Question

If $f\left(x\right)=\frac{\sin x}{e^x}$ in $\left[0,\mathrm{\pi }\right]$, then $f\left(x\right)$

• A. satisfies Rolle's theorem and $c=\frac{\mathrm{\pi }}{4}$, so that $f\text{'}\left(\frac{\mathrm{\pi }}{4}\right)=4$
• B. does not satisfy Rolle's theorem but $f\text{'}\left(\frac{\mathrm{\pi }}{4}\right)>0$
• C. satisfies Rolle's theorem but $f\text{'}\left(\frac{\mathrm{\pi }}{4}\right)=0$
• D. satisfies LaGrange's Mean value theorem but $f\text{'}\left(\frac{\mathrm{\pi }}{4}\right)\ne 0$

Answer 1

The correct option is C

satisfies Rolle's theorem but $f\text{'}\left(\frac{\mathrm{\pi }}{4}\right)=0$

Explanation for the correct option.

Find the correct relation:

Given,

$f\left(x\right)=\frac{\sin x}{e^x}$

$\begin{array}{rcl}f\left(0\right)& =& \frac{\sin 0}{e^0}\\ & =& 0\end{array}$and

$\begin{array}{rcl}f\left(\mathrm{\pi }\right)& =& \frac{\mathrm{sin\pi }}{e^{\mathrm{\pi }}}\\ & =& 0\end{array}$

$\Rightarrow f\left(0\right)=f\left(\mathrm{\pi }\right)=0$

Therefore, $f\left(x\right)$ is continuous in $\left[0,\mathrm{\pi }\right]$.

Since, the given function is continuous in its domain and is differentiable.

So, put $f\text{'}\left(x\right)=0$

Differentiate $f\left(x\right)$ w.r.t. $x$ and put equals to 0.

$\begin{array}{rcl}\frac{d}{dx}\left(\frac{\sin x}{e^x}\right)& =& 0\\ & \Rightarrow & \frac{e^x·\cos x-\sin x·e^x}{e^{2x}}=0\mathbf{[}\mathbf{\because }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\frac{\mathbf{u}}{\mathbf{v}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{v}{\displaystyle \frac{du}{dx}}\mathbf{-}\mathbf{u}{\displaystyle \frac{dv}{dx}}}{{\mathbf{v}}^{\mathbf{2}}}\mathbf{]}\\ & \Rightarrow & \cos x-\sin x=0\\ & \Rightarrow & \cos x=\sin x\end{array}$

$x=\frac{\mathrm{\pi }}{4}$

Therefore, $f\text{'}\left(\frac{\mathrm{\pi }}{4}\right)=0$.

Hence, the correct option is C.