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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm ${f}'\left( x \right)=x{{e}^{x-a}}, \forall x\in \mathbb{R}$ và $f\left( 0 \right)=-{{e}^{-a}}-1$ ( với $a$ là tham số thực). Khi $\int\limits_{0}^{a}{f\left( x \right)dx=4}$, khẳng định nào dưới đây đúng?
A. $a\in \left( -2;-1 \right)$.
B. $a\in \left( -1;0 \right)$.
C. $a\in \left( 0;1 \right)$.
D. $a\in \left( 1;2 \right)$.
A. $a\in \left( -2;-1 \right)$.
B. $a\in \left( -1;0 \right)$.
C. $a\in \left( 0;1 \right)$.
D. $a\in \left( 1;2 \right)$.
Ta có: ${f}'\left( x \right)=x{{e}^{x-a}}, \forall x\in \mathbb{R}$
Suy ra $f\left( x \right)=\int{{f}'\left( x \right)dx=\int{x{{e}^{x-a}}dx}}$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={{e}^{x-a}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v={{e}^{x-a}} \\
\end{aligned} \right.$.
Ta có $f\left( x \right)=\int{x{{e}^{x-a}}dx=x{{e}^{x-a}}}-\int{{{e}^{x-a}}dx=x{{e}^{x-a}}-{{e}^{x-a}}+C}$.
Do $f\left( 0 \right)=-{{e}^{-a}}-1$ ta có $-{{e}^{-a}}+C=-{{e}^{-a}}-1\Leftrightarrow C=-1$.
Suy ra $f\left( x \right)=x{{e}^{x-a}}-{{e}^{x-a}}-1$.
Khi đó $\int\limits_{0}^{a}{f\left( x \right)dx=\int\limits_{0}^{a}{\left( x{{e}^{x-a}}-{{e}^{x-a}}-1 \right)}}dx$
$=\int\limits_{0}^{a}{x{{e}^{x-a}}dx}-\int\limits_{0}^{a}{{{e}^{x-a}}dx-\int\limits_{0}^{a}{dx}}$ $=\left. \left( x{{e}^{x-a}}-{{e}^{x-a}} \right) \right|\begin{matrix}
a \\
0 \\
\end{matrix}-\left. {{e}^{x-a}} \right|\begin{matrix}
a \\
0 \\
\end{matrix}-\left. x \right|\begin{matrix}
a \\
0 \\
\end{matrix}=2{{e}^{-a}}-2$.
Theo giả thiết $\int\limits_{0}^{a}{f\left( x \right)dx=4}$, suy ra $2{{e}^{-a}}-2=4\Leftrightarrow a=-\ln 3\approx -1,098$.
Suy ra $f\left( x \right)=\int{{f}'\left( x \right)dx=\int{x{{e}^{x-a}}dx}}$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={{e}^{x-a}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v={{e}^{x-a}} \\
\end{aligned} \right.$.
Ta có $f\left( x \right)=\int{x{{e}^{x-a}}dx=x{{e}^{x-a}}}-\int{{{e}^{x-a}}dx=x{{e}^{x-a}}-{{e}^{x-a}}+C}$.
Do $f\left( 0 \right)=-{{e}^{-a}}-1$ ta có $-{{e}^{-a}}+C=-{{e}^{-a}}-1\Leftrightarrow C=-1$.
Suy ra $f\left( x \right)=x{{e}^{x-a}}-{{e}^{x-a}}-1$.
Khi đó $\int\limits_{0}^{a}{f\left( x \right)dx=\int\limits_{0}^{a}{\left( x{{e}^{x-a}}-{{e}^{x-a}}-1 \right)}}dx$
$=\int\limits_{0}^{a}{x{{e}^{x-a}}dx}-\int\limits_{0}^{a}{{{e}^{x-a}}dx-\int\limits_{0}^{a}{dx}}$ $=\left. \left( x{{e}^{x-a}}-{{e}^{x-a}} \right) \right|\begin{matrix}
a \\
0 \\
\end{matrix}-\left. {{e}^{x-a}} \right|\begin{matrix}
a \\
0 \\
\end{matrix}-\left. x \right|\begin{matrix}
a \\
0 \\
\end{matrix}=2{{e}^{-a}}-2$.
Theo giả thiết $\int\limits_{0}^{a}{f\left( x \right)dx=4}$, suy ra $2{{e}^{-a}}-2=4\Leftrightarrow a=-\ln 3\approx -1,098$.
Đáp án A.