2022 Paper 1 Question 6

Translate the curve $y=\frac{1}{x}$ by $a$ units in the positive $x\text{-axis}$ direction.
Scale the resulting curve by a factor of $a^2+k$ parallel to the $y\text{-axis}.$
Translate the resulting curve by $a$ units in the positive $y\text{-axis}$ direction.

$f^{-1}(x)=\frac{xa+k}{x-a}.$

$f^2(x)=x.$

$f^{2023}(1)=\frac{a+k}{1-a}.$

$$\frac{ax+k}{x-a}=a+\frac{a^2+k}{x-a}$$

• Translate the curve $y=\frac{1}{x}$ by $a$ units in the positive $x\text{-axis}$ direction
• Scale the resulting curve by a factor of $a^2+k$ parallel to the $y\text{-axis}$
• Translate the resulting curve by $a$ units in the positive $y\text{-axis}$ direction

$$\begin{array}{c}\text{Let }y=\frac{ax+k}{x-a}\\ y(x-a)=ax+k\\ yx-ya=ax+k\\ yx-ax=ya+k\\ x(y-a)=ya+k\\ x=\frac{ya+k}{y-a}\\ f^{-1}(x)=\frac{xa+k}{x-a}■\end{array}$$

We observe from part (b) that $f(x)=f^{-1}(x).$ Hence $$\begin{array}{rl}{f}^2(x)& =ff(x)\\ & =f{f}^{-1}(x)\\ & =x∎\end{array}$$

$$\begin{array}{rl}{f}^{2023}(1)& =f^{2021}ff(1)\\ & =f^{2021}f{f}^{-1}(1)\\ & =f^{2021}(1)\\ & =\cdots \\ & =f(1)\\ & =\frac{a+k}{1-a}∎\end{array}$$