Question

For $x\in (0,\frac{5\mathrm{\pi }}{2})$, define $f\left(x\right)={\int }_0^x\sqrt{t}\sin tdt$​​. Then $f$ has:-

• A. Local maximum at $\mathrm{\pi }$ and $2\mathrm{\pi }$
• B. Local minimum at $\mathrm{\pi }$ and local maximum at $2\mathrm{\pi }$
• C. local maximum at $\pi$ and a local minimum at $2\pi$
• D. local minimum at $\pi$ and $2\pi$

Answer 1

The correct option is C

local maximum at $\pi$ and a local minimum at $2\pi$

Find the maximum and minimum value of $f$:

Given, $f\left(x\right)={\int }_0^x\sqrt{t}\sin tdt$

$\Rightarrow$ $f\text{'}\left(x\right)={[\sqrt{t}\sin t]}_0^x$

$\Rightarrow$ $f\text{'}\left(x\right)=\sqrt{x}\sin x-0$

$\Rightarrow$ $f\text{'}\left(x\right)=\sqrt{x}\sin x$

Equate $f\text{'}\left(x\right)$ to zero to get maximum and minimum values.

$\Rightarrow$ $f\text{'}\left(x\right)=0$

$\Rightarrow$$\sqrt{x}\sin x=0forx=0,\mathrm{\pi },2\mathrm{\pi },4\mathrm{\pi }....$

as $\mathrm{x}\in (0,\frac{5\mathrm{\pi }}{2})$ on number line we create points

Between $(0,\mathrm{\pi })$

$\mathrm{f}\text{'}\left(\mathrm{x}\right)=+\mathrm{ve}$

Between $(\mathrm{\pi },2\mathrm{\pi })$

$\mathrm{f}\text{'}\left(\mathrm{x}\right)=-\mathrm{ve}$

Between $(2\mathrm{\pi },\frac{5\mathrm{\pi }}{2})$

$\mathrm{f}\text{'}\left(\mathrm{x}\right)=+\mathrm{ve}$

$\therefore$the equation has local maximum at $\mathit{\pi }$ and a local minimum at $\mathbf{2}\mathit{\pi }$

Hence, Option "C" is correct .