Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

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