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JEE Mains · Maths · STD 11 - 12. limits

मान लीजिए कि [t] t से कम या उसके बराबर सबसे बड़ा पूर्णांक है। तब \(\mathrm{p} \in \mathbf{N}\) का न्यूनतम मान जिसके लिए \(\lim _{x \rightarrow 0^{+}}(x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots+\left[\frac{\mathrm{p}}{x}\right]\right)-x^2(\left[\frac{1}{x^2}\right]+\left[\frac{2^2}{x^2}\right]\) \(+\ldots+\left[\frac{9^2}{x^2}\right])) \geq 1\) = __________

  1. A 21
  2. B 22
  3. C 23
  4. D 24
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Answer & Solution

Correct Answer

(D) 24

Step-by-step Solution

Detailed explanation

\(\lim _{x \rightarrow 0^{+}}(x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots . .+\left[\frac{p}{x}\right]\right)-x^2(\left[\frac{1}{x^2}\right]+\) \(\left[\frac{2^2}{x^2}\right]+\left[\frac{9^2}{x^2}\right])) \geq 1 \)…
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