JEE MAINS 2025 Session 1 and 2 (January and April) Questions and Solutions

JEE MAINS 2025 January Session

 

Question: 1

The following is a medium-level question from JEE Mains 2025, but don’t be misled by its simplicity — it’s deeply conceptual and relies on the fundamental definition of the Greatest Integer Function (GIF). Interestingly, the structure of this problem closely resembles that of the Sandwich Theorem in limits and also requires a similar logical approach for its solution.

JEE Mains January 29, 2025; Shift-1

The minimum value of p \in N such that

\lim\limits_{x \to 0} \left(x \left(\left[ \frac {1}{x} \right] +\left[ \frac {2}{x} \right]+...+\left[ \frac {p}{x} \right]\right)\right)

-x^{2} \left(\left(\left[ \frac {1^{2}}{x^{2}} \right] +\left[ \frac {2^{2}}{x^{2}} \right]+...+\left[ \frac {9^{2}}{x^{2}} \right]\right)\right)\ge 1

 is equal to (where [t]   is greatest integer less than or equal to t ).

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