A) \[\tan 1\]
B) \[\frac{1}{2}\tan 1\]
C) \[\frac{1}{2}\sec 1\]
D) \[\frac{1}{2}\text{cosec}1\]
Correct Answer: B
Solution :
\[\underset{n\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{{{n}^{2}}}{{\sec }^{2}}\frac{1}{{{n}^{2}}}+\frac{2}{{{n}^{2}}}{{\sec }^{2}}\frac{4}{{{n}^{2}}}+\frac{3}{{{n}^{2}}}{{\sec }^{2}}\frac{9}{{{n}^{2}}}+.....+\frac{1}{n}{{\sec }^{2}}1 \right]\]is equal to \[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{\frac{r}{{{n}^{2}}}{{\sec }^{2}}\frac{{{r}^{2}}}{{{n}^{2}}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{r=1}^{n}{\frac{r}{n}{{\sec }^{2}}\frac{{{r}^{2}}}{{{n}^{2}}}}\] Given limit is equal to the value of integral \[\int_{0}^{1}{x{{\sec }^{2}}{{x}^{2}}dx}\] =\[\frac{1}{2}\int_{0}^{1}{2x{{\sec }^{2}}{{x}^{2}}dx}=\frac{1}{2}\int_{0}^{1}{{{\sec }^{2}}t\ dt}\], [Put \[{{x}^{2}}=t\]] \[=\frac{1}{2}[\tan \ t]_{0}^{1}=\frac{1}{2}\tan 1\].
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