A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\pi \]
D) \[\frac{3\pi }{2}\]
Correct Answer: B
Solution :
\[\because \]\[f\,(x)={{e}^{x}}.\,\,\sin x\] |
\[f\,'(x)={{e}^{x}}\sin x+{{e}^{x}}\cos x\] |
\[f''(x)={{e}^{x}}\sin x+2{{e}^{x}}\cos x-{{e}^{x}}\sin x\] |
\[=2{{e}^{x}}\cos x=0\] |
\[\Rightarrow \]\[x=\frac{\pi }{2},\] \[\frac{3\pi }{2}\] |
\[\therefore \]\[f\,(x)\] is maximum at \[x=\pi /2\] |
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