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WMA11/01 Jun 2019, Question 8: Integration: Undoing Differentiation

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 9 marks

The question, in short

A curve y = f(x), x > 0, goes through P(4, 1) and has gradient 4√x − 2 − 8/(3x²). Find the normal's equation at P as ax + by + c = 0 with integers, then find f(x).

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

Part (a) · 4 marks

Work out the gradient at P, use it to get the perpendicular slope, then form the line through P as ax + by + c = 0 with integer a, b, c.

  1. 1f′(4) = 4 × 2 − 2 − 8/(3 × 16) = 6 − 1/6 = 35/6
  2. 2Gradient of normal = −6/35
  3. 3y − 1 = −(6/35)(x − 4)
  4. 435y − 35 = −6x + 24, so 6x + 35y − 59 = 0

Answer6x + 35y − 59 = 0

Part (b) · 5 marks

Integrate the given gradient expression and use the point P(4, 1) to determine the constant, giving f(x).

  1. 1f′(x) = 4x^(1/2) − 2 − (8/3)x^(−2)
  2. 2f(x) = (8/3)x^(3/2) − 2x + (8/3)x^(−1) + c
  3. 3At (4, 1): 1 = (8/3)(8) − 8 + 8/12 + c = 14 + c
  4. 4c = −13, so f(x) = (8/3)x^(3/2) − 2x + 8/(3x) − 13

Answerf(x) = (8/3)x^(3/2) − 2x + 8/(3x) − 13

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