If a = _ n → ∞ ∑ _ k = 1 ^ n 2n n ^ 2 + k ^ 2 and f(x) = (1 - x)/(1...

Given, $a = \lim_{n \to \infty}\mathop{∑}_{k = 1}^{n}\dfrac{2n}{{n}^{2} + {k}^{2}}$

$⇒ a = \dfrac{1}{n}\mathop{∑}_{k = 1}^{n}\dfrac{2}{1 + {\left(\dfrac{k}{n}\right)}^{2}} = {∫}_{0}^{1}\dfrac{2}{1 + {x}^{2}}dx$

$⇒ a = 2{\left[\tan^{- 1}x\right]}_{0}^{1}$

$⇒ a = 2\left[\dfrac{\pi}{4} - 0\right] = \dfrac{\pi}{2}$

Now, $f\left(x\right) = \sqrt{\dfrac{1 - \cos x}{1 + \cos x}}$

$⇒ f\left(x\right) = \tan \left(\dfrac{x}{2}\right); x \in \left(0, 1\right)$

So, $f\left(\dfrac{a}{2}\right) = f\left(\dfrac{\pi}{4}\right)$

$⇒ f\left(\dfrac{\pi}{4}\right) = \sqrt{2} - 1$

And ${f}^{'}\left(\dfrac{\pi}{4}\right) = \dfrac{1}{2}\sec^{2}\left(\dfrac{\pi}{8}\right) = \dfrac{\sqrt{2}}{\sqrt{2} + 1}$

So, ${f}^{'}\left(\dfrac{\pi}{4}\right) = \sqrt{2}f\left(\dfrac{\pi}{4}\right)$