Question

If $f\left(x\right)$ is a polynomial and satisfies the equations $f\left(x\right)(f\left(\frac{1}{x}\right)-1)-f\left(\frac{1}{x}\right)=0$ 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. $0$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (A)

$0$

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

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

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

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

At $x=1$

$f\left(1\right)f\left(1\right)=f\left(1\right)+f\left(1\right)$$f\left(1\right)=0,f\left(1\right)=2$

Minimum value of $f\left(x\right)+f\left(\frac{1}{x}\right)=f\left(1\right)+f\left(1\right)=2\times 0=0$

For $x>0$

$f\left(x\right)+f\left(\frac{1}{x}\right)$ will be minimum at $x=1$ and minimum value will be $0$