A) \[\frac{4}{\pi }\]
B) \[\frac{2}{\pi }\]
C) \[\frac{3}{\pi }\]
D) \[\frac{5}{\pi }\]
Correct Answer: A
Solution :
\[\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\left[ {{\sec }^{2}}\frac{\pi }{4n}+{{\sec }^{2}}\frac{2\pi }{4n}+...+{{\sec }^{2}}\frac{n\pi }{4n} \right]\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{r=1}^{n}{{{\sec }^{2}}}\frac{r\pi }{4n}\] \[=\int_{0}^{1}{{{\sec }^{2}}\frac{\pi x}{4}dx}\] \[=\frac{4}{\pi }\left[ \tan \frac{\pi x}{4} \right]_{0}^{1}=\frac{4}{\pi }\]
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