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0606/12 Feb/Mar 2026, Question 8: Tangents, Normals and Stationary Points (Further)

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

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The question, in short

Curve y = xe^(2x+1): differentiate it, find where the tangent at x = 1/2 crosses both axes, and integrate the curve expression.

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

Part (a) · 3 marks

Differentiate y = xe^(2x+1) with respect to x.

  1. 1d/dx(e^(2x+1)) = 2e^(2x+1)
  2. 2Product rule: dy/dx = x × 2e^(2x+1) + 1 × e^(2x+1)
  3. 3dy/dx = 2xe^(2x+1) + e^(2x+1)

Answerdy/dx = 2xe^(2x+1) + e^(2x+1)

Part (b) · 5 marks

Find the tangent line at x = 1/2 and give exact coordinates where it meets the x-axis and y-axis.

  1. 1At x = 1/2: dy/dx = 2(1/2)e² + e² = 2e²
  2. 2At x = 1/2: y = (1/2)e²
  3. 3L: y − (1/2)e² = 2e²(x − 1/2)
  4. 4x = 0: y = (1/2)e² − e² = −(1/2)e²
  5. 5y = 0: x − 1/2 = −1/4, so x = 1/4

Answer(0, −(1/2)e²) and (1/4, 0)

Part (c) · 4 marks

Using the derivative from part a, find the integral of xe^(2x+1) with respect to x.

  1. 1From (a): ∫ (2xe^(2x+1) + e^(2x+1)) dx = xe^(2x+1)
  2. 2∫ 2xe^(2x+1) dx = xe^(2x+1) − ∫ e^(2x+1) dx
  3. 3∫ 2xe^(2x+1) dx = xe^(2x+1) − (1/2)e^(2x+1)
  4. 4∫ xe^(2x+1) dx = (1/2)(xe^(2x+1) − (1/2)e^(2x+1)) + c

Answer(1/2)(xe^(2x+1) − (1/2)e^(2x+1)) + c

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Also searched as: 0606/12/F/M/26 · 0606_m26_qp_12 · 0606 m26 12 · question 8.
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