $y = \frac{x + 4}{x - 3}$ .
[3 marks]
Answer
C5
$f (x) = 3 x - 1$ and $g (x) = \frac{1}{x + 2}$ .
(a)
Find $g f (3)$ .
[2 marks]
Answer
(b)
Find $f g (x)$ . Give your answer as a single fraction.
[2 marks]
Answer
(c)
Which value of $x$ cannot be used as an input to $g$ ?
[1 mark]
Answer
Total for C5: 5 marks
C6
$f (x) = \frac{2 x + 7}{3}$ . Find $f^{- 1} (x)$ .
[2 marks]
Answer
C7
Use the iteration formula $x_{n + 1} = 3 5 x_{n} + 3$ with $x_{0} = 2$ .
(a)
Work out $x_{1}$ . Give your answer to 4 decimal places.
[1 mark]
Answer
(b)
Work out $x_{3}$ . Give your answer to 4 decimal places.
[2 marks]
Answer
Total for C7: 3 marks
C8
Show that the equation $x^{2} - 7 x + 2 = 0$ can be rearranged to give $x = 7 - \frac{2}{x}$ .
[2 marks]

Exam-style questions

Grades 6–9

Answer every question in the space given. Show your working: most of the marks are for method.

D1
Show that $\frac{3}{x + 2} - \frac{2}{x - 1}$ can be written as $\frac{x - 7}{( x + 2 ) ( x - 1 )}$ .
[3 marks]

D2
Ben says, " $\frac{x + 6}{x + 3}$ simplifies to $2$ , because $6 \div 3 = 2$ ."
Ben is wrong. Explain why.
[2 marks]
Answer
D3
$f (x) = x + 3$ and $g (x) = 2 x^{2}$ .
Solve $f g (x) = g f (x)$ . You must show your working.
[4 marks]
Answer
D4
$f (x) = 4 - 3 x$ .
Solve $f (x) = f^{- 1} (x)$ .
[3 marks]
Answer
D5
The equation $x^{3} + 4 x - 9 = 0$ has one solution.
(a)
Show that the solution lies between $1$ and $2$ .
[2 marks]
(b)
Show that the equation can be rearranged to give $x = 3 9 - 4 x$ .
[1 mark]
(c)
Use the iteration formula $x_{n + 1} = 3 9 - 4 x_{n}$ with $x_{0} = 1.5$ to find $x_{1}$ , $x_{2}$ and $x_{3}$ . Give each answer to 4 decimal places.
[3 marks]
Answer $x_{1} = \dots\dots$ $x_{2} = \dots\dots$ $x_{3} = \dots\dots$
Total for D5: 6 marks
D6
Make $t$ the subject of the formula $s = \frac{a + b t}{1 - t}$ .
[4 marks]
Answer
D7
$f (x) = x^{2} + 6 x + 4$ for $x \geq - 3$ .
Find $f^{- 1} (x)$ .
[4 marks]
Answer
D8
This question is about the sum $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \frac{1}{4 \times 5}$ .
(a)
Show that $\frac{1}{n} - \frac{1}{n + 1} = \frac{1}{n ( n + 1 )}$ .
[2 marks]
(b)
Hence work out the value of the sum.
[1 mark]
Answer
Total for D8: 3 marks
D9
A sequence is given by $x_{n + 1} = 2 + \frac{3}{x _{n}}$ with $x_{1} = 1$ .
(a)
Work out $x_{2}$ , $x_{3}$ and $x_{4}$ . Give $x_{4}$ to 2 decimal places.
[2 marks]
Answer $x_{2} = \dots\dots$ $x_{3} = \dots\dots$ $x_{4} = \dots\dots$
(b)
The values get closer and closer to a number $L$ . By solving $L = 2 + \frac{3}{L}$ , find $L$ .
[2 marks]
Answer
Total for D9: 4 marks
D10
$f (x) = 3 x + k$ , where $k$ is a constant. $f f (2) = 32$ .
Find the value of $k$ .
[3 marks]
Answer

Extension

Grade 9 and beyond

No route is given. Spend five minutes on a problem before you look at a hint. The hints are on the last page of the worksheet, and you can use one at a time.

E1
$f (x) = \frac{x + 5}{x - 1}$ , $x \neq = 1$ .
Show that $f$ is its own inverse.
[3 marks]
Hint 1 · What to try
Find $f^{- 1} (x)$ in the usual way and compare it with $f (x)$ .
Hint 2 · The first line
Write $y (x - 1) = x + 5$ , so $x y - y = x + 5$ .
Hint 3 · The full method
$x (y - 1) = y + 5$ , so $x = \frac{y + 5}{y - 1}$ . That gives $f^{- 1} (x) = \frac{x + 5}{x - 1}$ , which is $f (x)$ .

E2
Simplify fully $\frac{x ^{4} - 16}{x ^{3} + 2 x ^{2} + 4 x + 8}$ .
[3 marks]
Hint 1 · What to try
Factorise the top as a difference of two squares, twice if you can.
Hint 2 · The first line
$x^{4} - 16 = (x^{2} + 4) (x + 2) (x - 2)$ . Factorise the bottom in pairs.
Hint 3 · The full method
$x^{3} + 2 x^{2} + 4 x + 8 = x^{2} (x + 2) + 4 (x + 2) = (x + 2) (x^{2} + 4)$ , so the fraction is $x - 2$ .
Answer
E3
Make $x$ the subject of $y = \frac{x + a}{x - a}$ .
[3 marks]
Hint 1 · What to try
Square both sides first.
Hint 2 · The first line
$y^{2} (x - a) = x + a$ , so $x y^{2} - a y^{2} = x + a$ .
Hint 3 · The full method
$x (y^{2} - 1) = a (y^{2} + 1)$ , so $x = \frac{a ( y ^{2} + 1 )}{y ^{2} - 1}$ .
Answer
E4
$f (x) = x^{2} - 2 x$ and $g (x) = 2 x + 1$ .
Solve $f g (x) = g f (x)$ . Give your answers in surd form.
[3 marks]
Hint 1 · What to try
Work out each composite function and set them equal.
Hint 2 · The first line
$f g (x) = (2 x + 1)^{2} - 2 (2 x + 1) = 4 x^{2} - 1$ and $g f (x) = 2 x^{2} - 4 x + 1$ .
Hint 3 · The full method
$2 x^{2} + 4 x - 2 = 0$ , so $x^{2} + 2 x - 1 = 0$ and $x = - 1 \pm 2$ .
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can simplify, add, subtract, multiply and divide algebraic fractions.
I can make a letter the subject of a formula when it appears more than once.
I can use function notation, work out composite functions and find an inverse function.
I can show that an equation has a solution in an interval and use an iteration formula to find it.

Answers and mark scheme

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