Question

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

• A. satisfies Rollers 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 Lagranges Mean value theorem but $ f'\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'(x)=\frac{{{e}^{x}}(\cos \,x-\sin \,x)}{{{e}^{x}}} $
Put $ f'(x)=0 $
$ \Rightarrow $ $ \cos \,x\,-\,\sin \,x=0 $
$ \Rightarrow $ $ x=\frac{\pi }{4} $
$ \therefore $ $ f'\left( \frac{\pi }{4} \right)=0 $