(THPT Võ Nguyên Giáp – Quảng Bình – 2022) Cho hàm số \(y = f\left( x \right)\) liên tục trên \(\mathbb{R}\) và \(f’\left( x \right) = – {x^3} + 6{x^2} – 32\). Khi đó hàm số \(g\left( x \right) = f\left( {{x^2} – 3x} \right)\) nghịch biến trên khoảng

A. \(\left( { – \infty ; + \infty } \right)\).

B. \(\left( {1; + \infty } \right)\).

C. \(\left( {2; + \infty } \right)\).

D. \(\left( { – \infty ;1} \right)\).

Lời giải:

Chọn C

Ta có:

+) \(f'(t) = – (t + 2){(t – 4)^2}\).

+) \(g'(x) = {\left[ {f({x^2} – 3x)} \right]^\prime } = ({x^2} – 3x)’.f'({x^2} – 3x) = (2x – 3).f'({x^2} – 3x)\).

Suy ra:

*) \(g'(x) = 0 \Leftrightarrow (2x – 3).f'({x^2} – 3x) = 0 \Leftrightarrow \left[ \begin{array}{l}2x – 3 = 0\\f'({x^2} – 3x) = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}2x = 3\\{x^2} – 3x = – 2\\{x^2} – 3x = 4\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{3}{2}\\x = \pm 1\\x = 2\\x = 4\end{array} \right.\)

*) \(g'(x) < 0 \Leftrightarrow (2x – 3).f'({x^2} – 3x) < 0 \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}2x – 3 < 0\\f'({x^2} – 3x) > 0\end{array} \right.\\\left\{ \begin{array}{l}2x – 3 > 0\\f'({x^2} – 3x) < 0\end{array} \right.\end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x < \frac{3}{2}\\{x^2} – 3x < – 2\end{array} \right.\\\left\{ \begin{array}{l}x > \frac{3}{2}\\{x^2} – 3x > – 2\end{array} \right.\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x < \frac{3}{2}\\1 < x < 2\end{array} \right.\\\left\{ \begin{array}{l}x > \frac{3}{2}\\\left[ \begin{array}{l}x < 1\\x > 2\end{array} \right.\end{array} \right.\end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}1 < x < \frac{3}{2}\\x > 2\end{array} \right.\).