Let f(x) be a polynomial function satisfying $f(x)+f\left(\frac{1}{x}\right)=f(x) f\left(\frac{1}{x}\right)$ for all $x \neq 0$. If f(5) = 126 and a, b, c are in G.P., then f'(a), f'(b), f'(c) are in

GP

$f(x)= \pm x^n+1, n \in N$

We have, $f(5)=126>0$. Therefore,

$f(x)=x^n+1$

$\Rightarrow f(5)=5^n+1 $

$\Rightarrow 126=5^n+1 \Rightarrow 5^n=125 \Rightarrow 5^n=5^3 \Rightarrow n=3$

∴  $f(x)=x^3+1$

$\Rightarrow f'(a)=3 x^2$  for all  $x \neq 0$

$\Rightarrow f'(a)=3 a^2, f'(b)=3 b^2$  and  $f'(c)=3 c^2$

Now,

a, b, c are in G.P.

⇒ $3 a^2, 3 b^2, 3 c^2$ are in G.P. 

⇒ f'(a), f'(b), f'(c) are in G.P.