Let [t] be the greatest integer less than or equal to t. Then the least value of for which is equal to _________. [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let [t] be the greatest integer less than or equal to t. Then the least value of $p \in N$ for which $\lim_{x \to 0^{+}} (x ([\frac{1}{x}] + [\frac{2}{x}] + . . . + [\frac{p}{x}]) - x^{2} ([\frac{1}{x^{2}}] + [\frac{2^{2}}{x^{2}}] + . . . + [\frac{9^{2}}{x^{2}}])) \geq 1$ is equal to _________. [2025]

Ans.
(24)

We have,

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

$\Rightarrow (1 + 2 + . . . + p) - (1^{2} + 2^{2} + . . . + 9^{2}) \geq 1$

$\Rightarrow \frac{p (p + 1)}{2} - \frac{9.10.19}{6} \geq 1$

$\Rightarrow p (p + 1) \geq 572 \Rightarrow p \geq 24$ .

Want Us to Email you About Special Offers & Updates?