Question

If $f\left(x\right)$ is a polynomial function satisfying
$f\left(x\right)f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$ and $f\left(1\right)=2$ then the number of such functions possible is/are

• A. $1$
• B. $2$
• C. more than $2$ but finite
• D. infinitely many

Answer 1

The correct option is D infinitely many

$f\left(x\right)f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)\Rightarrow f\left(x\right)f\left(\frac{1}{x}\right)-f\left(x\right)=f\left(\frac{1}{x}\right)\Rightarrow f\left(x\right)=\frac{f\left(\frac{1}{x}\right)}{f\left(\frac{1}{x}\right)-1}\cdots \left(1\right)$
$f\left(x\right)f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)\Rightarrow f\left(x\right)f\left(\frac{1}{x}\right)-f\left(\frac{1}{x}\right)=f\left(x\right)\Rightarrow f\left(\frac{1}{x}\right)=\frac{f\left(x\right)}{f\left(x\right)-1}\cdots \left(2\right)$

Multiplying $1$ and $2$,
$f\left(x\right)\cdot f\left(\frac{1}{x}\right)=\frac{f\left(x\right)\cdot f\left(\frac{1}{x}\right)}{\left(f\right(x)-1)(f\left(\frac{1}{x}\right)-1)}$
$\Rightarrow \left(f\right(x)-1)(f\left(\frac{1}{x}\right)-1)=1$

As $x$ and $\frac{1}{x}$ are reciprocal of each other.
So, the possible function will be,
$f\left(x\right)-1=\pm x^n\Rightarrow f\left(x\right)=1\pm x^n\Rightarrow f\left(1\right)=1\pm 1^n=2\Rightarrow 1^n=1$

Hence, there are infinite number of possible functions.