Question

The value of $\underset{n\to \infty }{lim}\frac{1}{n}\{{\sec }^2\frac{\pi }{4n}+{\sec }^2\frac{2\pi }{4n}+.....{\sec }^2\frac{n\pi }{4n}\}$ is

• A. ${\log }_{c}2$
• B. $\frac{\pi }{2}$
• C. $\frac{4}{\pi }$
• D. $e$

Answer 1

The correct option is C $\frac{4}{\pi }$

${\int }_a^{b}cf\left(x\right)dx=\underset{n\to }{lim}\sum _{i=1}^{n}cf(x_i\ast )\Delta x$
Or, ${\int }_a^{b}cf\left(x\right)dx=c\underset{n\to }{lim}\sum _{i=1}^{n}f(x_i\ast )\Delta x$

Thus Or, ${\int }_a^{b}cf\left(x\right)dx=c{\int }_a^{b}f\left(x\right)dx$

As can be seen above limit of sum tending to infinity can be defined in terms of definite integration
Similarly we can write above equation of limit as definite integral as :
${\int }_0^{1}se{c}^2\frac{\pi x}{4}dx(\frac{tan\frac{\pi x}{4}}{\frac{\pi }{4}}{)}_0^1=\frac{4}{\pi }$
Therefore Answer is $C$