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4PM1/01R Jun 2019, Question 4: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 10 marks

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The question, in short

Given f(x) = e^(3x) √(1 + 2x), differentiate to a stated form, then find the normal at x = 0 as ax + by + c = 0 with integer coefficients.

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Part (a) · 4 marks

Show that differentiating f(x) = e^(3x) √(1 + 2x) gives f′(x) = 2e^(3x)(2 + 3x)/√(1 + 2x).

  1. 1Product rule: f′(x) = 3e^(3x)(1 + 2x)^(1/2) + e^(3x) × ½ × 2(1 + 2x)^(−1/2)
  2. 2= [3e^(3x)(1 + 2x) + e^(3x)]/√(1 + 2x)
  3. 3= e^(3x)(4 + 6x)/√(1 + 2x) = 2e^(3x)(2 + 3x)/√(1 + 2x)

Answerf′(x) = 2e^(3x)(2 + 3x)/√(1 + 2x) shown

Part (b) · 6 marks

Find the equation of the normal to y = f(x) at x = 0, giving it as ax + by + c = 0 with integers a, b, c.

  1. 1f′(0) = 2(2)/1 = 4
  2. 2Gradient of normal = −1/4
  3. 3f(0) = e⁰√1 = 1
  4. 4y − 1 = −¼(x − 0)
  5. 5x + 4y − 4 = 0

Answerx + 4y − 4 = 0

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