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WMA11/01 Oct 2024, Question 8: Integrating with Two Conditions

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 7 marks

The question, in short

Curve C has y = f(x) with point P at x = 3. Given f′(x) = 4x² + kx + 3 and the normal at P is y = −(1/24)x + 5, show k = −5 then find f(x).

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Part (a) · 3 marks

Use the normal's gradient at x = 3 to find the tangent gradient, equate to f′(3), and show that k = −5.

  1. 1Normal gradient is −1/24, so the tangent gradient at P is 24
  2. 2f′(3) = 4(9) + 3k + 3 = 24
  3. 33k + 39 = 24, so 3k = −15
  4. 4k = −5

AnswerShown: tangent gradient = 24, so f′(3) = 36 + 3k + 3 = 24 ⇒ 3k = −15 ⇒ k = −5

Part (b) · 4 marks

Integrate f′(x) with k = −5 and use the coordinates of P to determine the constant, giving f(x).

  1. 1f(x) = ∫ (4x² − 5x + 3) dx = (4/3)x³ − (5/2)x² + 3x + c
  2. 2At P, x = 3 and y = −3/24 + 5 = 39/8
  3. 3(4/3)(27) − (5/2)(9) + 9 + c = 39/8, so 36 − 22.5 + 9 + c = 39/8
  4. 422.5 + c = 4.875, so c = −141/8
  5. 5f(x) = (4/3)x³ − (5/2)x² + 3x − 141/8

Answerf(x) = (4/3)x³ − (5/2)x² + 3x − 141/8

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Also searched as: WMA11 October 2024 · WMA11/01 · wma11 oct24 · IAS Maths Pure Mathematics 1 October 2024 · question 8.
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