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, $f(x)=e^x\sin x$
$\Rightarrow f^{\prime }(x)=e^x\cos x+\sin x\cdot e^x$
and $f^{\prime \prime }(x)=-\sin x\cdot e^x+\cos x\cdot e^x+\cos x\cdot e^x+\sin x\cdot e^x$
Now, for maximum or minimum slope put
$(f^{\prime }(x){)}^{\prime }=0\Rightarrow f^{\prime \prime }(x)=0$
$\Rightarrow 2\cos x\cdot e^x=0$
$\Rightarrow \cos x=0$
$\Rightarrow x=\frac{\pi }{2}$
Also, $f^{\prime \prime \prime }(x)=-2\sin x\cdot e^x+2\cos x\cdot e^x=$ negative
$\therefore$ Slope is maximum at $x=\pi /2$.