Let $f\left(x\right)=\ln \left(2x-x^2\right)+\sin \frac{\pi x}{2}$. Then which one

of the following options is not correct ?

Clearly, $f \left(x\right)$ is defined for all $x\in (0,2)$

Also, $f(1+\alpha )=f(1-\alpha )\text{ for all }\alpha \in (0,1)$

So, $y=f \left(x\right)$ is symmetrical about the line $x=1$.

Clearly, $2x-x^2$ and $\sin \frac{\pi x}{2}$ attain the maximum value 1 at $x=1$.

Therefor , $f\left(x\right)=\ln \left(2x-x^2\right)+\sin \frac{\pi x}{2}$ attains its maximum

value at   $x=1$.  Also,

Maximum value $=f\left(1\right)=\ln (2-1)+\sin \frac{\pi }{2}=1$.

We observe that $f \left(x\right)$ does not exist.

Hence, potion (b) is not correct.