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