Đáp án đúng: A
Giải chi tiết:Ta có \(g\left( x \right) = f\left( x \right) + \dfrac{1}{3}{x^3} - x - 2\) \(\begin{array}{l} \Rightarrow g'\left( x \right) = f'\left( x \right) + {x^2} - 1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {x + 1} \right){\left( {x - 1} \right)^2}\left( {x - 2} \right) + {x^2} - 1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {{x^2} - 1} \right)\left( {\left( {x - 1} \right)\left( {x - 2} \right) + 1} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {{x^2} - 1} \right)\left( {{x^2} - 3x + 3} \right)\end{array}\).Cho \(g'\left( x \right) = 0 \Leftrightarrow x = \pm 1 \in \left[ { - 1;2} \right]\).Ta có BBT:Dựa vào BBT ta thấy \(\mathop {\min }\limits_{\left[ { - 1;2} \right]} g\left( x \right) = g\left( 1 \right) = f\left( 1 \right) - \dfrac{8}{3}\).Chọn A
