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
Answer & explanation
Correct answer: option 2
$f(x)= \pm x^n+1, n \in N$
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.