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0606/13 May/Jun 2026, Question 8: Functions: One-One, Inverses and Restricted Domains

Cambridge IGCSE Additional Mathematics 0606 · Paper 13 (Non-calculator) · Non-calculator · 8 marks

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The question, in short

For f(x) = e^(2x+1) on ℝ, find the inverse function's range, domain and expression, then sketch f and f⁻¹ with exact axis intercepts.

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

Part (a) · 1 mark

State the range of f⁻¹.

  1. 1The range of f⁻¹ is the domain of f
  2. 2Domain of f is x ∈ ℝ, so range of f⁻¹ is ℝ

Answerf⁻¹(x) ∈ ℝ (all real numbers)

Part (b) · 1 mark

Find the domain of f⁻¹.

  1. 1The domain of f⁻¹ is the range of f
  2. 2e^(2x+1) > 0 for all x, so the domain of f⁻¹ is x > 0

Answerx > 0

Part (c) · 2 marks

Find an expression for f⁻¹(x).

  1. 1y = e^(2x+1)
  2. 2ln y = 2x + 1
  3. 3x = (1/2)(ln y − 1)
  4. 4f⁻¹(x) = (1/2)(ln x − 1)

Answerf⁻¹(x) = (1/2)(ln x − 1)

Part (d) · 4 marks

Sketch y = f(x) and y = f⁻¹(x) on the axes, giving exact coordinates of axis intercepts.

  1. 1f(0) = e¹ = e, so y = f(x) crosses the y-axis at (0, e)
  2. 2f(x) > 0 and f(x) → 0 as x → −∞: asymptote y = 0
  3. 3Reflect in y = x to get y = f⁻¹(x)
  4. 4f⁻¹ crosses the x-axis at (e, 0), with asymptote x = 0

Answery = f(x): exponential curve above the x-axis, asymptotic to the negative x-axis, through (0, e); y = f⁻¹(x): its reflection in y = x, a log curve to the right of the y-axis, asymptotic to the negative y-axis, through (e, 0); no other intercepts

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Also searched as: 0606/13/M/J/26 · 0606_s26_qp_13 · 0606 s26 13 · question 8.
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