Cho hàm số \(f\left( x \right) = {\ln ^2}\left( {{...

C

- Hướng dẫn giải

Tập xác định: $D = \mathbb{R}$

$f'\left( x \right) = \frac{{4x - 4}}{{{x^2} - 2x + 4}}\ln \left( {{x^2} - 2x + 4} \right)$

$f'\left( x \right) > 0 \Leftrightarrow \left( {4x - 4} \right)\ln \left( {{x^2} - 2x + 4} \right) > 0 \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x - 1 > 0\\\ln \left( {{x^2} - 2x + 4} \right) > 0\end{array} \right.\\\left\{ \begin{array}{l}x - 1 < 0\\\ln \left( {{x^2} - 2x + 4} \right) < 0\end{array} \right.\end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x > 1\\{x^2} - 2x + 4 > 1\end{array} \right.\\\left\{ \begin{array}{l}x < 1\\{x^2} - 2x + 4 < 1\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x > 1\\{x^2} - 2x + 3 > 0\end{array} \right.\\\left\{ \begin{array}{l}x < 1\\{x^2} - 2x + 3 < 0\end{array} \right.\left( {VN} \right)\end{array} \right. \Leftrightarrow x > 1$