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Câu hỏi: Cho hàm số y = f(x). Đồ thị của hàm số $g\left( 2 \right)<g\left( -1 \right)<g\left( 0 \right)$ như hình bên. Đặt $g\left( x \right)={{x}^{3}}-3f\left( x \right)$. Mệnh đề nào dưới đây đúng?
A. $g\left( 0 \right)<g\left( -1 \right)<g\left( 2 \right)$.
B. $g\left( 2 \right)<g\left( -1 \right)<g\left( 0 \right)$.
C. $g\left( 2 \right)<g\left( 0 \right)<g\left( -1 \right)$.
D. $g\left( -1 \right)<g\left( 0 \right)<g\left( 2 \right)$.
A. $g\left( 0 \right)<g\left( -1 \right)<g\left( 2 \right)$.
B. $g\left( 2 \right)<g\left( -1 \right)<g\left( 0 \right)$.
C. $g\left( 2 \right)<g\left( 0 \right)<g\left( -1 \right)$.
D. $g\left( -1 \right)<g\left( 0 \right)<g\left( 2 \right)$.
$3{{S}_{1}}=3\int\limits_{-1}^{0}{\left( {{x}^{2}}-f'\left( x \right) \right)dx}=\left( {{x}^{3}}-3f\left( x \right) \right)\left| \begin{aligned}
& 0 \\
& -1 \\
\end{aligned} \right.=g\left( 0 \right)-g\left( -1 \right)>0\Rightarrow g\left( 0 \right)>g\left( -1 \right)$.
$3{{S}_{2}}=3\int\limits_{0}^{2}{\left( f'\left( x \right)-{{x}^{2}} \right)dx}=\left( 3f\left( x \right)-{{x}^{3}} \right)\left| \begin{aligned}
& 2 \\
& 0 \\
\end{aligned} \right.=g\left( 0 \right)-g\left( 2 \right)>0\Rightarrow g\left( 0 \right)>g\left( 2 \right)$.
Mà S1 < S2 nên $g\left( 0 \right)-g\left( -1 \right)<g\left( 0 \right)-g\left( 2 \right)\Leftrightarrow g\left( -1 \right)>g\left( 2 \right)$.
Vậy $g\left( 2 \right)<g\left( -1 \right)<g\left( 0 \right)$.
& 0 \\
& -1 \\
\end{aligned} \right.=g\left( 0 \right)-g\left( -1 \right)>0\Rightarrow g\left( 0 \right)>g\left( -1 \right)$.
$3{{S}_{2}}=3\int\limits_{0}^{2}{\left( f'\left( x \right)-{{x}^{2}} \right)dx}=\left( 3f\left( x \right)-{{x}^{3}} \right)\left| \begin{aligned}
& 2 \\
& 0 \\
\end{aligned} \right.=g\left( 0 \right)-g\left( 2 \right)>0\Rightarrow g\left( 0 \right)>g\left( 2 \right)$.
Mà S1 < S2 nên $g\left( 0 \right)-g\left( -1 \right)<g\left( 0 \right)-g\left( 2 \right)\Leftrightarrow g\left( -1 \right)>g\left( 2 \right)$.
Vậy $g\left( 2 \right)<g\left( -1 \right)<g\left( 0 \right)$.
Đáp án B.