1. Chapter 1 – Inequalities: Manipulate inequalities Determine the critical values of an inequality Find solutions of algebraic inequalities. Edexcel Further Pure 2

2. Example 1: Solve x 2 – 5x – 18 ≤ 6 Take 6 from both side x 2 – 5x – 24 ≤ 0 ( x – 8 )( x + 3 ) ≤ 0 Factorise -3 +8 y x – 3 ≤ x ≤ 8 Think graph Chapter 1 – Inequalities: C1 Recap

3. Example 2: Solve x 2 – x - 24 ≥ 6 Take 6 from both side x 2 – x – 30 ≥ 0 ( x – 6 )( x + 5 ) ≥ 0 Factorise -5 +6 y x – 5 ≥ x ≥ 6 Think graph Chapter 1 – Inequalities: C1 Recap

4. Example 3:Solve(x + 5)(x – 4)(x + 1) > 0 -5 -1 +4 y x -5 4 Try this:Solve(x – 1) 2 (x + 4)(x – 3) < 0 Chapter 1 – Inequalities: C1 Recap

5. Chapter 1 – Inequalities: Algebraic Fractions Solve the inequality: 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 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. -2 +2 y x

6. Chapter 1 – Inequalities: Algebraic Fractions Solve the inequality: 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.

7. Exercise 1A, Page 4 Solve the following inequalities: Questions 5, 6, 7 and 8. Extension Task: Questions 13, 14 and 15.

8. -5 -4 -3 -2 -1 0 1 2 3 4 5 x y 5 4 3 2 1 -2 -3 -4 -5 2x – 4 = x + 1 x = 5 – ( 2x – 4) = x + 1 – 2x + 4 = x + 1 – 3x = – 3 x = 1 1 ≤ x ≤ 5 Solve  2x – 4  ≤ x + 1 Chapter 1 – Inequalities: Modulus on one side

9. -5 -4 -3 -2 -1 0 1 2 3 4 5 x y 8 7 6 5 4 3 2 1 -2 Solve:  x 2 – 2  < 2x + 1 x 2 – 2 = 2x + 1 x 2 – 2x – 3 = 0 ( x – 3 )( x + 1 ) = 0 x = 3 – ( x 2 – 2 ) = 2x + 1 x 2 + 2x – 1 = 0 x = -1 + √2 -1 + √2 < x < 3 Chapter 1 – Inequalities: Modulus on one side

10. Solve  5x – 2  ≤  3x + 1  Square both sides (5x – 2) 2 ≤ (3x + 1) 2 ( 2x – 3 )( 8x – 1 ) ≤ 0Factorise 1 / 8 1 1 / 2 y x 1 / 8 ≤ x ≤ 1 1 / 2 Subtract (3x + 1) 2 (5x – 2) 2 – (3x + 1) 2 ≤ 0 Simplify [ (5x – 2) – (3x + 1) ][ (5x – 2) + (3x + 1) ] ≤ 0 Chapter 1 – Inequalities: Modulus on both sides

11. Exercise 1B, Page 8 Solve the following inequalities: Questions 1, 3 and 4. Extension Task: Question 9.

12. 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 (Total 7 marks)

13. Exam Answers 1. (a) Shape B1 Points on axis B1 (2) (b) -2x + 3 = 5x – 1 M1 x = 4/7 A1 x > 4/7 A1 (3) (Total 5 marks)

14. 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)