COLUMN-I		COLUMN-II
A)	Let $f\left(x\right)=\min (\left|x\right|,1-\left|x\right|,1/4),\forall x\in R$, then value of  $\underset{-1}{\overset{1}{\int }}f\left(x\right)dx$ is equal to	P)	59/6
B)	Let $f\left(x\right)=\underset{0}{\overset{x}{\int }}(t^2-t+1)dt,\forall x\in [3,4]$, then the difference between the greatest and the least value of the function is	Q)	1/14
C)	The integral  ${\int }_{\pi /4}^{5\pi /4}\left(\right|\cos t|\sin t+|\sin t\left|\cos t\right)dt$ has the value equal to	R)	0
D)	$\underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{\sqrt{n}}{\sqrt{r}{(3\sqrt{r}+4\sqrt{n})}^2}$ is equal to	S)	3/8

A)  $\underset{-1}{\overset{1}{\int }}f\left(x\right)dx=2\underset{0}{\overset{1}{\int }}f\left(x\right)dx=2(\underset{0}{\overset{1/4}{\int }}xdx+\underset{1/4}{\overset{3/4}{\int }}\frac{1}{4}dx+\underset{3/4}{\overset{1}{\int }}(1-x)dx)$

$=2(\frac{1}{2}\times \frac{1}{4}\times \frac{1}{4}+\frac{1}{4}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{4}\times \frac{1}{4})=\frac{3}{8}$
B)  $f\text{'}\left(x\right)=x^2-x+1>0\forall x\in R$
 $f\left(x\right)$  is increasing in $x\in (3,4)$  so minimum at x = 3 and maximum at x = 4
 $\min f\left(x\right)=\underset{0}{\overset{3}{\int }}(t^2-t+1)dt=[\frac{t^3}{3}-\frac{t^2}{2}+t]=\frac{15}{2}$
$$\begin{gathered} \max f\left(x\right)=\underset{0}{\overset{4}{\int }}(t^2-t+1)dt=[\frac{t^3}{3}-\frac{t^2}{2}+t]=\frac{52}{3} \\ \therefore \max f\left(x\right)-\min f\left(x\right)=\frac{59}{6} \end{gathered}$$
$I=\underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{\sqrt{n}}{\sqrt{r}.n{(3\sqrt{\frac{r}{n}}+4)}^2}=\underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{1}{\sqrt{\frac{r}{n}}{(3\sqrt{\frac{r}{n}}+4)}^2}.\frac{1}{n}=\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{x}{(3\sqrt{x}+4)}^2}dx$
c) Conceptual
d)  
Put   $z=3\sqrt{x}+4\Rightarrow I=\frac{2}{3}\underset{4}{\overset{7}{\int }}\frac{dz}{z^2}=\frac{1}{14}$