If a=limn→∞ ∑k=1n 2nn2+k2 and f(x)=1−cos⁡x1+cos⁡x x∈(0,1) then

a=1n∑k=1n 21+kn2=∫01 21+x2dx=2Tan-1x01=2π4=π2and f(x)=tan⁡x2;x∈(0,1)fπ4=2−1f'π4=12sec2⁡π8=22+1f'π4=2fπ4

If a=limn→∞ ∑k=1n 2nn2+k2 and f(x)=1−cos⁡x1+cos⁡x x∈(0,1) then

a=1n∑k=1n 21+kn2=∫01 21+x2dx=2Tan-1x01=2π4=π2and f(x)=tan⁡x2;x∈(0,1)fπ4=2−1f'π4=12sec2⁡π8=22+1f'π4=2fπ4

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

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

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

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

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

answer is C.

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Detailed Solution

$\begin{array}{l} a = \frac{1}{n} \sum_{k = 1}^{n} \frac{2}{1 + {(\frac{k}{n})}^{2}} = \int_{0}^{1} \frac{2}{1 + x^{2}} dx = 2 {(T a n^{- 1} x)}_{0}^{1} = 2 (\frac{π}{4}) = \frac{π}{2} \\ and f (x) = \tan (\frac{x}{2}); x \in (0, 1) \\ f (\frac{π}{4}) = \sqrt{2} - 1 \\ f' (\frac{π}{4}) = \frac{1}{2} \sec^{2} (\frac{π}{8}) = \frac{\sqrt{2}}{\sqrt{2} + 1} \\ f' (\frac{π}{4}) = \sqrt{2} f (\frac{π}{4}) \end{array}$