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

Question

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]

Smart Achievers · Accepted Answer

(24)

We have,

$p \in N$

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

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

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

Solution

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Solution

Q. Let [t] be the greatest integer less than or equal to t. Then the least value of p∈N for which limx→0+(x([1x]+[2x]+...+[px])–x2([1x2]+[22x2]+...+[92x2]))≥1 is equal to _________. [2025]

Ans. (24) We have, limx→0+(x([1x]+[2x]+...+[px])–x2([1x2]+[22x2]+...+[92x2]))≥1 ⇒ (1+2+...+p)–(12+22+...+92)≥1 ⇒ p(p+1)2–9.10.196≥1 ⇒ p(p+1)≥572 ⇒ p≥24.

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