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0606/22 Feb/Mar 2025, Question 7: 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) = 2e^x + a for x ≥ 0 and g(x) = √(x − 1) for x ≥ 1, find a so gf exists, then solve gf(x) = 3 with a = 5.

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Part (a) · 2 marks

Find the smallest integer a such that the composite function gf is defined for every x ≥ 0.

  1. 1Range of f: f(x) ≥ 2 + a
  2. 2gf exists when the range of f lies in the domain of g: 2 + a ≥ 1
  3. 3Least a = −1

Answera = −1

Part (b) · 3 marks

With a = 5, solve gf(x) = 3, giving the answer to 3 decimal places.

  1. 1gf(x) = √((2e^x + 5) − 1) = 3
  2. 22e^x + 4 = 9
  3. 3e^x = 2.5
  4. 4x = ln 2.5 = 0.916

Answerx = 0.916

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Also searched as: 0606/22/F/M/25 · 0606_m25_qp_22 · 0606 m25 22 · question 7.
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