JEE Main 2024 (Online) 30th January Morning ShiftQuestion 1
The value of $9\int_{0}^{9}\left[\sqrt{\frac{10x}{x+1}}\right]dx$, where $[t]$ denotes the greatest integer less than or equal to $t$, is $\_\_\_\_\_\_$ .
Definite Integration
Answer$155$
Solution
Concept. $g(x)=\sqrt{\frac{10x}{x+1}}$ increases from $0$ to $3$ on $[0,9]$, so $[g]$ is a step function; find where it jumps. Formulas used. (1) $[g(x)]\ge k\iff\dfrac{10x}{x+1}\ge k^2\iff x\ge\dfrac{k^2}{10-k^2}$ (for $k^2<10$)(2) $\int_0^9[g]\,dx=\sum_{k\ge1}\big(9-x_k\big)$, where $x_k$ is the point from which $[g]\ge k$Step 1 - Range. $g(0)=0$, $g(9)=\sqrt{\frac{90}{10}}=3$, and $\frac{10x}{x+1}=10-\frac{10}{x+1}$ is increasing. Step 2 - Jump points. By (1): $k=1$: $x_1=\frac19$; $k=2$: $x_2=\frac46=\frac23$; $k=3$: $x_3=9$.So $[g]=0$ on $[0,\frac19)$, $1$ on $[\frac19,\frac23)$, $2$ on $[\frac23,9)$. Step 3 - Integrate. By (2), $$\int_0^9[g]\,dx=\Big(9-\frac19\Big)+\Big(9-\frac23\Big)+0=18-\frac79=\frac{155}{9}.$$ Step 4 - Value. $9\cdot\frac{155}9=155$. Answer: 155