A-Level Maths (Edexcel 9MA0) · Pure Year 2 · Chapter 2: Functions and graphs · Lesson 1

The modulus function

112 marks

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Date

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Quick questions to start.

A1
Needed today
Solve the inequality $5 - 2 x > 3 x + 20$ .
[2 marks]
Answer
A2
Needed today
The line $y = 3 x - 12$ meets the $x$ -axis at $A$ and the $y$ -axis at $B$ . Write down the coordinates of $A$ and of $B$ .
[2 marks]
Answer
A3
Common slip
Given that $a = - 5$ and $b = 3$ , work out $∣ a + b ∣$ , $∣ a ∣ + ∣ b ∣$ and $∣ a - b ∣$ .
[3 marks]
Answer
A4
Common slip
Find both values of $x$ for which $∣ x - 3∣ = 5$ .
[2 marks]
Answer

Example, then your turn

A-Level

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

Example 1

Sketch the graph of $y = ∣2 x - 5∣$ . Give the coordinates of the points where the graph meets the axes.

Hint Sketch

y = 2 x - 5

first, then reflect the part below the

x

-axis in the

x

-axis.

B1
Sketch the graph of $y = ∣4 - 3 x ∣$ . Give the coordinates of the points where the graph meets the axes.
[3 marks]
Answer

Example 2

Solve $∣ x - 4∣ = 2 x + 1$ .

Hint Solve

x - 4 = 2 x + 1

and

- (x - 4) = 2 x + 1

, then put each answer back into the original equation.

B2
Solve $∣2 x + 3∣ = 3 x - 1$ .
[5 marks]
Answer

Example 3

Solve $∣3 x - 1∣ = ∣ x + 5∣$ .

Hint Either the insides are equal or one is minus the other. Squaring both sides gives the same two answers.

B3
Solve $∣2 x + 7∣ = ∣ x - 4∣$ .
[4 marks]
Answer

Example 4

Find the set of values of $x$ for which $∣2 x - 3∣ < x + 3$ .

Hint Find where the graphs of

y = ∣2 x - 3∣

and

y = x + 3

cross, then use a sketch to see where the V is below the line.

B4
Find the set of values of $x$ for which $∣ x + 2∣ > 2 x - 1$ .
[5 marks]
Answer

Practice

A-Level

C1
Rosa solves $∣ x - 6∣ = 3 x + 2$ . She writes: " $x - 6 = 3 x + 2$ gives $x = - 4$ , and $- (x - 6) = 3 x + 2$ gives $x = 1$ . So $x = - 4$ or $x = 1$ ." Find her mistake and give the correct solution.
[4 marks]
Answer

C2
Kai says that $∣ x + 3∣ = ∣ x ∣ + 3$ for every real number $x$ . Show that Kai is wrong, and find the set of values of $x$ for which his statement is true.
[4 marks]
Answer
C3
Solve the inequality $∣5 - 2 x ∣ \geq 9$ .
[4 marks]
Answer
C4
The graph of $y = ∣4 x - a ∣$ , where $a$ is a positive constant, meets the $y$ -axis at $(0, 12)$ . (a) Find $a$ and the coordinates of the vertex. (b) Solve $∣4 x - a ∣ = x$ .
[5 marks]
Answer
C5
By squaring both sides, find the set of values of $x$ for which $∣ x - 1∣ < ∣ x + 5∣$ .
[4 marks]
Answer
C6
Find the set of values of $x$ for which $∣2 x + 1∣ \leq ∣ x - 3∣$ .
[5 marks]
Answer

Exam-style questions

Edexcel exam style

D1
$f (x) = ∣2 x - 8∣$
(a)
Sketch the graph of $y = f (x)$ , giving the coordinates of any points where the graph meets the coordinate axes.
[3 marks]
(b)
Solve $f (x) = 12 - x$ .
[3 marks]
Answer
(c)
Hence find the set of values of $x$ for which $f (x) > 12 - x$ .
[3 marks]
Answer
Total for D1: 9 marks

