2009 Paper 2 Question 3

Answers

$f^{- 1} (x) = \frac{a x}{b x - a} .$
$f^{2} (x) = x .$
$R_{f^{2}} = (- \infty, \frac{a}{b}) \cup (\frac{a}{b}, \infty) .$

(b)

The composite function $f g$ does not exist as $R_{g} \subseteq̸ D_{f} .$

(c)

$x = 0 or \frac{2 a}{b} .$

Solutions

(a)

$\begin{matrix} Let y = \frac{a x}{b x - a} \\ y (b x - a) = a x \\ y b x - y a = a x \\ y b x - a x = y a \\ x (y b - a) = y a \\ x = \frac{y a}{y b - a} \\ f^{- 1} (x) = \frac{x a}{x b - a} ■ \end{matrix}$ $\begin{aligned} f^{2} (x) & = f f (x) \\ = f f^{- 1} (x) \\ = x ∎ \end{aligned}$

$R_{f^{2}} = (- \infty, \frac{a}{b}) \cup (\frac{a}{b}, \infty) ∎$

(b)

$\begin{aligned} R_{g} & = (- \infty, 0) \cup (0, \infty) \\ D_{f} & = (- \infty, \frac{a}{b}) \cup (\frac{a}{b}, \infty) \end{aligned}$

Since $a$ and $b$ are non-zero,

$R_{g} \subseteq̸ D_{f}$

so the composite function $f g$ does not exist $∎$

(c)

$\begin{matrix} f^{- 1} (x) = x \\ \frac{a x}{b x - a} = x \\ a x = x^{2} b - x a \\ x^{2} b - 2 x a = 0 \\ x (x b - 2 a) = 0 \end{matrix}$

$\begin{aligned} x & = 0 ■ & or & x b - 2 a & = 0 \\ x b & = 2 a \\ x & = \frac{2 a}{b} ■ \end{aligned}$