Đáp án:
$D.\ (2;4)$
Giải thích các bước giải:
$\quad g(x)= f\left(\dfrac{5x}{x^2 + 4}\right)$
$\Rightarrow g'(x)= - \dfrac{5(x^2-4)}{(x^2 + 4)^2}\cdot f'\left(\dfrac{5x}{x^2 + 4}\right)$
$\Rightarrow g'(x)= - \dfrac{5(x^2-4)}{(x^2 + 4)^2}\cdot \dfrac{5x}{x^2 + 4}\cdot \left(\dfrac{5x}{x^2 + 4}-1\right)^2\cdot \left(\dfrac{5x}{x^2 + 4}- 2\right)$
Hàm số đồng biến $\Leftrightarrow g'(x) > 0$
$\Leftrightarrow \dfrac{x^2-4}{(x^2 + 4)^2}\cdot \dfrac{x}{x^2 + 4}\cdot \left(\dfrac{5x}{x^2 + 4}-1\right)^2\cdot \left(\dfrac{5x}{x^2 + 4}- 2\right) < 0$
$\Leftrightarrow x(x^2 - 4)> 0$
$\Leftrightarrow x\in (-2;0)\cup (2;+\infty)$
