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JEE MAIN 2025
29-01-2025 SHIFT-1
Question
Let [t] be the greatest integer less than or equal to t. Then the least value of $p\in \mathbb{N}$ for which $\mathop {\lim }\limits_{x \to {0^ + }} \left( {x\left( {\left[ {\frac{1}{x}} \right] + \left[ {\frac{2}{x}} \right] + \ldots + \left[ {\frac{{\rm{p}}}{x}} \right]} \right) - {x^2}\left( {\left[ {\frac{1}{{{x^2}}}} \right] + \left[ {\frac{{{2^2}}}{{{x^2}}}} \right] + \ldots + \left[ {\frac{{{9^2}}}{{{x^2}}}} \right]} \right)} \right) \ge 1$ is equal to ________.
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Solution
$\mathop {\lim }\limits_{x \to {0^ + }} \left( {x\left( {\left[ {\frac{1}{x}} \right] + \left[ {\frac{2}{x}} \right] + \ldots . + \left[ {\frac{p}{x}} \right]} \right) - {x^2}\left( {\left[ {\frac{1}{{{x^2}}}} \right] + \left[ {\frac{{{2^2}}}{{{x^2}}}} \right] + \left[ {\frac{{{9^2}}}{{{x^2}}}} \right]} \right)} \right) \ge 1$
$\quad (1 + 2 + \ldots \ldots + p) - \left( {{1^2} + {2^2} + \ldots {{.9}^2}} \right) \ge 1$
$\frac{{p(p + 1)}}{2} - \frac{{9.10.19}}{6} \ge 1$
$p(p + 1) \ge 572$
Least natural value of p is 24
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