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

(C) $\sqrt{2} \mathrm{f}\left(\frac{\mathrm{a}}{2}\right)=\mathrm{f}^{\prime}\left(\frac{\mathrm{a}}{2}\right)$

(A) $2 \sqrt{2} \mathrm{f}\left(\frac{\mathrm{a}}{2}\right)=\mathrm{f}^{\prime}\left(\frac{\mathrm{a}}{2}\right)$

(B) $f\left(\frac{a}{2}\right) f^{\prime}\left(\frac{a}{2}\right)=\sqrt{2}$

(D) $\mathrm{f}\left(\frac{\mathrm{a}}{2}\right)=\sqrt{2} \mathrm{f}^{\prime}\left(\frac{\mathrm{a}}{2}\right)$

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Definite Integrals

a = lim_{n \to \infty} \sum_{k = 1}^{n} \frac{2 n}{n ^{2} + k ^{2}}

and

f (x) = \frac{1 - c o s x}{1 + c o s x}, x \in (0, 1)

, then :

Answer: (C)

The limit is a Riemann sum:

a = n \to \infty lim \frac{1}{n} k = 1 \sum n \frac{2}{1 + ( k / n ) ^{2}} = \int_{0}^{1} \frac{2}{1 + x ^{2}} d x = 2 tan^{- 1} (1) = \frac{π}{2}

For

f (x) = \frac{1 - c o s x}{1 + c o s x} = \frac{2 s i n ^{2} ( x /2 )}{2 c o s ^{2} ( x /2 )} = tan (x /2)

.

Thus

f (π /4) = tan (π /8)

and

f^{'} (x) = \frac{1}{2} sec^{2} (x /2)

, so

f^{'} (π /4) = \frac{1}{2} sec^{2} (π /8)

.

Using

sec^{2} (π /8) = 1 + tan^{2} (π /8)

and the identity

tan (π /8) = 2 - 1

f^{'} (π /4) = \frac{1}{2} (1 + (2 - 1)^{2}) = \frac{1}{2} (2 (2 - 2)) = 2 - 2

And

2 f (π /4) = 2 (2 - 1) = 2 - 2

.

Therefore

2 f (a /2) = f^{'} (a /2)

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option