Question

If $f\left(x\right)$ is a polynomial function satisfying $f\left(x\right)f\left(\frac{1}{x}\right)=f\left(x\right)+\left(\frac{1}{x}\right)$ and $f\left(3\right)=28$, then $f\left(4\right)=$

• A. $63$
• B. $65$
• C. $66$
• D. $27$

Answer 1

The correct option is B $65$

$\text{solution:}f\left(x\right)\cdot f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$

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

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

$\text{Now replace}x\text{'}\mathrm{by}\frac{1}{x}$

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

$\text{Multiplying equation (1) and (2)}$

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

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

Let $f\left(x\right)-1=g\left(x\right)$

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

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

$\Rightarrow g\left(x\right)\cdot g\left(\frac{1}{x}\right)=1$4 (using equation 3)

so we can say that $g\left(x\right)$ must be a

function of type $x^n$,

$g\left(x\right)=x^n$

so $f\left(x\right)-1=x^n$

$\Rightarrow f\left(x\right)=x^n+1$

we know that $f\left(3\right)=28$

$\Rightarrow f\left(3\right)=3^n+1$

$\Rightarrow 28=3^n+1$

$\Rightarrow n=3$

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

$f\left(4\right)=4^3+1=65$

Ansuer: option (B)