JEE Main 2025 Jan 29 Shift 1, Mathematics Q23: Methods to Evaluate Limits

JEE Main2025Jan 29, Shift 1Mathematics

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

$x \to 0^{+} lim (x ([\frac{1}{x}] + [\frac{2}{x}] + \dots + [\frac{p}{x}]) - x^{2} ([\frac{1}{x ^{2}}] + [\frac{2 ^{2}}{x ^{2}}] + \dots + [\frac{9 ^{2}}{x ^{2}}])) \geq 1$

is equal to ______.

Your answer

Solution

Use the standard bound $t - 1 < [t] \leq t$ , which gives $x [k / x] \to k$ and $x^{2} [k^{2} / x^{2}] \to k^{2}$ as $x \to 0^{+} .$

So $x \sum_{k = 1}^{p} [k / x] \to \sum_{k = 1}^{p} k = \frac{p ( p + 1 )}{2}$ and $x^{2} \sum_{k = 1}^{9} [k^{2} / x^{2}] \to \sum_{k = 1}^{9} k^{2} = \frac{9 \cdot 10 \cdot 19}{6} = 285.$

The limit equals $\frac{p ( p + 1 )}{2} - 285.$ Require $\frac{p ( p + 1 )}{2} - 285 \geq 1$ , i.e. $p (p + 1) \geq 572.$

Test $p = 23$ : $23 \cdot 24 = 552 < 572.$ Test $p = 24$ : $24 \cdot 25 = 600 \geq 572.$

Least natural value: $p = 24.$

Concept behind this question

Methods to Evaluate LimitsNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 29 Shift 1, Mathematics Q23: Methods to Evaluate Limits

Concept behind this question

More previous year questions on Methods to Evaluate Limits

Practice more Mathematics