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4MA1/1HR Jun 2023, Question 21: Gradient of a Curve at a Point

Pearson Edexcel International GCSE Mathematics A 4MA1 · Paper 1HR (Higher) · Calculator · 6 marks

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The question, in short

For the curve y = x³ − 2x² − 9x + 15, find the derivative and then the x-values where the gradient is positive, to 3 significant figures.

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

Part (a) · 2 marks

Differentiate y = x³ − 2x² − 9x + 15 to obtain dy/dx.

  1. 1Differentiate x³: 3x²
  2. 2Differentiate −2x²: −4x
  3. 3Differentiate −9x: −9
  4. 4Derivative of constant 15 is 0
  5. 5So dy/dx = 3x² − 4x − 9

Answer3x² − 4x − 9

Part (b) · 4 marks

Solve the inequality for positive gradient, giving x values to 3 significant figures with clear working.

  1. 1Set derivative greater than 0: 3x² − 4x − 9 > 0
  2. 2Solve 3x² − 4x − 9 = 0 using quadratic formula
  3. 3x = [4 ± √((−4)² − 4×3×(−9))] / (2×3)
  4. 4x = [4 ± √(16 + 108)] / 6
  5. 5x = [4 ± √124] / 6
  6. 6√124 = √(4×31) = 2√31 ≈ 11.1355
  7. 7x = (4 + 11.1355)/6 ≈ 15.1355/6 ≈ 2.5226
  8. 8x = (4 − 11.1355)/6 ≈ −7.1355/6 ≈ −1.1892
  9. 9Critical values: x ≈ −1.189 and x ≈ 2.523
  10. 10The parabola 3x² − 4x − 9 opens upwards (coefficient of x² positive)
  11. 11So gradient is positive when x < −1.189 or x > 2.523
  12. 12Rounded to 3 significant figures: x < −1.19 or x > 2.52

Answerx < −1.19 or x > 2.52

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