A function f is defined by \(f(x) = \frac{{2x - 3}}{{x - 1}},{\text{ }}x \ne 1\).

(a)     Find an expression for \({f^{ - 1}}(x)\).

(b)     Solve the equation \(\left| {{f^{ - 1}}(x)} \right| = 1 + {f^{ - 1}}(x)\).

(a)     Note: Interchange of variables may take place at any stage.

for the inverse, solve for x in

\(y = \frac{{2x - 3}}{{x - 1}}\)

\(y(x - 1) = 2x - 3\)     M1

\(yx - 2x = y - 3\)

\(x(y - 2) = y - 3\)     (A1)

\(x = \frac{{y - 3}}{{y - 2}}\)

\( \Rightarrow {f^{ - 1}}(x) = \frac{{x - 3}}{{x - 2}}\,\,\,\,\,(x \ne 2)\)     A1

Note: Do not award final A1 unless written in the form \({f^{ - 1}}(x) =  \ldots \)

(b)     \( \pm {f^{ - 1}}(x) = 1 + {f^{ - 1}}(x)\) leads to

\(2\frac{{x - 3}}{{x - 2}} = - 1\)     (M1)

\(x = \frac{8}{3}\)     A1

[6 marks]

Many candidates gained the correct answer to part (a), although a significant minority left the answer in the form \(y = \ldots {\text{ or }}x =  \ldots \) rather than \({f^{ - 1}}(x) =  \ldots \). Only the better candidates were able to make significant progress in part (b).