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0606/12 Oct/Nov 2025, Question 9: Integrating eˣ, Trig and (ax+b)ⁿ

Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 12 marks

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The question, in short

Given f″(x) = (2x + 5)^(−3/2) and gradient 2/3 at (2, 2), locate the stationary point and classify it.

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

Part (a) · 11 marks

Integrate twice using the gradient condition at (2, 2), then solve f′(x) = 0 to get the stationary point's coordinates.

  1. 1f′(x) = ∫ (2x + 5)^(−3/2) dx = −(2x + 5)^(−1/2) + c
  2. 2f′(2) = 2/3: −1/3 + c = 2/3 ⇒ c = 1, so f′(x) = 1 − (2x + 5)^(−1/2)
  3. 3f(x) = ∫ f′(x) dx = x − (2x + 5)^(1/2) + d
  4. 4f(2) = 2: 2 − 3 + d = 2 ⇒ d = 3, so f(x) = x − √(2x + 5) + 3
  5. 5f′(x) = 0: (2x + 5)^(−1/2) = 1 ⇒ 2x + 5 = 1 ⇒ x = −2
  6. 6y = −2 − 1 + 3 = 0

Answer(−2, 0)

Part (b) · 1 mark

Use the second derivative or a sign test to state whether the stationary point is a maximum or minimum.

  1. 1f″(−2) = (2(−2) + 5)^(−3/2) = 1^(−3/2) = 1
  2. 2f″(−2) > 0, so the stationary point is a minimum

AnswerMinimum point, since f″(−2) = 1 > 0

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