A) \[\pi /8\]
B) \[\pi /6\]
C) \[\pi /4\]
D) \[\pi /3\]
Correct Answer: A
Solution :
\[f(x)={{e}^{-2x}}\sin 2x\] Þ \[{f}'(x)=2{{e}^{-2x}}(\cos 2x-\sin 2x)\] Now,\[{f}'(c)=0\] Þ\[\cos 2c-\sin 2c=0\]Þ\[\tan 2c=1\]Þ\[c=\frac{\pi }{8}\].You need to login to perform this action.
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