(i)     Show that ${f^{ - 1}}(x) = \ln x - 3$ .

(ii)    Write down the domain of ${f^{ - 1}}$ .

Solve the equation ${f^{ - 1}}(x) = \ln \frac{1}{x}$ .

(i) interchanging x and y (seen anywhere)     M1

e.g. $x = {{\rm{e}}^{y + 3}}$

correct manipulation     A1

e.g. $\ln x = y + 3$ , $\ln y = x + 3$

${f^{ - 1}}(x) = \ln x - 3$     AG     N0

(ii) $x > 0$     A1     N1 

[3 marks]

collecting like terms; using laws of logs     (A1)(A1)

e.g. $\ln x - \ln \left( {\frac{1}{x}} \right) = 3$ , $\ln x + \ln x = 3$ , $\ln \left( {\frac{x}{{\frac{1}{x}}}} \right) = 3$ , $\ln {x^2} = 3$

simplify     (A1)

e.g. $\ln x = \frac{3}{2}$ ,  ${x^2} = {{\rm{e}}^3}$

$x = {{\rm{e}}^{\frac{3}{2}}}\left( { = \sqrt {{{\rm{e}}^3}} } \right)$     A1     N2

[4 marks]

Many candidates interchanged the $x$ and $y$ to find the inverse function, but very few could write down the correct domain of the inverse, often giving $x \ge 0$ , $x > 3$ and "all real numbers" as responses.

Where students attempted to solve the equation in (b), most treated $\ln x - 3$ as $\ln (x - 3)$ and created an incorrect equation from the outset. The few who applied laws of logarithms often carried the algebra through to completion.