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WMA11/01 Jan 2023, Question 11: Integrating with Two Conditions

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 8 marks

The question, in short

Use a curve's second derivative 4x + 1/√x, a point P at x = 4 and its tangent y = 3x + 4 to find a normal and the curve.

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

Part (a) · 2 marks

Find the normal line's equation at P, using the given tangent gradient at x = 4.

  1. 1At x = 4 the tangent gives y = 3 × 4 + 4 = 16, so P is (4, 16)
  2. 2Tangent gradient 3, so normal gradient = −1/3
  3. 3y − 16 = −⅓(x − 4), i.e. x + 3y − 52 = 0

Answery − 16 = −⅓(x − 4), i.e. x + 3y − 52 = 0

Part (b) · 6 marks

Integrate twice from f″(x) = 4x + 1/√x, applying the point and tangent conditions at x = 4 to determine f(x).

  1. 1f′(x) = ∫ (4x + x^(−1/2)) dx = 2x² + 2x^(1/2) + c
  2. 2f′(4) = 3: 32 + 4 + c = 3, so c = −33
  3. 3f′(x) = 2x² + 2x^(1/2) − 33
  4. 4f(x) = ⅔x³ + (4/3)x^(3/2) − 33x + d
  5. 5f(4) = 16: 128/3 + 32/3 − 132 + d = 16, so d = 284/3
  6. 6f(x) = ⅔x³ + (4/3)x^(3/2) − 33x + 284/3

Answerf(x) = ⅔x³ + (4/3)x^(3/2) − 33x + 284/3

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