A) \[{{\tan }^{-1}}(2)\]
B) \[\frac{\pi }{2}\]
C) \[{{\tan }^{-1}}(3)\]
D) \[\frac{\pi }{4}\]
Correct Answer: A
Solution :
Given, \[\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{n}{{{n}^{2}}+{{1}^{2}}}+\frac{n}{{{n}^{2}}+{{2}^{2}}}+...+\frac{1}{5n} \right)\] \[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{2n}{\frac{n}{{{n}^{2}}+{{r}^{2}}}=}\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{2n}{\frac{1}{n\left( 1+\frac{{{r}^{2}}}{{{n}^{2}}} \right)}}\] \[=\int\limits_{0}^{1}{\left( \frac{1}{1+{{x}^{2}}} \right)dx}[Putting\frac{r}{n}=x,\]we get\[\frac{1}{n}dx=dx\]and\[r=1\Rightarrow x=0,r=2n\Rightarrow x=2]\] \[=\left[ {{\tan }^{-1}}x \right]_{0}^{2}={{\tan }^{-1}}2\]
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