Question

If $f\left(x\right)$ be 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 $f\left(4\right)=65$. Then, find $f\left(6\right)$.

Answer 1

$f\left(x\right)f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$

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

$\Rightarrow$ $f\left(x\right)=\frac{f\left(1/x\right)}{f\left(1/x\right)-1}$ (i)

Also, $f\left(x\right)f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$

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

$\Rightarrow$ $f\left(\frac{1}{x}\right)=\frac{f\left(x\right)}{f\left(x\right)-1}$ (ii)

On multiplying Eqs. (i) and (ii), we get,

$f\left(x\right)f\left(\frac{1}{x}\right)=\frac{f\left(1/x\right)f\left(x\right)}{\{f\left(1/x\right)-1\}\{f\left(x\right)-1\}}$

$\Rightarrow$ $(f\left(\frac{1}{x}\right)-1)(f\left(x\right)-1)=1$ (iii)

Since, $f\left(x\right)$ is polynomial function, so $\left(f\right(x)-1)$ and $f\left(\frac{1}{x}\right)-1$ are reciprocals of each other. Also, $x$ and $\frac{1}{x}$ are reciprocals of each other.

Thus, Eq. (iii) can hold only when

$f\left(x\right)-1=\pm x^n$, where $n\in R$

$\therefore$ $f\left(x\right)=\pm x^n+1$ but $f\left(4\right)=65$

$\Rightarrow$ $\pm 4^n+1=65$ $\Rightarrow$ $4^n=64$

$\Rightarrow$ $4^n=4^3$ $(\because 4^n>0)$

$\Rightarrow$ $n=3$

So, $f\left(x\right)=x^3+1$

Hence, $f\left(6\right)=6^3+1=217$
Ans: 217