JEE Advanced 2021 — Paper 2 — Question 25

$$\begin{gathered} I=\int_{0}^{10}\left[\sqrt{\frac{10 x}{x+1}}\right] d x {\left[\sqrt{\frac{10 x}{x+1}}\right]=n} \end{gathered}$$

$\Rightarrow \frac{n^{2}}{10-n^{2}} \leq x<\ \frac{(n+1)^{2}}{10-(n+1)^{2}} \text { where } n \in I$

For $\mathrm{n}=0,0 \leq \mathrm{x}<1 / 9$

$\mathrm{n}=1 ; 1 / 9 \leq \mathrm{x}<2 / 3$

$\mathrm{n}=2 ; 2 / 3 \leq \mathrm{x}<9, \mathrm{n}=3, \mathrm{x} \geq 9$

$\Rightarrow \quad I=\int_{0}^{1 / 9} 0 \cdot d x+\int_{1 / 9}^{2 / 3} 1 \cdot d x+\int_{2 / 3}^{9} 2 \cdot d x+\int_{9}^{10} 3 \cdot d x$ $=\left(\frac{2}{3}-\frac{1}{9}\right)+2\left(9-\frac{2}{3}\right)+3(10-9)=\frac{182}{9}=9 \mathrm{I}=182$