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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)