Simplify Functions-1!

In response to challenge master note and Sandeep Bhardwaj Sir

First of all, rearranging the given condition we can also write:

$\Large f(\frac{1}{x})=\Large \frac{f(x)}{f(x)-1}$ ..................... (1)

Let $g(x)=f(x)-1$ be a polynomial function then we have

$g(x)=f(x)-1$ and $g(\frac{1}{x})=f(\frac{1}{x})-1$

Rewriting equation (1) in terms of $g(x)$ we have,

$\Large g(\frac{1}{x})+1= \Large \frac{g(x)+1}{g(x)}$

$\implies \Large g(\frac{1}{x})+1=1+ \Large \frac{1}{g(x)}$

$\implies \Large g(\frac{1}{x})=\Large \frac{1}{g(x)}$

Now $g(x)$ is a polynomial and the above condition can be true only when the polynomial has a single variable term(without any coefficient). Because if there are multiple variable terms in $g(x)$ having coefficients other than 1 attached to them then the reciprocal of the polynomial can never be equal to the actual polynomial for all values of $x$ in the domain of $g(x)$ . Hence the polynomial should be of the form:

$g(x)=\pm x^n$

$\implies f(x)-1=\pm x^n$

$\implies f(x)= 1\pm x^n$

Now see Tanishq Varshney's solution below for complete answer.

Any function satisfying

$f(x).f(\frac{1}{x})=f(x)+f(\frac{1}{x})$ is $f(x)=1 \pm x^{n}$

given $f(10)=1001$

$1+10^{n}=1001$

$n=3$

$f(x)=1+x^{3}$

$f(20)=1+(20)^{3}$

$f(20)=8001$