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WMA11/01 Jun 2025, Question 8: Integration: Undoing Differentiation

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 11 marks

The question, in short

Curve y = f(x), x > 0, passes through P(4, 12), with f′(x) = 3√x + kx² and tangent at P given as y = 10x + c. Show k = 1/4, find c, then f″(4) and f(x).

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

Part (ai) · 2 marks

Use the tangent gradient at P together with f′(x) to show that the constant k equals 1/4.

  1. 1The tangent at P has gradient 10, so f′(4) = 10
  2. 23√4 + k(4²) = 10
  3. 36 + 16k = 10, so k = 4/16 = 1/4

AnswerShown: f′(4) = 10 gives 3(2) + 16k = 10, so k = 1/4

Part (aii) · 2 marks

Use the point P(4, 12) on the tangent line to find the value of c.

  1. 1P(4, 12) lies on y = 10x + c
  2. 212 = 10(4) + c, so c = −28

Answerc = −28

Part (b) · 3 marks

Differentiate f′(x) and evaluate f″(x) at P using k = 1/4.

  1. 1f′(x) = 3x^(1/2) + (1/4)x²
  2. 2f″(x) = (3/2)x^(−1/2) + (1/2)x
  3. 3f″(4) = (3/2)(1/2) + 2 = 3/4 + 2 = 11/4

Answerf″(4) = 11/4

Part (c) · 4 marks

Integrate f′(x) and use the point P to determine the constant, giving f(x).

  1. 1f(x) = ∫(3x^(1/2) + (1/4)x²) dx = 2x^(3/2) + (1/12)x³ + d
  2. 2At P(4, 12): 12 = 2(8) + (1/12)(64) + d
  3. 312 = 16 + 16/3 + d, so d = −28/3
  4. 4f(x) = 2x^(3/2) + (1/12)x³ − 28/3

Answerf(x) = 2x^(3/2) + (1/12)x³ − 28/3

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Also searched as: WMA11 June 2025 · WMA11/01 · wma11 jun25 · IAS Maths Pure Mathematics 1 June 2025 · question 8.
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