If a = lim_{n to infty} sum_{k=1}^{n} frac{2n | vedclass

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(C) First,evaluate $a$: $a = \lim_{n 	o \infty} \sum_{k=1}^{n} \frac{2}{n(1+(k/n)^2)} = \int_{0}^{1} \frac{2}{1+x^2} dx = 2[	an^{-1} x]_0^1 = 2(\fra

Vedclass · Accepted Answer

$\sqrt{2} f \left(\frac{a}{2}\right) = f'\left(\frac{a}{2}\right)$

Vedclass · Answer

(C) First,evaluate $a$: $a = \lim_{n 	o \infty} \sum_{k=1}^{n} \frac{2}{n(1+(k/n)^2)} = \int_{0}^{1} \frac{2}{1+x^2} dx = 2[	an^{-1} x]_0^1 = 2(\frac{\pi}{4}) = \frac{\pi}{2}$. Next,simplify $f(x)$: $f(x) = \sqrt{\frac{2\sin^2(x/2)}{2\cos^2(x/2)}} = 	an(x/2)$ for $x \in (0, 1)$. Then $f'(x) = \frac{1}{2} \sec^2(x/2)$. We need to evaluate at $x = a/2 = \pi/4$: $f(\pi/4) = 	an(\pi/8) = \sqrt{2}-1$. $f'(\pi/4) = \frac{1}{2} \sec^2(\pi/8) = \frac{1}{2} (1 + 	an^2(\pi/8)) = \frac{1}{2} (1 + (\sqrt{2}-1)^2) = \frac{1}{2} (1 + 2 + 1 - 2\sqrt{2}) = \frac{4-2\sqrt{2}}{2} = 2-\sqrt{2} = \sqrt{2}(\sqrt{2}-1) = \sqrt{2} f(\pi/4)$. Thus,$f'(\frac{a}{2}) = \sqrt{2} f(\frac{a}{2})$.

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

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