A) \[\frac{\pi }{4}\]
B) \[\frac{3\pi }{4}\]
C) \[\frac{5\pi }{6}\]
D) \[\frac{\pi }{2}\]
Correct Answer: B
Solution :
Given, \[f(x)={{e}^{x}}\sin x\] Now, \[f(0)={{i}^{0}}\sin 0=0\] and \[f(\pi )={{e}^{\pi }}\sin \pi =0\] \[\therefore \] \[f(0)=f(\pi )\] It is also continuous in the interval \[[0,\,\,\pi ]\] and also differentiable\[(0,\,\,\pi )\]. Now, \[f(x)={{e}^{x}}\sin x\] \[\Rightarrow \] \[f'(x)={{e}^{x}}\cos x+{{e}^{x}}\sin x\] Put \[f'(x)=0\] \[\Rightarrow \] \[{{e}^{x}}\cos x+{{e}^{x}}\sin x=0\] \[\Rightarrow \] \[{{e}^{x}}(\cos x+\sin x)=0\] \[\Rightarrow \] \[\tan x=-1\] \[\Rightarrow \] \[x=\frac{3\pi }{4}\] \[\therefore \]=\[c=\frac{3\pi }{4}\]
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