JEE Mathematics 2024 Question with Solution

Given: Evaluate

$$9 \int_{0}^{9} \left\lfloor \sqrt{\frac{10x}{x+1}} \right\rfloor \, dx$$

where $\lfloor t \rfloor$ denotes the greatest integer less than or equal to $t$.

Find: The numerical value of the given expression.

Let

$$f(x) = \sqrt{\frac{10x}{x+1}}$$

We write

$$\frac{10x}{x+1} = 10 - \frac{10}{x+1}$$

which is increasing for $x \ge 0$. Therefore, $f(x)$ is also increasing on $[0,9]$.

Now check the endpoint values:

$$f(0) = \sqrt{\frac{10 \cdot 0}{0+1}} = 0$$ $$f(9) = \sqrt{\frac{10 \cdot 9}{9+1}} = \sqrt{9} = 3$$

So $\left\lfloor f(x) \right\rfloor$ can take the values $0,1,2$ on the interval.

Find the break points where $f(x)$ becomes an integer.

For $f(x)=1$:

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

For $f(x)=2$:

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

Since $f(x)$ is increasing,

• for $0 \le x < \frac{1}{9}$, we have $0 \le f(x) < 1$, so

$$\left\lfloor \sqrt{\frac{10x}{x+1}} \right\rfloor = 0$$

• for $\frac{1}{9} \le x < \frac{2}{3}$, we have $1 \le f(x) < 2$, so

$$\left\lfloor \sqrt{\frac{10x}{x+1}} \right\rfloor = 1$$

• for $\frac{2}{3} \le x < 9$, we have $2 \le f(x) < 3$, so

$$\left\lfloor \sqrt{\frac{10x}{x+1}} \right\rfloor = 2$$

Let

$$I = \int_{0}^{9} \left\lfloor \sqrt{\frac{10x}{x+1}} \right\rfloor \, dx$$

Then

$$I = \int_{0}^{1/9} 0 \, dx + \int_{1/9}^{2/3} 1 \, dx + \int_{2/3}^{9} 2 \, dx$$

Now evaluate:

$$I = 0 + \left(\frac{2}{3} - \frac{1}{9}\right) + \left(2 \cdot 9 - 2 \cdot \frac{2}{3}\right)$$ $$I = \frac{5}{9} + \left(18 - \frac{4}{3}\right)$$ $$I = \frac{5}{9} + \frac{50}{3}$$ $$I = \frac{5}{9} + \frac{150}{9} = \frac{155}{9}$$

Therefore,

$$9I = 9 \cdot \frac{155}{9} = 155$$

So the value of the given expression is $155$.

The answer key does not match the solution. Using the solution, the numerical answer is $155$.