0606/23 Oct/Nov 2025, Question 2: Functions: One-One, Inverses and Restricted Domains

Part (ai) · 1 mark

On the given axes, sketch the graph of y = f⁻¹(x) using the shown curve y = f(x).

1. 1Reflect each point (a, b) of y = f(x) to (b, a)
2. 2The curve starts on the y-axis, rises slowly at first then becomes steep, and crosses y = f(x) on the line y = x at about (4, 4)

AnswerReflection of y = f(x) in the line y = x: starts on the positive y-axis just above the origin, rises slowly then steeply, meeting y = f(x) again at roughly (4, 4)

Part (aii) · 1 mark

State how the graph of f⁻¹(x) is related geometrically to the graph of f(x).

1. 1Swapping x and y reflects a graph in the line y = x
2. 2The graph of f⁻¹(x) is the reflection of the graph of f(x) in the line y = x

AnswerReflection in the line y = x

Part (bi) · 3 marks

For g(x) = e^(√(x − 2)), x ≥ 2, find an expression for g⁻¹(x).

1. 1y = e^(√(x − 2))
2. 2ln y = √(x − 2)
3. 3(ln y)² = x − 2, so x = (ln y)² + 2
4. 4g⁻¹(x) = (ln x)² + 2

Answerg⁻¹(x) = (ln x)² + 2

Part (bii) · 1 mark

Write down the range of the inverse function g⁻¹.

1. 1The range of g⁻¹ is the domain of g
2. 2g⁻¹(x) ≥ 2

Answerg⁻¹(x) ≥ 2

Part (biii) · 2 marks

Given h(x) = 1/x² + 2 for x > 0, find gh(x) in its simplest form.

1. 1gh(x) = g(1/x² + 2) = e^(√(1/x² + 2 − 2))
2. 2= e^(√(1/x²)) = e^(1/x) since x > 0

Answergh(x) = e^(1/x)

More past-paper questions on functions: one-one, inverses and restricted domains

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