Question

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

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

Answer 1

Answer: (C)

Given, $f(x)=\frac{\sin x}{e^x}$
$f(0)=0,f(\pi )=0$
$f(x)$ is continuous in $[0,\pi ].$
since, every exponential function and trigonometric functions is continuous in their domain and it is differentiable the open interval.
Now $f^{\prime }(x)=\frac{e^x(\cos x-\sin x)}{e^x}$
Put $f^{\prime }(x)=0$
$\Rightarrow$ $\cos x-\sin x=0$
$\Rightarrow$ $x=\frac{\pi }{4}$
$\therefore$ $f^{\prime }\left(\frac{\pi }{4}\right)=0$