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\[=\frac{1}{2}\,\tan 1\]You need to login to perform this action.
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