Question

Minimum value of $(\sin x{)}^{\sin x}$ is $(0<x<\frac{\pi }{2})$

• A. $(e{)}^{\frac{-1}{e}}$
• B. $1$
• C. $\frac{\pi }{2}$
• D. $\frac{1}{e}$

Answer 1

The correct option is A $(e{)}^{\frac{-1}{e}}$

According to question......

$\begin{array}{c}f\left(x\right)=(sinx{)}^{sinx}\\ y=(sinx{)}^{sinx}\\ \log y=\sin x.\log \left(\sin x\right)----now,differentiateofthisequ.\\ \frac{1}{y}.y^{{}^{\prime }}=\cos x.\log \left(\sin x\right)+\sin x.\frac{1}{\sin x}\cos x\\ y^{{}^{\prime }}=(\sin {)}^{\sin x}\left[\cos x(\log \sin x+1)\right]\\ \frac{df}{dx}=(\sin {)}^{\sin x}\left[\cos x(\log \sin x+1)\right]|formax\&min\\ \frac{df}{dx}=0\end{array}$

$\begin{array}{c}then,\\ \cos x=0,or\log \sin x+1=0\\ x=\frac{\pi }{2},\sin x=\frac{1}{e}\\ Again,\\ differentiate,\\ \frac{d^{2}f}{d{x}^2}=(\sin {)}^{\sin x}\left[\cos x(\log \sin x+1)\right]{}^2+(\sin {)}^{\sin x}\left[\left[-\sin x(\log \sin x\right]+\frac{\cos x.\cos x}{\sin x}\right]\\ {\left(\frac{d^{2}f}{d{x}^2}\right)}_{\frac{\pi }{2}}=\left(1\right)\left[-1+0\right]=-1<0(maximum)\end{array}$

$\begin{array}{c}{\left(\frac{d^{2}f}{d{x}^2}\right)}_{x={\sin }^{-1}\frac{1}{e}}={\left(\frac{1}{e}\right)}^{\frac{1}{e}}.\left(\frac{e^2-1}{e^2}\right)e=(e-\frac{1}{e}).{\left(\frac{1}{e}\right)}^{\frac{1}{e}}>0\\ \therefore f_{min}={\left(\frac{1}{e}\right)}^{\frac{1}{e}}=(e{)}^{-\frac{1}{e}}\\ sothatthecorrectoptionisA.\end{array}$