Let $[t]$ be the greatest integer less than or equal to $t$. Then the least value of $p \in N$ for which $\lim _{x \rightarrow 0^{+}}\left(x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots \ldots+\left[\frac{p}{x}\right]\right)-x^{2}\left(\left[\frac{1}{x^{2}}\right]+\left[\frac{2^{2}}{x^{2}}\right]+\ldots .+\left[\frac{9^{2}}{x^{2}}\right]\right)\right) \geq 1$ is equal to ______.

Limits XII — JEE Main · 2025 · 29 Jan 2025 · Shift 1 Question with Solution

Let

[t]

be the greatest integer less than or equal to

t

. Then the least value of

p \in N

for which

lim_{x \to 0^{+}} (x ([\frac{1}{x}] + [\frac{2}{x}] + \dots\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 ______.

Answer: 24

x \to 0^{+}

x [k / x] = k - x {k / x} \to k

and

x^{2} [k^{2} / x^{2}] = k^{2} - x^{2} {k^{2} / x^{2}} \to k^{2}

, since the fractional-part terms vanish. So the limit equals

k = 1 \sum p k - k = 1 \sum 9 k^{2} = \frac{p ( p + 1 )}{2} - 285.

The condition is

\frac{p ( p + 1 )}{2} - 285 \geq 1 ⟹ p (p + 1) \geq 572.

Checking:

23 \times 24 = 552 < 572

and

24 \times 25 = 600 \geq 572

.

The least such

p

is 24.

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Integer answer