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Mathematics · 2025

JEE Main · 29 January 2025, Shift 1 · Q24

Let [t] be the greatest integer less than or equal to t. Then the least value of p ∈ N for which lim_x → 0^+…

Let [t] be the greatest integer less than or equal to t. Then the least value of $\displaystyle \mathrm{p} \in \mathbf{N}$ for which $$\lim _{x \rightarrow 0^{+}}\left(x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots+\left[\frac{\mathrm{p}}{x}\right]\right)-x^2\left(\left[\frac{1}{x^2}\right]+\left[\frac{2^2}{x^2}\right]+\ldots+\left[\frac{9^2}{x^2}\right]\right)\right) \geq 1 \text { is equal to } $$ $\displaystyle \_\_\_\_$.
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.