A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\pi \]
D) \[\frac{3\pi }{2}\]
Correct Answer: B
Solution :
We have,\[f(x)={{e}^{x}}\sin x\] \[\Rightarrow \] \[f(x)={{e}^{x}}\cos x+\sin x\cdot {{e}^{x}}\] and \[f(x)=-\sin x\cdot {{e}^{x}}+\cos x\cdot {{e}^{x}}+\cos x\cdot {{e}^{x}}\] \[+\sin x\cdot {{e}^{x}}\] Now, for maximum or minimum slope put \[(f(x))=0\Rightarrow f(x)=0\] \[\Rightarrow \] \[2\cos x\cdot {{e}^{x}}=0\] \[\Rightarrow \] \[\cos x=0\Rightarrow x=\frac{\pi }{2}\] Also, \[f(x)=-2\sin x\cdot {{e}^{x}}+2\cos x\cdot {{e}^{x}}=\]negative \[\therefore \]Slope is maximum at\[x=\pi /2\].
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