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4PM1/01R Jun 2025, Question 9: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 9 marks

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The question, in short

Differentiate f(x) = (1 + x²)⁴, then use this to find the gradient of the normal to y = sin 3x/(1 + x²)⁴ at the point where x = 2π/3.

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Part (a) · 2 marks

Show that differentiating (1 + x²)⁴ gives 8x(1 + x²)³.

  1. 1Chain rule: f′(x) = 4(1 + x²)³ × 2x
  2. 2= 8x(1 + x²)³

Answerf′(x) = 8x(1 + x²)³ shown

Part (b) · 7 marks

For y = sin 3x/(1 + x²)⁴ at x = 2π/3, show the normal's gradient equals −(1/3)(1 + (2π/3)²)⁴.

  1. 1Quotient rule: dy/dx = [3cos 3x(1 + x²)⁴ − 8x(1 + x²)³ sin 3x]/(1 + x²)⁸
  2. 2= [3(1 + x²)cos 3x − 8x sin 3x]/(1 + x²)⁵
  3. 3At x = 2π/3: cos 2π = 1, sin 2π = 0
  4. 4dy/dx = 3(1 + (2π/3)²)/(1 + (2π/3)²)⁵ = 3/(1 + (2π/3)²)⁴
  5. 5Normal gradient = −1/(dy/dx) = −(1/3)(1 + (2π/3)²)⁴

AnswerGradient of normal = −(1/3)(1 + (2π/3)²)⁴ shown

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