Question

Let $f$ be a function defined by $f\left(x\right)=\frac{x-5}{x-3};x\ne 3$, $f^k\left(x\right)$ denote the composition of $f$ with itself taken $k$ times i.e $f^3\left(x\right)=f\left(f\right(f\left(x\right)\left)\right)$
Then

• A. $f^{2019}\left(2020\right)=\frac{6055}{2019}$
• B. $f^{2020}\left(2021\right)=2021$
• C. $f^{2022}\left(2019\right)=\frac{4033}{2017}$
• D. $f^{2021}\left(2020\right)=\frac{2015}{2019}$

Answer 1

Answer: (A), (B), (C)

$f^{2022}\left(2019\right)=\frac{4033}{2017}$

$f\left(x\right)=\frac{x-5}{x-3}\Rightarrow f\left(f\right(x\left)\right)=\frac{\frac{x-5}{x-3}-5}{\frac{x-5}{x-3}-3}=\frac{2x-5}{x-2}$
$\therefore f^2\left(x\right)=\frac{2x-5}{x-2}{f}^3\left(x\right)=f\left(f^2\right(x\left)\right)=\frac{\frac{2x-5}{x-2}-5}{\frac{2x-1}{x-2}-3}=\frac{3x-5}{x-1}$
Now,
$f^4\left(x\right)=f\left(f^3\right(x\left)\right)=\frac{\frac{3x-5}{x-1}-5}{\frac{3x-5}{x-1}-3}=x$

$\therefore f^4\left(x\right)=x\Rightarrow f^5\left(x\right)=f\left(x\right),f^6\left(x\right)=f^2\left(x\right),f^7\left(x\right)=f^3\left(x\right)\cdots \therefore f^{4\alpha }\left(x\right)=x,f^{4\alpha +1}\left(x\right)=f\left(x\right),f^{4\alpha +2}\left(x\right)=f^2\left(x\right),f^{4\alpha +3}\left(x\right)=f^3\left(x\right),\alpha \in N$
Now,
$f^{2019}\left(2020\right)=f^3\left(2020\right)=\frac{6015}{2019}$
$f^{2020}\left(2021\right)=2020$
$f^{2022}\left(2019\right)=f^2\left(2019\right)=\frac{4033}{2017}$
$f^{2021}\left(2020\right)=f\left(2020\right)=\frac{2015}{2017}$