Let f(x) = $min\left\{\left[x-1\right],\left[x-2\right],....,\left[x-10\right]\right\}$  where [t] denotes the greatest integer $\le$  t. Then ${\displaystyle \underset{0}{\overset{10}{\int }}f\left(x\right)dx+{\displaystyle \underset{0}{\overset{10}{\int }}{\left(f\left(x\right)\right)}^{2}dx+{\displaystyle \underset{0}{\overset{10}{\int }}\left|f\left(x\right)\right|dx}}}$  is equal to………..

$3-2^x-2^{1-x}\ge 0put{2}^x=t$

$3-t-\frac{2}{t}\ge 0\Rightarrow t^2-3t+2\le 0$

$\left(t-1\right)\left(t-2\right)\le 0\Rightarrow t\in \left[1,2\right]$

? P? = 6!/2! = 360

Kindly consider the following Image 

19.00
Only '2' in range → 1 function
one element out of 1, 3, 4 is in range with '2'
number of ways = ³C? × (3!/(2!1!)) × 2! = 18
(Select one from 1, 3, 4 and distribute among a, b, c)
Total function = 1 + 18 = 19

$f\left(x\right)=\frac{1-x}{1+x},0<x<1$ (on simplification)

$f\text{'}\left(x\right)=\frac{-2}{{\left(1+x\right)}^2}$

${\left(1-x\right)}^{2}f\text{'}\left(x\right)=-2{\left(f\left(x\right)\right)}^2$