0606/22 May/Jun 2026, Question 8: Functions: One-One, Inverses and Restricted Domains

Part (a) · 1 mark

State the set of values that f(x) = 1 + 2ˣ takes for x ≥ −1.

1. 1f is increasing, so its least value is f(−1) = 1 + 2⁻¹ = 1.5
2. 2Range: f(x) ≥ 1.5

Answerf(x) ≥ 1.5

Part (bi) · 1 mark

Give the domain of the inverse function f⁻¹.

1. 1Domain of f⁻¹ = range of f
2. 2x ≥ 1.5

Answerx ≥ 1.5

Part (bii) · 2 marks

Derive a formula for f⁻¹(x) in terms of x.

1. 1y = 1 + 2ˣ
2. 2Swap: x = 1 + 2ʸ, so 2ʸ = x − 1
3. 3y = log₂(x − 1)
4. 4f⁻¹(x) = log₂(x − 1)

Answerf⁻¹(x) = log₂(x − 1)

Part (biii) · 4 marks

Sketch y = f(x) and y = f⁻¹(x) on the axes, stating axis intercepts.

1. 1f(−1) = 1.5, so y = f(x) starts at (−1, 1.5)
2. 2f(0) = 1 + 1 = 2: y-intercept (0, 2)
3. 3f increases ever more steeply (exponential)
4. 4Reflect in y = x: y = f⁻¹(x) starts at (1.5, −1) and crosses the x-axis at (2, 0)

Answery = f(x): increasing exponential curve starting at (−1, 1.5), through (0, 2); y = f⁻¹(x): its reflection in y = x, starting at (1.5, −1), through (2, 0); intercepts (0, 2) and (2, 0)

Part (ci) · 1 mark

Explain why the composite function gf can be formed.

1. 1gf(x) = g(f(x)) needs every output of f to be a valid input of g
2. 2Range of f is f(x) ≥ 1.5, which lies inside the domain of g, x ≥ −1

AnswerThe range of f (f ≥ 1.5) is a subset of the domain of g (x ≥ −1)

Part (cii) · 1 mark

Obtain an expression for gf(x) using f and g.

1. 1gf(x) = g(1 + 2ˣ)
2. 2= (1 + 2ˣ)(1 + 2ˣ + k)

Answergf(x) = (1 + 2ˣ)(1 + 2ˣ + k)

Part (ciii) · 2 marks

Given gf(3) = 126, determine the constant k.

1. 1f(3) = 1 + 8 = 9
2. 2gf(3) = 9(9 + k) = 126
3. 39 + k = 14, k = 5

Answerk = 5

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

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