Column I		Column II
(i)	$\underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{n}{\left(n+r\right)\sqrt{r\left(2n+r\right)}}$ is	(A)	0
(ii)	$\underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{1}{\sqrt{2rn-r^2}}$  is	(B)	$\frac{\pi }{4}$
(iii)	${\int }_{\log \frac{1}{2}}^{\log 2}\sin \left(\frac{e^x-1}{e^x+1}\right)dx$ is	(C)	$\frac{\pi }{2}$
(iv)	If ${\int }_0^{\infty }\frac{\sin x}{x}dx=\frac{\pi }{2},$ then ${\int }_0^{\infty }\frac{{\sin }^{3}x}{x}dx$ is equal to	(D)	$\frac{\pi }{3}$
		(E)	$\frac{\pi }{6}$

We have,  $T_r=rthterm=\frac{n}{\left(n+r\right)\sqrt{r\left(2n+r\right)}}$

Put   $h=\frac{1}{n}$

So,  $T_r=\frac{\sqrt{h}}{\left(1+rh\right)\sqrt{r\left(2+rh\right)}}$

$=\frac{h}{\left(1+rh\right)\sqrt{rh\left(2+rh\right)}}$

Therefore, required sum

$={\int }_0^1\frac{dx}{\left(1+x\right)\sqrt{x\left(2+x\right)}}$

$={\int }_0^1\frac{dx}{\left(1+x\right)\sqrt{{\left(1+x\right)}^2-1}}$

$={\left[{\sec }^{-1}\left(1+x\right)\right]}_0^1$

$={\sec }^{-1}2-se{c}^{-1}1$

$=\frac{\pi }{3}-0=\frac{\pi }{3}$

(ii)we have,

rth term $=\frac{1}{\sqrt{2rn-r^2}}$

$=\frac{\sqrt{h}}{\sqrt{2r-r^{2}h}}$, Where $h=\frac{1}{n}$

$=\frac{h}{\sqrt{2rh-r^2{h}^2}}$

Therefore, required sum is ${\int }_0^1\frac{dx}{\sqrt{2x-x^2}}$

$={\int }_0^1\frac{dx}{\sqrt{1-{\left(1-x\right)}^2}}$

$=-{\left[{\sin }^{-1}\left(1-x\right)\right]}_0^1$

$=-{\sin }^{-1}0+{\sin }^{-1}1$

$=\frac{\pi }{2}$

(iii) we have,  $l={\int }_{\log 1/2}^{\log 2}\sin \left[\frac{e^x-1}{e^x+1}\right]dx$

$={\int }_{-\log 2}^{\log 2}\sin \left[\frac{e^x-1}{e^x+1}\right]dx$

Let $f\left(x\right)=\sin \left[\frac{e^x-1}{e^x+1}\right]$

$f\left(-x\right)=\sin \left(\frac{e^x-1}{e^x+1}\right)$

$f\left(-x\right)=-f\left(x\right)$

Hence, is an odd function.

So, $I=0$

(iv) Given that,  ${\int }_0^{\infty }\frac{\sin x}{x}dx=\frac{\pi }{2}$

$\therefore {\int }_0^{\infty }\frac{{\sin }^{3}x}{x}dx={\int }_0^{\infty }\left(\frac{\frac{3}{4}\sin x-\frac{1}{4}\sin 3x}{x}\right)dx$

$\frac{3}{4}{\int }_0^{\infty }\frac{\sin x}{x}dx-\frac{1}{4}{\int }_0^{\infty }\frac{\sin 3x}{x}dx$

$\frac{3}{4}{\int }_0^{\infty }\frac{\sin x}{x}dx-\frac{1}{4}{\int }_0^{\infty }\frac{\sin u}{u}du\left(letu=3x\right)$

$=\frac{3}{4}.\frac{\pi }{2}-\frac{1}{4}.\frac{\pi }{2}=\frac{\pi }{4}$