2023 Paper 1 Question 7

Answers

$R_{f} = [0, \infty) .$

$f^{2}$ does not exist as $R_{f} \subseteq̸ D_{f} .$

Greatest value of $a = - 2 .$

$f^{- 1} (x) = \frac{3 x + 4}{x - 2} .$
$D_{f^{- 1}} = R_{f} = [0, 2) .$

Solutions

$R_{f} = [0, \infty) ∎$

$\begin{aligned} R_{f} & = [0, \infty) \\ D_{f} & = (- \infty, 3) \cup (3, \infty) \end{aligned}$

Since $R_{f} \subseteq̸ D_{f},$ the composite function $f^{2}$ does not exist $∎$

Greatest value of $a = - 2 ∎$

$Let y = | \frac{2 x + 4}{3 - x} |$

Since $\frac{2 x + 4}{3 - x} < 0$ for $x \leq - 2,$ $\begin{matrix} y = - \frac{2 x + 4}{3 - x} \\ y (3 - x) = - 2 x - 4 \\ 3 y - y x = - 2 x - 4 \\ y x - 2 x = 3 y + 4 \\ x (y - 2) = 3 y + 4 \\ x = \frac{3 y + 4}{y - 2} \end{matrix}$

$f^{- 1} (x) = \frac{3 x + 4}{x - 2} ■$

$\begin{aligned} D_{f^{- 1}} & = R_{f} \\ = [0, 2) ■ \end{aligned}$