Question

If $f\left(x\right)$ is a polynomial and satisfies the equation $f\left(x\right)(f\left(\frac{1}{x}\right)-1)-f\left(\frac{1}{x}\right)=0$ and $f\left(3\right)=10$ then the minimum value of $f\left(x\right)+f\left(\frac{1}{x}\right),(x>0)$, is

• A. $3$
• B. $4$
• C. $0$
• D. $2$

Answer 1

Answer: (B)

$4$

$f\left(x\right)(f\left(\frac{1}{x}\right)-1)-f\left(\frac{1}{x}\right)=0$

$f\left(x\right)f\left(\frac{1}{x}\right)-f\left(x\right)-f\left(\frac{1}{x}\right)=0$

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

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

$f\left(3\right)=10$

$f\left(\frac{1}{3}\right)=\frac{10}{10-1}=\frac{10}{9}$

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

Minimum value is when $x=1$

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

$f\left(1\right)=2$

$\therefore f\left(1\right)+f\left(1\right)=2+2=4$