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WMA11/01 Jan 2024, Question 10: The Derivative as a Limit

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

The question, in short

Curve C has y = f(x), x > 0, with P(2, 8√2) on C, f′(x) = 4√(x³) + k/x² and f″(x) = 0 at P. Find k, then find f(x).

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

Part (a) · 4 marks

Use f″(x) = 0 at P(2, 8√2) to find the exact value of the constant k.

  1. 1f′(x) = 4x^(3/2) + kx⁻²
  2. 2f″(x) = 6x^(1/2) − 2kx⁻³
  3. 3f″(2) = 6√2 − 2k/8 = 0
  4. 4k/4 = 6√2 ⇒ k = 24√2

Answerk = 24√2

Part (b) · 4 marks

Integrate f′(x) with your k, use P to find the constant, and give f(x) in simplest form.

  1. 1f(x) = ∫ (4x^(3/2) + 24√2 x⁻²) dx = (8/5)x^(5/2) − 24√2 x⁻¹ + c
  2. 2At P(2, 8√2): (8/5) × 2^(5/2) = (8/5) × 4√2 = (32/5)√2
  3. 38√2 = (32/5)√2 − 12√2 + c
  4. 4c = 8√2 + 12√2 − (32/5)√2 = (68/5)√2
  5. 5f(x) = (8/5)x^(5/2) − 24√2/x + (68/5)√2

Answerf(x) = (8/5)x^(5/2) − 24√2/x + (68/5)√2

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Also searched as: WMA11 January 2024 · WMA11/01 · wma11 jan24 · IAS Maths Pure Mathematics 1 January 2024 · question 10.
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