Let f(x)=ln ≤ft(2 x-x^2)+sin (π x)/2. Then which one of the following…

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

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

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

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

Therefore, $f(x)=\ln \left(2 x-x^2\right)+\sin \frac{\pi x}{2}$ attains its maximum value at $x=1$.

Also,

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

We observe that $f(x) \rightarrow-\infty$ as $x \rightarrow 0^{+}$or, $x \rightarrow 2^{-}$.

Therefore, minimum value of f(x) does not exist.

Hence, option (b) is not correct.