Mathematics · Limits, Continuity and Differentiability

JEE Main 2025 — 29 January, Morning Shift — Question 67

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

Then the least value of p∈Np \in N for which lim⁡x→0+(x([1x]+[2x]+…..+[px])−x2([1x2]+[22x2]+….+[92x2]))≥1\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]+\ldots .+\left[\frac{9^{2}}{x^{2}}\right]\right)\right) \geq 1

is equal to \qquad .

Answer: 24

Numerical answer — enter this value.

Step-by-step solution

lim⁡x→0+(x([1x]+[2x]+…..+[px])−x2([1x2]+[22x2]+..+[92x2]))≥1\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

(1+2+……+p)−(12+22+…+92)≥1(1+2+\ldots \ldots+p)-\left(1^{2}+2^{2}+\ldots +9^{2}\right) \geq 1 p(p+1)2−9.10.196≥1\frac{\mathrm{p}(\mathrm{p}+1)}{2}-\frac{9.10 .19}{6} \geq 1

p(p+1)≥572\mathrm{p}(\mathrm{p}+1) \geq 572

Least natural value of pp is 24

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Indeterminate forms & its solving methods