Verified Solution • 2025–26 Session moderate QJEEM25Ma1569: JEE Main 2025 Mathematics - Class 12 JEEM25Ma1569 Solution: Let [.] be the greatest integer less than or equal to t. Then… Let [.] be the greatest integer less than or equal to t. Then the least value of p ∈ N for which amp;limx→0+(x([1x]+[2x]+…..+[px])−x2([1x2]+[22x2]+[92x2]))≥1amp;(1+2+……+p)−(12+22+….92)≥1\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}amp;x→0+lim(x([x1]+[x2]+…..+[xp])−x2([x21]+[x222]+[x292]))≥1amp;(1+2+……+p)−(12+22+….92)≥1 is equal to _______. Step-by-Step Verified Solution amp;limx→0+(x([1x]+[2x]+…..+[px])−x2([1x2]+[22x2]+[92x2]))≥1amp;(1+2+……+p)−(12+22+….92)≥1\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}amp;x→0+lim(x([x1]+[x2]+…..+[xp])−x2([x21]+[x222]+[x292]))≥1amp;(1+2+……+p)−(12+22+….92)≥1 amp;amp;p(p+1)2−9.10.196≥1amp;p(p+1)≥572amp; Least natural value of p is 24\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}amp;amp;2p(p+1)−69.10.19≥1amp;p(p+1)≥572amp; Least natural value of p is 24 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring