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 \[f'\left( \frac{\pi }{4} \right)\ne 0\]
Correct Answer: C
Solution :
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 \[f'(x)=0\] \[\Rightarrow \] \[\cos \,x\,-\,\sin \,x=0\] \[\Rightarrow \] \[x=\frac{\pi }{4}\] \[\therefore \] \[f'\left( \frac{\pi }{4} \right)=0\]
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