JEE Main 12 Jan 2019 Evening Question 23
Question: $ \underset{n\to \infty }{\mathop{\lim }},( \frac{n}{n^{2}+1^{2}}+\frac{n}{n^{2}+2^{2}}+\frac{n}{n^{2}+3^{2}}+….+\frac{1}{5n} ) $ is equal to [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ {{\tan }^{-1}}(2) $
B) $ \frac{\pi }{2} $
C) $ {{\tan }^{-1}}(3) $
D) $ \frac{\pi }{4} $
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Answer:
Correct Answer: A
Solution:
- Given, $ \underset{n\to \infty }{\mathop{\lim }},( \frac{n}{n^{2}+1^{2}}+\frac{n}{n^{2}+2^{2}}+…+\frac{1}{5n} ) $
$ \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( 1+\frac{r^{2}}{n^{2}} )}} $
$ =\int\limits_0^{1}{( \frac{1}{1+x^{2}} )dx} $
$[Putting\frac{r}{n}=x, $ we get $ \frac{1}{n}dx=dx $ and $ r=1\Rightarrow x=0,r=2n\Rightarrow x=2] $ $ =[ {{\tan }^{-1}}x ]_0^{2}={{\tan }^{-1}}2 $
