Đáp án C

Ta có \(g'\left( x \right) = f'\left( x \right) - x\).

\(\int\limits_{ - 2}^0 {g'\left( x \right)d{\rm{x}}} = g\left( 0 \right) - g\left( { - 2} \right) = \int\limits_{ - 2}^0 {\left[ {f'\left( x \right) - x} \right]d{\rm{x}}} > 0 \Rightarrow g\left( 0 \right) - g\left( { - 2} \right) > 0\).

\(\int\limits_0^1 {g'\left( x \right)d{\rm{x}}} = \int\limits_0^1 {\left[ {f'\left( x \right) - x} \right]d{\rm{x}}} < 0 \Rightarrow g\left( 1 \right) - g\left( 0 \right) < 0 \Rightarrow g\left( 1 \right) < g\left( 0 \right)\)

Mặt khác, ta có \(g\left( 0 \right) - g\left( 1 \right) = \int\limits_0^1 {\left[ {x - f'\left( x \right)} \right]d{\rm{x}}} > 0\).

Từ hình vẽ, ta có \(g\left( 0 \right) - g\left( { - 2} \right) > g\left( 0 \right) - g\left( 1 \right) \Rightarrow g\left( 1 \right) > g\left( { - 2} \right)\).

Vậy \(g\left( { - 2} \right) < g\left( 1 \right) < g\left( 0 \right)\).