Google
Câu hỏi: Cho $F\left( x \right)=\left( a{{x}^{2}}+bx+c \right){{e}^{-x}}$ là một nguyên hàm của hàm số $f\left( x \right)=\left( 2{{x}^{2}}-5x+2 \right){{e}^{-x}}$. Giá trị của $f\left[ F\left( 0 \right) \right]$ bằng
A. $-{{e}^{-1}}$
B. $20{{e}^{2}}$
C. $9e$
D. $3e$
A. $-{{e}^{-1}}$
B. $20{{e}^{2}}$
C. $9e$
D. $3e$
Ta có $f\left( x \right)={F}'\left( x \right)\Rightarrow \left( 2{{x}^{2}}-5x+2 \right){{e}^{-x}}=\left( 2ax+b \right){{e}^{-x}}-\left( a{{x}^{2}}+bx+c \right){{e}^{-x}}$
$\Rightarrow 2{{x}^{2}}-5x+2=\left( 2ax+b \right)-\left( a{{x}^{2}}+bx+c \right)=-a{{x}^{2}}+\left( 2a-b \right)x+b-c$
$\Rightarrow \left\{ \begin{aligned}
& -a=2 \\
& 2\text{a}-b=-5 \\
& b-c=2 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& a=-2 \\
& b=1 \\
& c=-1 \\
\end{aligned} \right.\Rightarrow F\left( x \right)=\left( -2{{\text{x}}^{2}}+x-1 \right){{e}^{-x}}\Rightarrow F\left( 0 \right)=-1$
$\Rightarrow f\left[ F\left( 0 \right) \right]=f\left( -1 \right)=9\text{e}$.
$\Rightarrow 2{{x}^{2}}-5x+2=\left( 2ax+b \right)-\left( a{{x}^{2}}+bx+c \right)=-a{{x}^{2}}+\left( 2a-b \right)x+b-c$
$\Rightarrow \left\{ \begin{aligned}
& -a=2 \\
& 2\text{a}-b=-5 \\
& b-c=2 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& a=-2 \\
& b=1 \\
& c=-1 \\
\end{aligned} \right.\Rightarrow F\left( x \right)=\left( -2{{\text{x}}^{2}}+x-1 \right){{e}^{-x}}\Rightarrow F\left( 0 \right)=-1$
$\Rightarrow f\left[ F\left( 0 \right) \right]=f\left( -1 \right)=9\text{e}$.
Đáp án C.