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Edexcel Further Pure 2

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 x 2 – 5 x – 18 ≤ 6. Take 6 from both side. x 2 – 5 x – 24 ≤ 0.

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Edexcel Further Pure 2

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

  2. Chapter 1 – Inequalities: C1 Recap Example 1: Solvex2 – 5x – 18 ≤ 6 Take 6 from both side x2 – 5x – 24 ≤ 0 ( x– 8 )( x + 3 ) ≤ 0 Factorise y Think graph x – 3 ≤ x ≤ 8 -3 +8

  3. Chapter 1 – Inequalities: C1 Recap Example 2: Solvex2 – x - 24 ≥ 6 Take 6 from both side x2– x – 30 ≥ 0 ( x– 6 )( x + 5 ) ≥ 0 Factorise y Think graph x – 5 ≥ x ≥ 6 -5 +6

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

  5. Chapter 1 – Inequalities: Algebraic Fractions 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 Solve the inequality: Do NOT multiply out but cancel out like terms. y Now we have a similar question seen in C1. Sketch the graph and find the values. x -2 +2

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

  9. Chapter 1 – Inequalities: Modulus on one side y Solve: x2 – 2  < 2x + 1 8 7 6 5 4 3 2 1 x2 – 2 = 2x + 1 x2 – 2x – 3 = 0 ( x – 3 )( x + 1 ) = 0 x = 3 – ( x2 – 2 ) = 2x + 1 x2 + 2x – 1 = 0 x x = -1 + √2 -5 -4 -3 -2 -1 0 1 2 3 4 5 -1 -2 -1 + √2 < x < 3

  10. 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 (5x– 2)2 –(3x+ 1)2 ≤ 0 [ (5x– 2)– (3x+ 1) ][(5x– 2) + (3x+ 1) ]≤ 0 Simplify y Factorise ( 2x– 3 )( 8x – 1 ) ≤ 0 x 1/8 11/2 1/8 ≤ x ≤ 11/2

  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) = 0x = 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 otherwisea 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)

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