Question

If $f\left(x\right)$ is a polynomial function satisfying the condition $f\left(x\right)\times f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$ and $f\left(2\right)=9$ then

• A. $2f\left(4\right)=3f\left(6\right)$
• B. $14f\left(1\right)=f\left(3\right)$
• C. $9f\left(3\right)=2f\left(5\right)$
• D. $f\left(10\right)=f\left(11\right)$

Answer 1

Answer: (B), (C)

$9f\left(3\right)=2f\left(5\right)$
$14f\left(1\right)=f\left(3\right)$

The polynomial which satisfies $f\left(x\right)f\left(1/x\right)=f\left(x\right)+f\left(1/x\right)$ is $\pm x^n+1$ (standard result)
Given that $f\left(2\right)=9\Rightarrow \pm 2^n+1=9\Rightarrow 2^n=8$
(-ve sign not possible here)
$\Rightarrow n=3$
Hence the function is $f\left(x\right)=x^3+1$
$\Rightarrow f\left(1\right)=2,f\left(3\right)=28,f\left(5\right)=126$
$f\left(4\right)=65,f\left(6\right)=217$
Using these, we see only option B and C are correct.