STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

We have,

$f(x) =(\sin x)^{\sin x}, x \in(0, \pi)$

$x =\frac{\pi}{2}$

$\Rightarrow \quad f\left(\frac{\pi}{2}\right) =1$

Hence, maximum value of $f(x)$ is $1.$

Taking log both sides

$\log f(x)=\sin x \log \sin x$

On differentiating w.r.t. $x$, we get $f^{\prime}(x)=f(x)[\cos x+\cos x \log \sin x]$ put $f^{\prime}(x)=0$

$\cos x(1+\log \sin x)=0$

$\cos x=0, \log \sin x=-1$

$x=\frac{\pi}{2}, \sin x=e^{-1}$

$\therefore$ Range of $f(x)$ is $\left[\left(\frac{1}{e}\right)^{1 / e}, 1\right]=\left[e^{-1 / e}, 1\right]$