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Câu hỏi: Cho hàm số $f\left( x \right)={{x}^{3}}+a{{x}^{2}}+bx+c$ với $a$, $b$, $c$ là các số thực. Đặt $g\left( x \right)=f\left( x \right)+{f}'\left( x \right)+{{f}'}'\left( x \right)$, biết $g\left( 0 \right)=2, g\left( 1 \right)=6$. Tính tích phân $\int\limits_{0}^{1}{\dfrac{6x-f\left( x \right)}{{{e}^{x}}}}\text{d}x$.
A. $-2$.
B. $6$.
C. $2$.
D. $4$.
A. $-2$.
B. $6$.
C. $2$.
D. $4$.
Ta có: $f\left( x \right)={{x}^{3}}+a{{x}^{2}}+bx+c$ $\Rightarrow {f}'\left( x \right)=3{{x}^{2}}+2ax+b,\ {{f}'}'\left( x \right)=6x+2a,\ {{{f}'}'}'\left( x \right)=6$.
Do $g\left( x \right)=f\left( x \right)+{f}'\left( x \right)+{{f}'}'\left( x \right)$ $\left( 1 \right)$
$\Rightarrow {g}'\left( x \right)={f}'\left( x \right)+{{f}'}'\left( x \right)+{{{f}'}'}'\left( x \right)$ $\left( 2 \right)$.
Từ $\left( 1 \right)$ và $\left( 2 \right)$ suy ra $g\left( x \right)=f\left( x \right)+{g}'\left( x \right)-{{{f}'}'}'\left( x \right)$
$\Rightarrow 6x-f\left( x \right)={g}'\left( x \right)-6-g\left( x \right)+6x$.
$\begin{aligned}
& \Rightarrow \dfrac{6x-f\left( x \right)}{{{e}^{x}}}=\dfrac{{g}'\left( x \right)-6-\left( g\left( x \right)-6x \right)}{{{e}^{x}}} \\
& \Rightarrow \dfrac{6x-f\left( x \right)}{{{e}^{x}}}=\dfrac{\left( {g}'\left( x \right)-6 \right){{e}^{x}}-\left( g\left( x \right)-6x \right){{e}^{x}}}{{{e}^{2x}}}={{\left( \dfrac{g\left( x \right)-6x}{{{e}^{x}}} \right)}^{\prime }} \\
& \Rightarrow \int\limits_{0}^{1}{\dfrac{6x-f\left( x \right)}{{{e}^{x}}}\text{d}x}=\int\limits_{0}^{1}{{{\left( \dfrac{g\left( x \right)-6x}{{{e}^{x}}} \right)}^{\prime }}\text{d}x}=\left. \dfrac{g\left( x \right)-6x}{{{e}^{x}}} \right|_{0}^{1}=\dfrac{g\left( 1 \right)-6}{{{e}^{1}}}-\dfrac{g\left( 0 \right)-0}{{{e}^{0}}}=-2 \\
\end{aligned}$
Do $g\left( x \right)=f\left( x \right)+{f}'\left( x \right)+{{f}'}'\left( x \right)$ $\left( 1 \right)$
$\Rightarrow {g}'\left( x \right)={f}'\left( x \right)+{{f}'}'\left( x \right)+{{{f}'}'}'\left( x \right)$ $\left( 2 \right)$.
Từ $\left( 1 \right)$ và $\left( 2 \right)$ suy ra $g\left( x \right)=f\left( x \right)+{g}'\left( x \right)-{{{f}'}'}'\left( x \right)$
$\Rightarrow 6x-f\left( x \right)={g}'\left( x \right)-6-g\left( x \right)+6x$.
$\begin{aligned}
& \Rightarrow \dfrac{6x-f\left( x \right)}{{{e}^{x}}}=\dfrac{{g}'\left( x \right)-6-\left( g\left( x \right)-6x \right)}{{{e}^{x}}} \\
& \Rightarrow \dfrac{6x-f\left( x \right)}{{{e}^{x}}}=\dfrac{\left( {g}'\left( x \right)-6 \right){{e}^{x}}-\left( g\left( x \right)-6x \right){{e}^{x}}}{{{e}^{2x}}}={{\left( \dfrac{g\left( x \right)-6x}{{{e}^{x}}} \right)}^{\prime }} \\
& \Rightarrow \int\limits_{0}^{1}{\dfrac{6x-f\left( x \right)}{{{e}^{x}}}\text{d}x}=\int\limits_{0}^{1}{{{\left( \dfrac{g\left( x \right)-6x}{{{e}^{x}}} \right)}^{\prime }}\text{d}x}=\left. \dfrac{g\left( x \right)-6x}{{{e}^{x}}} \right|_{0}^{1}=\dfrac{g\left( 1 \right)-6}{{{e}^{1}}}-\dfrac{g\left( 0 \right)-0}{{{e}^{0}}}=-2 \\
\end{aligned}$
Đáp án A.