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Câu hỏi: Cho hàm số $y=f\left( x \right)$ xác định và liên tục trên $\mathbb{R}$. Đồ thị của hàm số $y={f}'\left( x \right)$ như hình bên dưới.
Đặt $g\left( x \right)=f\left( x \right)-\dfrac{{{x}^{2}}}{2}-x+2022$. Mệnh đề nào dưới đây đúng?
A. $g\left( 2 \right)>g\left( -3 \right)>g\left( 0 \right)$.
B. $g\left( -3 \right)>g\left( 0 \right)>g\left( 2 \right)$.
C. $g\left( 2 \right)>g\left( 0 \right)>g\left( -3 \right)$.
D. $g\left( 0 \right)>g\left( 2 \right)>g\left( -3 \right)$.
$g\left( x \right)=f\left( x \right)-\dfrac{{{x}^{2}}}{2}-x+2022\Rightarrow {g}'\left( x \right)={f}'\left( x \right)-\left( x+1 \right)$.
${g}'\left( x \right)=0\Leftrightarrow {f}'\left( x \right)=x+1\Leftrightarrow \left[ \begin{aligned}
& x=-3 \\
& x=0 \\
& x=2 \\
\end{aligned} \right.$
Xét $\int\limits_{-3}^{0}{{g}'\left( x \right)}\text{d}x=\int\limits_{-3}^{0}{\left[ {f}'\left( x \right)-\left( x+1 \right) \right]}\text{d}x>0$ $\Leftrightarrow g\left( 0 \right)-g\left( -3 \right)>0\Leftrightarrow g\left( 0 \right)>g\left( -3 \right)$.
Tương tự, xét $\int\limits_{0}^{2}{{g}'\left( x \right)}\text{d}x=\int\limits_{0}^{2}{\left[ {f}'\left( x \right)-\left( x+1 \right) \right]}\text{d}x<0$ $\Leftrightarrow g\left( 2 \right)-g\left( 0 \right)<0\Leftrightarrow g\left( 2 \right)<g\left( 0 \right)$.
Xét $\int\limits_{-3}^{2}{{g}'\left( x \right)}\text{d}x=\int\limits_{-3}^{0}{\left[ {f}'\left( x \right)-\left( x+1 \right) \right]}\text{d}x+\int\limits_{0}^{2}{\left[ {f}'\left( x \right)-\left( x+1 \right) \right]}\text{d}x>0$ $\Leftrightarrow g\left( 2 \right)-g\left( -3 \right)>0\Leftrightarrow g\left( 2 \right)>g\left( -3 \right)$. Vậy ta có $g\left( 0 \right)>g\left( 2 \right)>g\left( -3 \right)$.
A. $g\left( 2 \right)>g\left( -3 \right)>g\left( 0 \right)$.
B. $g\left( -3 \right)>g\left( 0 \right)>g\left( 2 \right)$.
C. $g\left( 2 \right)>g\left( 0 \right)>g\left( -3 \right)$.
D. $g\left( 0 \right)>g\left( 2 \right)>g\left( -3 \right)$.
${g}'\left( x \right)=0\Leftrightarrow {f}'\left( x \right)=x+1\Leftrightarrow \left[ \begin{aligned}
& x=-3 \\
& x=0 \\
& x=2 \\
\end{aligned} \right.$
Xét $\int\limits_{-3}^{0}{{g}'\left( x \right)}\text{d}x=\int\limits_{-3}^{0}{\left[ {f}'\left( x \right)-\left( x+1 \right) \right]}\text{d}x>0$ $\Leftrightarrow g\left( 0 \right)-g\left( -3 \right)>0\Leftrightarrow g\left( 0 \right)>g\left( -3 \right)$.
Tương tự, xét $\int\limits_{0}^{2}{{g}'\left( x \right)}\text{d}x=\int\limits_{0}^{2}{\left[ {f}'\left( x \right)-\left( x+1 \right) \right]}\text{d}x<0$ $\Leftrightarrow g\left( 2 \right)-g\left( 0 \right)<0\Leftrightarrow g\left( 2 \right)<g\left( 0 \right)$.
Xét $\int\limits_{-3}^{2}{{g}'\left( x \right)}\text{d}x=\int\limits_{-3}^{0}{\left[ {f}'\left( x \right)-\left( x+1 \right) \right]}\text{d}x+\int\limits_{0}^{2}{\left[ {f}'\left( x \right)-\left( x+1 \right) \right]}\text{d}x>0$ $\Leftrightarrow g\left( 2 \right)-g\left( -3 \right)>0\Leftrightarrow g\left( 2 \right)>g\left( -3 \right)$. Vậy ta có $g\left( 0 \right)>g\left( 2 \right)>g\left( -3 \right)$.
Đáp án D.