22M.1.AHL.TZ2.10

pestleMathematicsAIHLPaper 122M· sl-2-2-functions-notation-domain-range-and-inverse-as-reflection, ahl-2-7-composite-functions-finding-inverse-function-incl-domain-restrictionsource ↗

The function $f (x) = \ln (\frac{1}{x - 2})$ is defined for $x > 2, x \in ℝ$ .

Find an expression for $f^{- 1} (x)$ . You are not required to state a domain.

[3]

Solve $f (x) = f^{- 1} (x)$ .

[2]

Markscheme / solution

$y = \ln (\frac{1}{x - 2})$

an attempt to isolate $x$ (or $y$ if switched) (M1)

$e^{y} = \frac{1}{x - 2}$

$x - 2 = e^{- y}$

$x = e^{- y} + 2$

switching $x$ and $y$ (seen anywhere) M1

$f^{- 1} (x) = e^{- x} + 2$ A1

[3 marks]

sketch of $f (x)$ and $f^{- 1} (x)$ (M1)

$x = 2.12 (2.12002 \dots)$ A1

[2 marks]

Examiners’ report

This question was well answered by most candidates.

[N/A]