Question

The point in the interval $ [0,\,2\pi ] $ , where $ f(x)={e}^{x}\sin x $ has maximum slope, is

• A. $ \frac{\pi }{4} $
• B. $ \frac{\pi }{2} $
• C. $ \pi $
• D. $ \frac{3\pi }{2} $

Answer 1

Answer: (B)

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 $x=\pi / 2$.