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).