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0606/11 May/Jun 2026, Question 7: Tangents, Normals and Stationary Points (Further)

Cambridge IGCSE Additional Mathematics 0606 · Paper 11 (Non-calculator) · Non-calculator · 9 marks

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The question, in short

Differentiate the quotient y = (4 + e^(2x))/(2 + eˣ) and use the zero-gradient condition to derive a quadratic in eˣ, then solve for the stationary point's x-coordinate.

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Part (a) · 5 marks

Differentiate the given quotient and show that setting dy/dx to zero leads to e^(2x) + 4eˣ − 4 = 0.

  1. 1Quotient rule: dy/dx = ((2 + eˣ)(2e^(2x)) − (4 + e^(2x))eˣ)/(2 + eˣ)²
  2. 2dy/dx = 0 when the numerator is 0
  3. 34e^(2x) + 2e^(3x) − 4eˣ − e^(3x) = 0
  4. 4e^(3x) + 4e^(2x) − 4eˣ = 0
  5. 5Divide by eˣ (eˣ ≠ 0): e^(2x) + 4eˣ − 4 = 0

Answere^(2x) + 4eˣ − 4 = 0 shown

Part (b) · 4 marks

Solve the resulting equation for eˣ and express the stationary point's x-coordinate as ln(a√b + c) with integers a, b, c.

  1. 1Let u = eˣ: u² + 4u − 4 = 0
  2. 2(u + 2)² = 8, so u = −2 ± 2√2
  3. 3eˣ > 0, so eˣ = 2√2 − 2
  4. 4x = ln(2√2 − 2)

Answerx = ln(2√2 − 2)

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Also searched as: 0606/11/M/J/26 · 0606_s26_qp_11 · 0606 s26 11 · question 7.
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