Q. If $f(x) = \sin{x/e^{x}}$ in [0, $\pi$], then $f(x)$
A) Satisfies Rolle’s Theorem and $c = \frac{\pi}{4}$, So that $f^{'\left( \frac{\pi}{4} \right)} = 4$
B) Does not satisfy Rolle’s Theorem but $f'\left( \frac{\pi}{4} \right) > 0$
C) Satisfies Rolle’s Theorem but $f'\left( \frac{\pi}{4} \right) = 0$
D) Satisfies Lagrange’s Mean Value Theorem but $f'\left( \frac{\pi}{4} \right) \neq 0$
Explanation
(c)
Given, $f(x) = \frac{\sin x}{e^{x}}$
Here, $f(0) = 0,\ \ f(\pi) = 0$
Also. $f(x)$Is continuous in $\lbrack 0,\pi\rbrack$Since, every exponential function and trigonometric function is continuous in their domain and it is differentiable in the open interval.
Now, $f^{'}(x) = \frac{e^{x}(\cos{x - \sin{x)}}}{e^{x}}$
Put $f^{'}(x) = 0$
$\Rightarrow \cos{x - \sin{x = 0 \Longrightarrow x = \frac{\pi}{4}}}$
$\therefore f^{'}\left( \frac{\pi}{4} \right) = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.