Question

If $f\left(x\right)$ is a polynomial function such that $f\left(x\right)f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$ and $f\left(3\right)=-80$ then $f\left(x\right)$ is equal to:

• A. $x^4+1$
• B. $x^4-1$
• C. $1-x^4$
• D. $-1-x^4$

Answer 1

Answer: (C)

$1-x^4$

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

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

$\Rightarrow$ $f\left(x\right)=\frac{f\left(1/x\right)}{f\left(1/x\right)-1}$ (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 $f\left(3\right)=-80$

$\Rightarrow$ $\pm 3^n+1=-80$ $\Rightarrow$ $3^n=81$

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

$\Rightarrow$ $n=4$

So, $f\left(x\right)=-x^4+1$