The function $\mathrm{f}$ is defined by
$$
f: x \rightarrow \frac{a x+k}{x-a}, \quad x \in \mathbb{R}, \quad x \neq a
$$
where $a$ and $k$ are constants.
(a) Describe fully a sequence of transformations which transforms the curve $y=\frac{1}{x}$ onto the curve $y=\mathrm{f}(x)$.
[4]
(b) Find $\mathrm{f}^{-1}(x)$.
[2]
(c) Hence, or otherwise, find $\mathrm{f}^2(x)$.
[1]
(d) Find $\mathrm{f}^{2023}(1)$ in terms of $a$ and $k$.
[2]
(a)
- First, translate the graph a units in the positive $x$ direction.
- Then, scale the graph with a scale factor of $\left(a^2+k\right)$ parallel to the $y$ axis.
- Lastly, translate the graph a units in the positive $y$ direction.
(b) $f^{-1}(x)=\frac{a x+k}{x-a}$, $x \in \mathbb{R}, x \neq a$
(c) $f^2(x)=x$
(d) $f^{2023}(1)=\frac{a+k}{1-a}$
(a) Observe that $f(x)=\frac{a x+k}{x-1}=a+\frac{a^2+k}{x-a}$
- First, translate the graph a units in the positive $x$ direction.
- Then, scale the graph with a scale factor of $\left(a^2+k\right)$ parallel to the $y$ axis.
- Lastly, translate the graph a units in the positive $y$ direction.
(b) Let $y=f(x)$
$$
\begin{array}{l}
y=\frac{a x+k}{x-1} \\
y(x-1)=a x+k \\
x=\frac{2 y+k}{y-1} \\
f^{-1}(x)=\frac{2 x+k}{x-1}, x \in \mathbb{R}, x \neq a
\end{array}
$$
(c) $f^2(x)=x$ since $\mathrm{f}$ is a self inverse function.
(d) $f^{2023}(1)=\frac{a+k}{1-a}$ since $\mathrm{f}$ is a self inverse function and 2023 is an odd number.