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4MA1/2H Jun 2018, Question 17: Gradient of a Curve at a Point

Pearson Edexcel International GCSE Mathematics A 4MA1 · Paper 2H (Higher) · Calculator · 6 marks

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The question, in short

Given y = x³ − 2x² − 15x + 5, differentiate it and find the x-values where the curve C has negative gradient.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 2 marks

Find dy/dx of y = x³ − 2x² − 15x + 5.

  1. 1Multiply by the power, then lower the power by 1
  2. 2x³ → 3x², −2x² → −4x, −15x → −15, 5 → 0
  3. 3dy/dx = 3x² − 4x − 15

Answer3x² − 4x − 15

Part (b) · 4 marks

Solve for the range of x where the gradient of C is negative.

  1. 1Negative gradient means dy/dx < 0: 3x² − 4x − 15 < 0
  2. 2Factorise: (3x + 5)(x − 3) < 0, so the ends are x = −5/3 and x = 3
  3. 3The quadratic is negative between its roots: −5/3 < x < 3

Answer−5/3 < x < 3

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