Câu hỏi 23

Ta có \(g\left( x \right)=f\left( x-{{x}^{2}} \right)\,\,\xrightarrow{{}}\,\,{g}'\left( x \right)=\left( 1-2x \right).{f}'\left( x-{{x}^{2}} \right);\,\,\forall x\in \mathbb{R}.\)

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

\( \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}1 - 2x > 0\\1 < x - {x^2} < 2\end{array} \right.\\\left\{ \begin{array}{l}1 - 2x < 0\\x - {x^2} \in \left( { - \,\infty ;1} \right) \cup \left( {2; + \,\infty } \right)\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x < \frac{1}{2}\\{x^2} - x + 1 < 0\\{x^2} - x + 2 > 0\end{array} \right.\\\left\{ \begin{array}{l}x > \frac{1}{2}\\\left[ \begin{array}{l}{x^2} - x + 1 > 0\\{x^2} - x + 2 < 0\end{array} \right.\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x < \frac{1}{2}\\VN\\VSN\end{array} \right.\\\left\{ \begin{array}{l}x > \frac{1}{2}\\\left[ \begin{array}{l}VSN\\VN\end{array} \right.\end{array} \right.\end{array} \right. \Leftrightarrow x > \frac{1}{2}.\)

Vậy hàm số \(y=g\left( x \right)\) nghịch biến trên khoảng \(\left( \frac{1}{2};+\,\infty  \right).\)

Chọn D