D2
Sam solves $∣3 x + 1∣ < 2 x + 6$ . He writes: " $3 x + 1 < 2 x + 6$ , so $x < 5$ . Also $- (3 x + 1) < 2 x + 6$ , so $- 5 x < 7$ , giving $x < - \frac{7}{5}$ . So the answer is $x < 5$ ."
Sam is wrong.
(a)
Identify the mistake Sam makes.
[2 marks]
(b)
Find the correct set of values of $x$ for which $∣3 x + 1∣ < 2 x + 6$ .
[4 marks]
Answer
(c)
Hence find how many integers satisfy $∣3 x + 1∣ < 2 x + 6$ .
[2 marks]
Answer
Total for D2: 8 marks
D3
Solve the equation $∣3 x - 2∣ = ∣ x ∣ + 4$ .
[6 marks]
Answer
D4
$g (x) = ∣ x - 4∣$ . The line $l$ has equation $y = k x + 2$ , where $k$ is a constant.
(a)
Sketch the graph of $y = g (x)$ , giving the coordinates of the points where it meets the axes.
[2 marks]
(b)
Find the value of $k$ for which $l$ passes through the vertex of the graph.
[2 marks]
Answer
(c)
Find the set of values of $k$ for which the equation $∣ x - 4∣ = k x + 2$ has exactly two solutions.
[4 marks]
Answer
(d)
State the number of solutions of $∣ x - 4∣ = k x + 2$ when $k = - 2$ . Give a reason.
[2 marks]
Total for D4: 10 marks
D5
The diagram shows part of the graph of $y = ∣ a x + b ∣$ , where $a$ and $b$ are constants and $a > 0$ . The graph meets the $y$ -axis at $(0, 12)$ and has its vertex at $(3, 0)$ .
(a)
Find the value of $a$ and the value of $b$ .
[2 marks]
Answer
(b)
Solve $∣ a x + b ∣ = 2 x + 3$ .
[3 marks]
Answer
(c)
Find the set of values of the constant $c$ for which $∣ a x + b ∣ = 2 x + c$ has two distinct solutions.
[4 marks]
Answer
Total for D5: 9 marks
D6
$h (x) = ∣ x + 2∣ + ∣ x - 4∣$
(a)
Show that $h (x) = 6$ for $- 2 \leq x \leq 4$ .
[2 marks]
(b)
Write $h (x)$ without modulus signs for $x > 4$ and for $x < - 2$ .
[2 marks]
Answer
(c)
Solve $h (x) = 10$ .
[3 marks]
Answer
(d)
Hence find the set of values of $x$ for which $h (x) < 8$ .
[2 marks]
Answer
Total for D6: 9 marks

Extension

Stretch

No route is given. The hints are at the end of the sheet. Use one at a time.

E1
The constant $k$ satisfies $k \geq 0$ . Find every value of $k$ for which the equation $∣2 x - 3∣ = k ∣ x + 1∣$ has exactly one real solution.
[9 marks]
Hint 1 · What to try
Because $k \geq 0$ , $k ∣ x + 1∣ = ∣ k (x + 1) ∣$ . So the equation means $2 x - 3 = k (x + 1)$ or $2 x - 3 = - k (x + 1)$ .
Hint 2 · The first line
The two cases give $x = \frac{3 + k}{2 - k}$ and $x = \frac{3 - k}{2 + k}$ . When can there be only one solution?
Hint 3 · The full method
Either the two answers are equal, or one of the cases has no solution at all. Look at $k = 0$ and at $k = 2$ .
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can sketch the graph of y = \|ax + b\| and give its vertex and intercepts.
I can solve an equation with a modulus and reject answers that do not work.
I can solve an inequality with a modulus, using a sketch to choose the regions.
I can explain why \|a + b\| is not always equal to \|a\| + \|b\|.

Answers and mark scheme

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