Edexcel Further Pure 2
Chapter 1 – Inequalities:
• Manipulate inequalities
• Determine the critical values of an inequality
• Find solutions of algebraic inequalities.
Chapter 1 – Inequalities: C1 Recap
Example 1: Solve
x2 – 5x – 18 ≤ 6
Take 6 from both side
x2 – 5x – 24 ≤ 0
( x – 8 )( x + 3 ) ≤ 0
Factorise
y
Think graph
–3 ≤ x ≤ 8
-3
+8
x
Chapter 1 – Inequalities: C1 Recap
Example 2: Solve
x2 – x - 24 ≥ 6
Take 6 from both side
x2 – x – 30 ≥ 0
( x – 6 )( x + 5 ) ≥ 0
Factorise
y
Think graph
–5≥ x ≥ 6
-5
+6
x
Chapter 1 – Inequalities: C1 Recap
Example 3:
Solve (x + 5)(x – 4)(x + 1) > 0
y
-5
-5 < x < -1
Try this:
or
-1
+4
x > 4
Solve (x – 1)2(x + 4)(x – 3) < 0
x
The natural step would
be to multiply both
sides by (x-2) but we
cannot be sure that
this is positive. So we
multiply both sides by
(x-2)2
Chapter 1 – Inequalities: Algebraic Fractions
Solve the inequality:
Do NOT multiply out
but cancel out like
terms.
Now we have a similar
question seen in C1.
y
Sketch the graph and
find the values.
-2
+2
x
The natural step would
be to multiply both
sides by (x+1)(x+3) but
we cannot be sure that
this is positive. So we
multiply both sides by
(x+1)2 (x+3)2
Do NOT multiply out
but cancel out like
terms.
Now we have a similar
question seen in C1.
Sketch the graph and
find the values.
Chapter 1 – Inequalities: Algebraic Fractions
Solve the inequality:
Exercise 1A, Page 4

Solve the following inequalities:
Questions 5, 6, 7 and 8.

Extension Task:
Questions 13, 14 and 15.
Chapter 1 – Inequalities: Modulus on one side
 2x – 4  ≤ x + 1
Solve
y
5
2x – 4
= x+1
4
x
3
=
5
2
– (2x – 4) = x + 1
1
-5 -4
-3
-2
-1
0
-1
-2
-3
-4
-5
1
2
3
4
5
x
– 2x + 4 = x + 1
– 3x
=
–3
x
=
1
1 ≤ x ≤ 5
Chapter 1 – Inequalities: Modulus on one side
Solve:  x2 – 2  < 2x + 1
y
8
7
x2 – 2 = 2x + 1
x2 – 2x – 3 = 0
( x – 3 )( x + 1 ) = 0
6
5
4
x=3
3
2
1
-5 -4
-3
-2
-1
0
-1
-2
( x2 – 2 ) = 2x + 1
x2 + 2x – 1 = 0
x = -1 + √2
x
–
1
2
3
4
5
-1 + √2 < x < 3
Chapter 1 – Inequalities: Modulus on both sides
Solve  5x – 2  ≤  3x + 1 
Square both
sides
(5x – 2)2 ≤ (3x + 1)2
Subtract
(3x + 1)2
Simplify
Factorise
(5x – 2)2 – (3x + 1)2 ≤ 0
[ (5x
– 2) – (3x + 1) ][ (5x – 2) + (3x + 1) ] ≤ 0
y
( 2x – 3 )( 8x – 1 ) ≤ 0
1/
1/
8
≤ x ≤ 11/2
8
11/2
x
Exercise 1B, Page 8

Solve the following inequalities:
Questions 1, 3 and 4.

Extension Task:
Question 9.
Exam Questions
1.
(a) Sketch, on the same axes, the graph with equation y = | 2x – 3 |, and the line
with equation y = 5x – 1.
(2)
(b) Solve the inequality | 2x – 3 | < 5x – 1.
(3)
(Total 5 marks)
2. Find the set of values of x for which
x 1
1

2x  3 x  3
(Total 7 marks)
Exam Answers
1.
(a)
y
y = 5x – 1
3
Shape B1
Points on axis B1
1
–1
y = | 2x – 3 |
12
(b) -2x + 3 = 5x – 1
x = 4/7
x > 4/7
(2)
M1
A1
A1
(3)
(Total 5 marks)
Exam Answers
5.
1.5 and 3 are ‘critical values’, e.g. used in solution, or both seen as asymptotes.
B1
(x + 1)(x – 3) = 2x – 3 → x(x – 4) = 0
x = 4, x = 0
M1,A1, A1
M1: Attempt to find at least one other critical value
0 < x < 1 ,3 < x < 4
M1,A1, A1
M1: An inequality using 1.5 or 3
First M mark can be implied by the two correct values, but otherwise
a method must be seen. (The method may be graphical, but either
(x =) 4 or (x =) 0 needs to be clearly written or used in this case).
Ignore ‘extra values’ which might arise through ‘squaring both sides’
methods.
≤ appearing: maximum one A mark penalty (final mark).
(Total 7 marks)