Edexcel Further Pure 2 Chapter 1 Inequalities Manipulate
- Slides: 14
Edexcel Further Pure 2 Chapter 1 – Inequalities: • Manipulate inequalities • Determine the critical values of an inequality • Find solutions of algebraic inequalities.
Chapter 1 – Inequalities: C 1 Recap Example 1: Solve x 2 – 5 x – 18 ≤ 6 Take 6 from both x 2 – 5 x – 24 ≤ 0 side ( x – 8 )( x + 3 ) ≤ 0 Factorise y Think graph – 3 ≤ x ≤ 8 -3 +8 x
Chapter 1 – Inequalities: C 1 Recap Example 2: Solve x 2 – x - 24 ≥ 6 Take 6 from both x 2 – x – 30 ≥ 0 side ( x – 6 )( x + 5 ) ≥ 0 Factorise y Think graph – 5≥ x ≥ 6 -5 +6 x
Chapter 1 – Inequalities: C 1 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 C 1. y Sketch the graph and find the values. -2 +2 x
The natural step Chapter 1 – Inequalities: would be to multiply both sides by (x+1)(x+3) but we Solve the inequality: 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 C 1. Sketch the graph and find the values. Algebraic Fractions
Exercise 1 A, 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 2 x – 4 ≤ x + 1 Solve y 5 2 x – 4 = x + 1 4 x 3 = 5 2 – (2 x – 4) = x + 1 1 -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 1 2 3 4 5 x – 2 x + 4 = x + 1 – 3 x = – 3 x = 1 1 ≤ x ≤ 5
Chapter 1 – Inequalities: Modulus on one side Solve: x 2 – 2 < 2 x + 1 y 8 7 x 2 – 2 = 2 x + 1 x 2 – 2 x – 3 = 0 ( x – 3 )( x + 1 ) = 0 6 5 4 x = 3 3 2 – ( x 2 1 -5 -4 -3 -2 -1 0 -1 -2 1 2 3 4 5 – 2 ) = 2 x + 1 x 2 + 2 x – 1 = 0 x = -1 + √ 2 x -1 + √ 2 < x < 3
Chapter 1 – Inequalities: Modulus on both sides Solve 5 x – 2 ≤ 3 x + 1 Square both sides (5 x – 2)2 ≤ (3 x + 1)2 Subtract (3 x + 1)2 Simplify Factorise (5 x – 2)2 – (3 x + 1)2 ≤ 0 [ (5 x – 2) – (3 x + 1) ][ (5 x – 2) + (3 x + 1) ] ≤ 0 y ( 2 x – 3 )( 8 x – 1 ) ≤ 0 1/ 1/ 8 ≤ x ≤ 11/2 8 11/2 x
Exercise 1 B, 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 = | 2 x – 3 |, and the line with equation y = 5 x – 1. (2) (b) Solve the inequality | 2 x – 3 | < 5 x – 1. (3) (Total 5 marks) 2. Find the set of values of x for which (Total 7 marks)
Exam Answers 1. (a) Shape B 1 Points on axis B 1 (2) (b) -2 x + 3 = 5 x – 1 x = 4/7 x > 4/7 M 1 A 1 (3) (Total 5 marks)
Exam Answers 5. 1. 5 and 3 are ‘critical values’, e. g. used in solution, or both seen as asymptotes. B 1 (x + 1)(x – 3) = 2 x – 3 → x(x – 4) = 0 x = 4, x = 0 M 1, A 1 M 1: Attempt to find at least one other critical value 0<x<1, 3<x<4 M 1, A 1 M 1: 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)