Let  $f\left(x\right)=\log (2x-x^2)+\sin \frac{\pi x}{2}$.  then which of the following is/are true

$f\left(x\right)=\log (2x-x^2)+\sin \frac{\pi x}{2}=\log (1-(x-1{)}^2)+\sin \frac{\pi x}{2}$

$f(1-x)=\log (1-(1-(x-1{)}^2))+\sin \frac{\pi (1-x)}{2}$

$=\log (1-x^2)+\cos \frac{\pi x}{2}$

Also,  $f(1+x)=\log (1-(1+(x-1{)}^2))+\sin \frac{\pi (1+x)}{2}$

$=\log (1-x^2)+\cos x\frac{\pi x}{2}$

Hence, function is symmetriacal about line $x=1$  ; Also , $f\left(1\right)=1$

Also , $f\left(1\right)=1$

Also, for domain of the function

$2x-x^2>0 \text{or }x\in (0,2)$

For $x>1, f\left(x\right)$ decrease.

Hence, x = 1 is point of maxima.

Also, maximum value of the function is

Also, $f\left(x\right)\to \mathrm{\infty }, \text{when }x\to 2$.

Hence, absolute minimum value of f does not exist