JEE MAIN - Mathematics (2025 - 29th January Morning Shift - No. 21)

Let [.] be the greatest integer less than or equal to t. Then the least value of p ∈ N for which

$ \lim\limits_{x \to 0^+} \left( x \left[ \frac{1}{x} \right] + \left[ \frac{2}{x} \right] + \ldots + \left[ \frac{p}{x} \right] \right) - x^2 \left( \left[ \frac{1}{x^2} \right] + \left[ \frac{2}{x^2} \right] + \ldots + \left[ \frac{9^2}{x^2} \right] \right) \geq 1 $ is equal to _______.

Answer
24

Explanation

$$\begin{aligned} & \lim _{x \rightarrow 0^{+}}\left(x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots . .+\left[\frac{p}{x}\right]\right)-x^2\left(\left[\frac{1}{x^2}\right]+\left[\frac{2^2}{x^2}\right]+\left[\frac{9^2}{x^2}\right]\right)\right) \geq 1 \\ & \quad(1+2+\ldots \ldots+p)-\left(1^2+2^2+\ldots .9^2\right) \geq 1 \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & \frac{p(p+1)}{2}-\frac{9.10 .19}{6} \geq 1 \\ & p(p+1) \geq 572 \end{aligned}\\ &\text { Least natural value of } p \text { is } 24 \end{aligned}$$

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