Let f be a differentiable function such that  $f\left(f\left(x\right)\right)=x$, where  $x\in [0,1]$. Also  $f\left(0\right)=1$, If the value of $\underset{0}{\overset{1}{\int }}{(x-f\left(x\right))}^{2022}dx=\frac{a}{b}$ (where a & b are co-prime), then value of 
(a + b) is equal to – 

Given $f\left(f\left(x\right)\right)=x$ put  $x=0\Rightarrow f\left(1\right)=0$
Now, $I=\underset{0}{\overset{1}{\int }}{(x-f\left(x\right))}^{2022}dx$  ….(i)
put  $x=f\left(t\right)\therefore dx=f\text{'}\left(t\right)dt$
So,  $I=-\underset{0}{\overset{1}{\int }}{(f\left(t\right)-t)}^{2022}f\text{'}\left(t\right)dt$
$\Rightarrow I=-\underset{0}{\overset{1}{\int }}{(t-f\left(t\right))}^{2022}f\text{'}\left(t\right)dt \mathrm{.........}\left(ii\right)$

(i) + (ii)

$2I=\underset{0}{\overset{1}{\int }}{(t-f\left(t\right))}^{2022}(1-f\text{'}\left(t\right))dt={\left[\frac{{(t-f\left(t\right))}^{2023}}{2023}\right]}_0^1=\frac{2}{2023}$So $I=\frac{1}{2023}$