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0606/22 Oct/Nov 2025, Question 9: Functions: One-One, Inverses and Restricted Domains

Cambridge IGCSE Additional Mathematics 0606 · Paper 22 (Calculator) · Calculator · 5 marks

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The question, in short

Given f(x) = ln(2x + 5) for x > a and g(x) = x² + 1, find the least a, the range of f, and solve fg(x) = 4 exactly.

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Part (a) · 1 mark

State the smallest possible value of the constant a for which f(x) = ln(2x + 5) is defined.

  1. 1ln needs 2x + 5 > 0
  2. 2x > −5/2, so the least value of a is −5/2

Answera = −5/2

Part (b) · 1 mark

Using your value of a, state the range of the function f.

  1. 1As x → −5/2, ln(2x + 5) → −∞; as x → ∞, ln(2x + 5) → ∞
  2. 2Range: f ∈ ℝ

Answerf ∈ ℝ (all real numbers)

Part (c) · 3 marks

Using your value of a, solve fg(x) = 4, giving answers in exact form.

  1. 1fg(x) = ln(2(x² + 1) + 5) = ln(2x² + 7)
  2. 2ln(2x² + 7) = 4, so 2x² + 7 = e⁴
  3. 3x² = (e⁴ − 7)/2
  4. 4x = ±√((e⁴ − 7)/2)

Answerx = ±√((e⁴ − 7)/2)

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