2013 Paper 2 Question 1

Answers

(a)

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

(b)

$g f (x) = \frac{- 3 - 3 x}{1 - x} .$
$(g f)^{- 1} (5) = 4 .$

Solutions

(a)

$\begin{aligned} R_{g} & = (- \infty, \infty) \\ D_{f} & = (- \infty, 1) \cup (1, \infty) \end{aligned}$

Since $R_{g} \subseteq̸ D_{f},$ the composite function $f g$ does not exist $∎$

(b)

$\begin{aligned} g f (x) & = g (\frac{2 + x}{1 - x}) \\ = 1 - 2 (\frac{2 + x}{1 - x}) \\ = 1 - \frac{2 (2 + x)}{1 - x} \\ = 1 - \frac{4 + 2 x}{1 - x} \\ = \frac{- 3 - 3 x}{1 - x} ∎ \end{aligned}$ $\begin{aligned} Let x & = (g f)^{- 1} (5) \\ g f (x) & = 5 \\ \frac{- 3 - 3 x}{1 - x} & = 5 \\ - 3 - 3 x & = 5 - 5 x \\ 2 x & = 8 \\ x & = 4 \end{aligned}$

$(g f)^{- 1} (5) = 4 ∎$