Question

If $a=\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{2n}{n^2+k^2}$ and $f(x)=\sqrt{\frac{1-\cos x}{1+\cos x}},x\in (0,1)$, then :

• A. $2\sqrt{2}f\left(\frac{a}{2}\right)=f^{\prime }\left(\frac{a}{2}\right)$
• B. $f\left(\frac{a}{2}\right)f^{\prime }\left(\frac{a}{2}\right)=\sqrt{2}$
• C. $\sqrt{2}f\left(\frac{a}{2}\right)=f^{\prime }\left(\frac{a}{2}\right)$
• D. $f\left(\frac{a}{2}\right)=\sqrt{2}{f}^{\prime }\left(\frac{a}{2}\right)$

Answer 1

Answer: (C)

$a=\frac{1}{n}\sum _{k=1}^n\frac{2}{1+{\left(\frac{k}{n}\right)}^2}={\int }_0^1\frac{2}{1+x^2}dx=\frac{\pi }{2}$
$f(x)=\tan \left(\frac{x}{2}\right);x\in (0,1)$
$f\left(\frac{\pi }{4}\right)=\sqrt{2}-1$
$f^{\prime }\left(\frac{\pi }{4}\right)=\frac{1}{2}{\sec }^2\left(\frac{\pi }{8}\right)=\frac{\sqrt{2}}{\sqrt{2}+1}$
$f^{\prime }\left(\frac{\pi }{4}\right)=\sqrt{2}f\left(\frac{\pi }{4}\right)$