Let f be a differentiable function satisfying $[f(x)]^n=f(n x)$ for all $x \in R$. Then, $f'(x) f(n x)$
Answer & explanation
Correct answer: option 3
We have,
$[f(x)]^n=f(n x)$ for all x
$\Rightarrow n[f(x)]^{n-1} f'(x)=n f'(n x)$
$\Rightarrow n[f(x)]^n f'(x)=n f(x) f'(n x)$ [Multiplying both sides by f(x)]
$\Rightarrow n f(n x) f'(x)=n f(x) f'(n x)$ [∵ [f(x)]n = f(nx)]
$\Rightarrow f(n x) f'(x)=f(x) f'(n x)$