A-Level Maths (Edexcel 9MA0) · Pure Year 2 · Chapter 2: Functions and graphs · Lesson 5

Solving modulus problems

113 marks

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Quick questions to start.

A1
Lesson 1
Solve $∣3 x + 2∣ = 7$ .
[2 marks]
Answer
A2
Lesson 4
Write down the coordinates of the vertex of $y = 4 - 3∣ x - 2∣$ , and the point where it meets the $y$ -axis.
[2 marks]
Answer
A3
From Year 1
Solve $x^{2} - x - 12 \geq 0$ .
[2 marks]
Answer
A4
Needed today
Solve $∣ x - 1∣ < 4$ .
[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

Solve $3∣ x - 2∣ - 4 = x$ .

Hint Split at

x = 2

, where the inside of the modulus is zero. Solve each case, then check each answer lies in its own case.

B1
Solve $2∣ x + 3∣ - 5 = 3 - x$ .
[4 marks]
Answer

Example 2

Solve $∣2 x - 3∣ + 1 = 3 x - 4$ .

Hint Write it as

∣2 x - 3∣ = 3 x - 5

. The right-hand side must be

\geq 0

, so any answer with

3 x - 5 < 0

is rejected.

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

Example 3

Solve $2∣ x - 3∣ > x$ .

Hint Find where

2∣ x - 3∣ = x

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

B3
Solve $3∣ x + 1∣ - 2 \leq x + 4$ .
[4 marks]
Answer

Example 4

Solve $∣ x^{2} - 2 x - 3∣ = 5$ .

Hint Solve

x^{2} - 2 x - 3 = 5

and

x^{2} - 2 x - 3 = - 5

. A case with no real roots gives no solutions.

B4
Solve $∣ x^{2} - 4 x ∣ = 4$ . Give your answers in exact form.
[4 marks]
Answer

Example 5

Solve $∣ x^{2} - 4∣ < 3 x$ .

Hint

∣ A ∣ < B

means

- B < A < B

. Solve each half, then take the values that satisfy both. Since

∣ x^{2} - 4∣ \geq 0

, any solution must have

3 x > 0

B5
Solve $∣ x^{2} - 9∣ < 4 x$ . Give your answer in exact form.
[5 marks]
Answer

Practice

A-Level

C1
Mia solves $4∣ x - 1∣ - 3 = 2 x - 7$ . She writes: " $4 (x - 1) = 2 x - 4$ gives $x = 0$ , and $- 4 (x - 1) = 2 x - 4$ gives $x = \frac{4}{3}$ . So $x = 0$ or $x = \frac{4}{3}$ ."
(a) Show that neither of Mia's values satisfies the equation.
(b) Explain, by considering the sign of $2 x - 4$ , why the equation has no solutions.
[5 marks]

C2
Ravi solves $∣2 x - 1∣ \geq x + 2$ . He finds the critical values $- \frac{1}{3}$ and $3$ and writes $- \frac{1}{3} \leq x \leq 3$ .
(a) Test $x = 0$ in the inequality to show Ravi is wrong.
(b) Write down the correct solution.
[4 marks]
Answer
C3
Solve $∣ x^{2} - 7 x + 6∣ = 6$ .
[4 marks]
Answer
C4
Solve $∣ x^{2} - 2 x - 4∣ < 4$ .
[5 marks]
Answer
C5
Solve $∣3 x - 2∣ = x^{2} - 2$ .
[5 marks]
Answer

Exam-style questions

Edexcel exam style

D1
The function $f$ is defined by $f (x) = 3∣ x - 2∣ - 6$ , $x \in R$ .
(a)
Sketch the graph of $y = f (x)$ . Give the coordinates of its vertex and of the points where it meets the axes.
[3 marks]
(b)
Solve $f (x) = x - 2$ .
[4 marks]
Answer
(c)
Hence solve $f (x) < x - 2$ .
[2 marks]
Answer
(d)
Find the value of $c$ for which $f (x) = x + c$ has exactly one solution.
[2 marks]
Answer
Total for D1: 11 marks

D2
Zara solves $∣ x^{2} - 3 x - 10∣ = 2 x$ . She solves $x^{2} - 3 x - 10 = 2 x$ and $x^{2} - 3 x - 10 = - 2 x$ , and gives all four roots as solutions.
Zara is wrong.
(a)
Explain why every solution must satisfy $x \geq 0$ .
[2 marks]
(b)
Find the exact solutions of the equation.
[5 marks]
Answer
Total for D2: 7 marks
D3
Find the set of values of $x$ for which $∣ x^{2} - 5 x ∣ > 2 x - 6$ . Give your answer in exact form.
[9 marks]
D4
The functions $f$ and $g$ are defined by $f (x) = 2∣ x + 1∣$ and $g (x) = ∣ x - 3∣ + 2$ , $x \in R$ .
(a)
On the same axes, sketch $y = f (x)$ and $y = g (x)$ . Give the coordinates of each vertex and of the points where each graph meets the $y$ -axis.
[3 marks]
(b)
Solve $f (x) = g (x)$ .
[4 marks]
Answer
(c)
Hence solve $f (x) \leq g (x)$ .
[2 marks]
Answer
Total for D4: 9 marks
D5
The function $h$ is defined by $h (x) = ∣ x^{2} + 2 x - 8∣$ , $x \in R$ .
(a)
Sketch the graph of $y = h (x)$ . Give the coordinates of the points where it meets the axes, and of its maximum point.
[3 marks]
(b)
Solve $h (x) = 5$ . Give your answers in exact form.
[4 marks]
Answer
(c)
Find the values of $k$ for which $h (x) = k$ has exactly two solutions.
[2 marks]
Answer
Total for D5: 9 marks
D6
The curve $y = ∣ x^{2} - 1∣$ and the line $y = x + 1$ are drawn on the same axes.
(a)
Find the $x$ -coordinates of the points where they meet.
[4 marks]
Answer
(b)
Hence solve $∣ x^{2} - 1∣ \leq x + 1$ .
[3 marks]
Answer
Total for D6: 7 marks

Extension

Stretch

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

E1
Find the set of values of $k$ for which the equation $∣ x^{2} - 4 x + 3∣ = k x$ has exactly four real solutions.
[9 marks]
Hint 1 · What to try
Sketch $y = ∣ x^{2} - 4 x + 3∣$ . It meets the $x$ -axis at $1$ and $3$ , and the part between them is reflected so it has a maximum at $(2, 1)$ . The line $y = k x$ passes through the origin.
Hint 2 · The first line
For $k > 0$ the line meets the curve once to the left of $x = 1$ and once to the right of $x = 3$ . The other two solutions must come from the reflected part.
Hint 3 · The full method
The largest $k$ is when the line touches the reflected part: $- (x^{2} - 4 x + 3) = k x$ has a repeated root, so its discriminant is zero.
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can solve an equation such as a\|x − b\| + c = mx + d by splitting it into two cases and rejecting answers that do not fit.
I can solve an inequality with a modulus by finding the crossing points and reading the solution from a sketch.
I can solve \|f(x)\| = k and \|f(x)\| < g(x) when f is a quadratic.
I can use a sketch to find the values of k that give a set number of solutions.

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