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4PM1/01 Jun 2021, Question 7: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 11 marks

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The question, in short

For the curve y = x/(x² + 4), use differentiation to locate stationary points, verify a given second derivative expression, and classify each stationary point.

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

Part (a) · 5 marks

Differentiate y = x/(x² + 4), set the derivative to zero, and state the coordinates of the stationary points.

  1. 1dy/dx = ((x² + 4) − x(2x))/(x² + 4)²
  2. 2= (4 − x²)/(x² + 4)²
  3. 3dy/dx = 0 ⇒ 4 − x² = 0 ⇒ x = ±2
  4. 4x = 2: y = 2/8 = 1/4; x = −2: y = −1/4

Answer(2, 1/4) and (−2, −1/4)

Part (b) · 4 marks

Differentiate again to obtain the stated form 2x(x² − 12)/(x² + 4)³, showing the required working.

  1. 1d²y/dx² = (−2x(x² + 4)² − (4 − x²) × 2(x² + 4)(2x))/(x² + 4)⁴
  2. 2Cancel (x² + 4): = (−2x(x² + 4) − 4x(4 − x²))/(x² + 4)³
  3. 3= (−2x³ − 8x − 16x + 4x³)/(x² + 4)³ = (2x³ − 24x)/(x² + 4)³
  4. 4= 2x(x² − 12)/(x² + 4)³

Answerd²y/dx² = 2x(x² − 12)/(x² + 4)³ shown

Part (c) · 2 marks

Use the second derivative to decide whether each stationary point is a maximum or minimum.

  1. 1x = 2: d²y/dx² = 4(−8)/512 = −1/16 < 0, so maximum
  2. 2x = −2: d²y/dx² = 1/16 > 0, so minimum

Answer(2, 1/4) is a maximum; (−2, −1/4) is a minimum

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