JEE Main 2024 — 30 January, Shift 1 — Question 25

Analyze the Function Let $f(x) = \sqrt{\frac{10x}{x+1}}$. We observe the behavior of $f(x)$ on the interval $[0, 9]$: At $x = 0$: $f(0) = \sqrt{\frac{0}{1}} = 0$. At $x = 9$: $f(9) = \sqrt{\frac{90}{10}} = \sqrt{9} = 3$. The function $f(x) = \sqrt{10 - \frac{10}{x+1}}$ is strictly increasing for $x \geq 0$. Therefore, the integer values that $[f(x)]$ can take on the interval $[0, 9]$ are $0, 1,$ and $2$. Note that while $f(9)=3$, the floor function only equals $3$ at that single point, which does not contribute to the value of the integral.

Find Critical Points of Discontinuity We solve for $x$ when $f(x)$ reaches integer values $1$ and $2$.

Case 1: $f(x) = 1$

$$\sqrt{\frac{10x}{x+1}} = 1 \implies \frac{10x}{x+1} = 1$$ $$10x = x + 1 \implies 9x = 1 \implies x = \frac{1}{9}$$

Case 2: $f(x) = 2$

$$\sqrt{\frac{10x}{x+1}} = 2 \implies \frac{10x}{x+1} = 4$$ $$10x = 4x + 4 \implies 6x = 4 \implies x = \frac{2}{3}$$

Evaluate the Definite Integral We split the integral into three parts based on the constant values of the floor function: For $x \in [0, 1/9)$, $[f(x)] = 0$. For $x \in [1/9, 2/3)$, $[f(x)] = 1$. For $x \in [2/3, 9]$, $[f(x)] = 2$.

Let $I = \int_{0}^{9} [f(x)] dx$:

$$I = \int_{0}^{1/9} 0 \, dx + \int_{1/9}^{2/3} 1 \, dx + \int_{2/3}^{9} 2 \, dx$$ $$I = 0 + \left( \frac{2}{3} - \frac{1}{9} \right) + 2 \left( 9 - \frac{2}{3} \right)$$ $$I = \left( \frac{6-1}{9} \right) + 2 \left( \frac{27-2}{3} \right)$$ $$I = \frac{5}{9} + \frac{50}{3} = \frac{5}{9} + \frac{150}{9} = \frac{155}{9}$$

Final Answer The expression requires $9 \times I$:

$$9 \times \frac{155}{9} = 155$$