(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.