Question

If $a = \lim_{{n \to \infty}} \sum_{{k=1}}^{n} \frac{2n}{{n^2 + k^2}}$and
$f(x) = \sqrt{\frac{1 - \cos x}{1 + \cos x}}, \quad x \in (0, 1)$ then

• A. $2\sqrt2f(\frac{a}{2})=f^′(\frac{a}{2})$
• B. $f(\frac{a}{2})f^′(\frac{a}{2})=\sqrt2$
• C. $\sqrt2f(\frac{a}{2})=f^′(\frac{a}{2})$
• D. $f(\frac{a}{2})=\sqrt2f^′(\frac{a}{2})$

Answer 1

Answer: (C)

$\sqrt2f(\frac{a}{2})=f^′(\frac{a}{2})$

Answer 2

The correct answer is (C) : $\sqrt2f(\frac{a}{2})=f^′(\frac{a}{2})$
$a = \lim_{{n \to \infty}} \sum_{{k=1}}^{n} \frac{2n}{{n^2 + k^2}}$
$a = \lim_{{n \to \infty}} \frac{1}{n} \sum_{{k=1}}^{n} \frac{2}{{1 + \left(\frac{k}{n}\right)^2}}$
$a = \int_{0}^{1} \frac{2}{1 + x^2} \,dx$ $=2\tan^{-1}(x) \int_{0}^{1} = \frac{\pi}{2}$
$f(x) = \sqrt{\frac{1 - \cos x}{1 + \cos x}}, \quad x \in (0, 1)$
$f(x)=\frac{1−\cos⁡x}{sin⁡x}=cosec x−cot⁡x$
$f'(x)=\cosec^2x−\cosec x\cot⁡x$
\begin{aligned} f\left(\frac{a}{2}\right) &= f\left(\frac{\pi}{4}\right) = \sqrt{2} - 1 \\\ f'\left(\frac{a}{2}\right) &= f'\left(\frac{\pi}{4}\right) = 2 - \sqrt{2} \end{aligned} \left. \begin{aligned} f'\left(\frac{a}{2}\right) &= \sqrt{2} \cdot f\left(\frac{a}{2}\right) \end{aligned} \right\\}