Let [t] be the greatest integer less than or equal to t. Then the least value of ( p ∈ N ) for which ( _ x arrow 0^ + ( x( [(1)/(x)]+[(2)/(x)]+ +[ p x ])- x^2([(1)/(x^2)]+[(2^2)/(x^2)]+ +[(9^2)/(x^2)])) ≥ 1) is equal to

Correct answer: 0 aligned & _ x arrow 0^ + (x([(1)/(x)]+[(2)/(x)]+ . .+[(p)/(x)])-x^2([(1)/(x^2)]+[(2^2)/(x^2)]+[(9^2)/(x^2)])) ≥ 1 & (1+2+ +p)-(1^2+2^2+ 9^2) ≥ 1 aligned aligned & (p(p+1))/(2)-(9.10 .19)/(6) ≥ 1 & p(p+1) ≥ 572 aligned Least natural value of p is 24

JEE Main 202529 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2025 · 29 January · Morning Shift · Mathematics · Limits

Let [t] be the greatest integer less than or equal to t. Then the...

Let [t] be the greatest integer less than or equal to t. Then the least value of (

\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) is equal to ________.

Correct answer

Step-by-step solution

\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} & \frac{p(p+1)}{2}-\frac{9.10 .19}{6} \geq 1 \\ & p(p+1) \geq 572 \end{aligned}

Least natural value of

p$ is 24

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