JEE Mains 2025 Paper Held on January 29, 2025; Shift-1

JEE MAINS 2025 January Session Questions and Solutions

Question: 1

Dear Readers, This question is a medium level question asked in JEE Mains 2025. But it is a very conceptual question based on fundamental property of the definition of Greatest Integer Function. This question looks similar to Sandwich Theorem on limits and it also uses the concept of solving Sandwich Theorem. So, let’s see the question and its simple and easy method to solve.

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 $t$ ).

Ans: 24 [JEE Mains January 29, 2025; Shift-1]

Definition of Greatest Integer Function or Formula: $x-1 \lt [x] \le x$

we use part of this definition,

$\because$ $\left[ \frac {1}{x} \right] \le \frac {1}{x}$

$\left[ \frac {2}{x} \right] \le \frac {2}{x}$

$\left[ \frac {3}{x} \right] \le \frac {3}{x}$

$\left[ \frac {p}{x} \right] \le \frac {p}{x}$

On adding all of the above, we get

$\left[ \frac {1}{x} \right] + \left[ \frac {2}{x} \right] +...+ \left[ \frac {p}{x} \right]$ $\le \frac {1}{x}+\frac {2}{x}+...+\frac {p}{x}$

Now, multiplying by $x$ on both sides, we get

$x \left(\left[ \frac {1}{x} \right] + \left[ \frac {2}{x} \right] +...+ \left[ \frac {p}{x} \right]\right)$ $\le x\left (\frac {1}{x}+\frac {2}{x}+...+\frac {p}{x} \right)$

$\therefore$

$x \left(\left[ \frac {1}{x} \right] + \left[ \frac {2}{x} \right] +...+ \left[ \frac {p}{x} \right]\right)$ $\le \left (1+2 +3 +...+p \right)$ ……(1)

Similarly, applying for another term in the question, we get

$\left[ \frac {1^{2}}{x^{2}} \right] \le \frac {1^{2}}{x^{2}}$

$\left[ \frac {2^{2}}{x^{2}} \right] \le \frac {2^{2}}{x^{2}}$

$\left[ \frac {3^{2}}{x^{2}} \right] \le \frac {3^{2}}{x^{2}}$

$\left[ \frac {9^{2}}{x^{2}} \right] \le \frac {9^{2}}{x^{2}}$

On adding all of the above, we get

$\left[ \frac {1^{2}}{x^{2}} \right] + \left[ \frac {2^{2}}{x^{2}} \right] +...+ \left[ \frac {9^{2}}{x^{2}} \right]$ $\le \frac {1^{2}}{x^{2}}+\frac {2^{2}}{x^{2}}+...+\frac {9^{2}}{x^{2}}$

Now, multiplying by $x^{2}$ on both sides, we get

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

$\therefore$

$x^{2} \left(\left[ \frac {1^{2}}{x^{2}} \right] + \left[ \frac {2^{2}}{x^{2}} \right] +...+ \left[ \frac {9^{2}}{x^{2}} \right]\right)$ $\le \left (1^{2}+2^{2} +3^{2} +...+9^{2} \right)$ ……(2)

Now, subtracting equations (1) and (2) we get (we can subtract the two equations because it is given in the question that the difference is more than 1 on left hand side)

$x \left(\left[ \frac {1}{x} \right] + \left[ \frac {2}{x} \right] +...+ \left[ \frac {p}{x} \right]\right)$ – $x^{2} \left(\left[ \frac {1^{2}}{x^{2}} \right] + \left[ \frac {2^{2}}{x^{2}} \right] +...+ \left[ \frac {9^{2}}{x^{2}} \right]\right)$ $\le \left (1+2 +3 +...+p \right)$ – $\left (1^{2}+2^{2} +3^{2} +...+9^{2} \right)$

$\because$

$\lim\limits_{x \to 0} \left(x \left(\left[ \frac {1}{x} \right] +\left[ \frac {2}{x} \right]+...+\left[ \frac {p}{x} \right]\right)\right)$ $-x^{2} \left(\left(\left[ \frac {1^{2}}{x^{2}} \right] +\left[ \frac {2^{2}}{x^{2}} \right]+...+\left[ \frac {9^{2}}{x^{2}} \right]\right)\right)\ge 1$

$\therefore$

$\left (1+2 +3 +...+p \right)$ – $\left (1^{2}+2^{2} +3^{2} +...+9^{2} \right) \ge 1$

Now, we get

$\frac{p\times(p+1)}{2}-\frac{9\times10\times19}{2}\ge1$

On simplifying we get

$\Longrightarrow$ ${p\times(p+1)} \ge 572$

Now by hit and trial, we can easily get $p=24$ and it is the minimum value of $p$ satisfying the inequality given in the question.

So, the final answer is $p=24$ .

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