question_answer

\[\underset{n\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{{{n}^{2}}}{{\sec }^{2}}\frac{1}{{{n}^{2}}}+\frac{2}{{{n}^{2}}}{{\sec }^{2}}\frac{4}{{{n}^{2}}} \right.\] \[\left. +....+\frac{n}{{{n}^{2}}}{{\sec }^{2}}1 \right]\] equals     AIEEE  Solved  Paper-2005

A) \[\frac{1}{2}\text{ }tan\text{ }1\] 

B)                        \[tan\text{ }1\]

C)        \[\frac{1}{2}cosec\text{ }1\]        

D)        \[\frac{1}{2}sec\text{ }1\]

Correct Answer: A

Solution :

Let \[A=\underset{n\to \infty }{\mathop{\lim }}\,\,\left( \frac{1}{{{n}^{2}}}{{\sec }^{2}}\,\frac{1}{{{n}^{2}}}\,+\frac{2}{{{n}^{2}}}\,{{\sec }^{2}}\,\frac{4}{{{n}^{2}}}\,+...+\frac{n}{{{n}^{2}}}\,{{\sec }^{2}}1 \right)\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{1}{n}\,\left[ \frac{1}{n}\,{{\sec }^{2}}\,{{\left( \frac{1}{n} \right)}^{2}}+\frac{2}{n}\,{{\sec }^{2}}\,{{\left( \frac{2}{n} \right)}^{2}}+...+\,\frac{n}{n}\,{{\sec }^{2}}{{\left( \frac{n}{n} \right)}^{2}} \right]\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{1}{n}\,\sum\limits_{r=1}^{n}{\left( \frac{r}{n} \right)\,{{\sec }^{2}}\,{{\left( \frac{r}{n} \right)}^{2}}}\] \[\therefore \,\,\,\,\,\,\,\,A=\int_{0}^{1}{x\,{{\sec }^{2}}\,({{x}^{2}})dx}\] Put  \[{{x}^{2}}=t\]  \[2x\,dx=dt\]  \[xdx=\frac{dt}{2}\] \[\therefore \,\,\,A=\frac{1}{2}\,\int_{0}^{1}{{{\sec }^{2}}\,t\,\,dt}\] \[=\frac{1}{2}[\tan ]_{0}^{1}\] x\[=\frac{1}{2}\,\tan 1\]