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0606/12 May/Jun 2025, Question 9: Functions: One-One, Inverses and Restricted Domains

Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 8 marks

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The question, in short

Given f(x) = 2ln(3x − 4) for x > a with an inverse, determine a, the range, the inverse function, and sketch both graphs showing intercepts and asymptotes.

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

Part (a) · 1 mark

State the smallest value of a that makes f one-to-one so its inverse exists.

  1. 1ln needs 3x − 4 > 0
  2. 2x > 4/3, so least a = 4/3

Answera = 4/3

Part (b) · 1 mark

Using your value of a, give the set of output values of f.

  1. 1As x → 4/3 from above, f(x) → −∞; as x → ∞, f(x) → ∞
  2. 2Range: all real numbers

Answerf(x) ∈ ℝ (all real numbers)

Part (c) · 2 marks

Using your value of a, derive a formula for f⁻¹(x).

  1. 1y = 2ln(3x − 4), so y/2 = ln(3x − 4)
  2. 23x − 4 = e^(y/2)
  3. 3x = (4 + e^(y/2))/3
  4. 4f⁻¹(x) = (4 + e^(x/2))/3

Answerf⁻¹(x) = (4 + e^(x/2))/3

Part (d) · 4 marks

Sketch y = f(x) and y = f⁻¹(x) for your a, labelling each, giving axis crossings and asymptote equations.

  1. 1f(x) = 0 when 3x − 4 = 1, so x-intercept 5/3; vertical asymptote x = 4/3
  2. 2f⁻¹(0) = (4 + 1)/3 = 5/3, so y-intercept 5/3; horizontal asymptote y = 4/3
  3. 3f⁻¹ is the reflection of f in y = x
  4. 4The curves cross each other twice

Answery = f(x): increasing log curve in the 1st and 4th quadrants, vertical asymptote x = 4/3, crossing the x-axis at 5/3. y = f⁻¹(x): increasing exponential curve in the 1st and 2nd quadrants, horizontal asymptote y = 4/3, crossing the y-axis at 5/3. The two curves are reflections in y = x and intersect twice in the 1st quadrant

Stuck on a step? Read the free lesson: Functions: One-One, Inverses and Restricted Domains

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