Question

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

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

Answer 1

The correct option is B

$\frac{\mathrm{\pi }}{2}$

Find the point with maximum slope in the given interval

Given: $f\left(x\right)=e^x\sin x$

On differentiating w.r.t to $x$,we get

$f\text{'}\left(x\right)=e^x\cos x+\sin x{e}^x$
Again, differentiating w.r.t. $x$, we get
$$\begin{gathered} f\text{'}\text{'}\left(x\right)=-e^x\sin x+\cos x{e}^x+e^x\cos x+e^x\sin x \\ =2e^x\cos x \end{gathered}$$
For maximum slope, put $f\prime \prime \left(x\right)=0$
$$\begin{gathered} \Rightarrow 2e^x\cos x=0 \\ \Rightarrow \cos x=0 \\ \Rightarrow x=\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2}\forall x\in \left[0,2\mathrm{\pi }\right] \end{gathered}$$

Now, $f\prime \prime \prime \left(x\right)=2[-e^x\sin x+e^x\cos x]$
At$x=\frac{\pi }{2},f\prime \prime \prime \left(x\right)<0$ and at $x=\frac{3\pi }{2},f\prime \prime \prime \left(x\right)>0$
$\therefore$Slope is maximum at $x=\frac{\pi }{2}$​.

Hence, option (B) is the correct answer.