Question

For $x \in\left(0, \frac{\pi}{2}\right)$, let $f_n(x)=\int n \sin 2 x\left(\sin ^{2 n-2} x-\cos ^{2 n-2} x\right) d x-\frac{1}{2^{n-1}}, n \in N$ and $f_n\left(\frac{\pi}{4}\right)=\frac{1}{2^{n-1}}$.
If $g(x)=\operatorname{Lim}_{n \rightarrow \infty} \sum_{n=1}^n f_n(x)$ then minimum value of $g(x)$, is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

We have, $f_n(x)=\int n \sin 2 x\left(\sin ^{2 n-2} x-\cos ^{2 n-2} x\right) d x-\frac{1}{2^{n-1}}$
$=2 n \int\left(\cos x \cdot \sin ^{2 n-1} x-\sin x \cos ^{2 n-1} x\right) d x-\frac{1}{2^{n-1}}$
$=2 n\left(\frac{\sin ^{2 n} x}{2 n}+\frac{\cos ^{2 n} x}{2 n}\right)-\frac{1}{2^{n-1}}+C=\sin ^{2 n} x+\cos ^{2 n} x-\frac{1}{2^{n-1}}+C$
$\text { As, } f _{ n }\left(\frac{\pi}{4}\right)=\frac{1}{2^{ n -1}} \Rightarrow \frac{1}{2^{ n }}+\frac{1}{2^{ n }}-\frac{1}{2^{ n -1}}+ C =\frac{1}{2^{ n -1}} \Rightarrow C =\frac{1}{2^{ n -1}} $
$\therefore f _{ n }( x )=\sin ^{2 n } x +\cos ^{2 n } x$
We have, $g(x)=\underset{n \rightarrow \infty}{\text{Lim}} \left(\displaystyle\sum_{n=1}^n\left(\sin ^{2 n} x+\cos ^{2 n} x\right)\right)$
$=\left(\sin ^2 x+\sin ^4 x+\sin ^6 x+\ldots \ldots \ldots\right)+\left(\cos ^2 x+\cos ^4 x+\cos ^6 x+\ldots\right. $
$=\frac{\sin ^2 x}{1-\sin ^2 x}+\frac{\cos ^2 x}{1-\cos ^2 x}=\tan ^2 x+\cot ^2 x=(\tan x-\cot x)^2+2$
Hence minimum value of $g(x)$ is 2 .