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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$ và thỏa mãn ${f}'\left( x \right)-3f\left( x \right)=\left( 2{{x}^{2}}+1 \right){{e}^{{{x}^{2}}+3x-1}},$ $\forall x\in \mathbb{R}$ và $f\left( 2 \right)=2{{e}^{9}}$. Biết $f\left( 1 \right)=a{{e}^{b}}$ với $a ,b\in \mathbb{N}$. Hệ thức nào sau đây đúng?
A. $a+b=5.$
B. $a-2b=-4.$
C. $a+3b=10.$
D. $a-b=-3.$
A. $a+b=5.$
B. $a-2b=-4.$
C. $a+3b=10.$
D. $a-b=-3.$
Ta có:
${f}'\left( x \right)-3f\left( x \right)=\left( 2{{x}^{2}}+1 \right){{e}^{{{x}^{2}}+3x-1}}\Leftrightarrow \dfrac{{f}'\left( x \right)-3f\left( x \right)}{{{e}^{3x}}}=\left( 2{{x}^{2}}+1 \right){{e}^{{{x}^{2}}-1}}$
$\Leftrightarrow {{\left( \dfrac{f\left( x \right)}{{{e}^{3x}}} \right)}^{\prime }}=\left( 2{{x}^{2}}+1 \right){{e}^{{{x}^{2}}-1}}\Leftrightarrow \int\limits_{1}^{2}{{{\left( \dfrac{f\left( x \right)}{{{e}^{3x}}} \right)}^{\prime }}\text{dx}}=\int\limits_{1}^{2}{\left( 2{{x}^{2}}+1 \right){{e}^{{{x}^{2}}-1}}\text{dx}}$
$\Leftrightarrow \left. \left( \dfrac{f\left( x \right)}{{{e}^{3x}}} \right) \right|_{1}^{2}=\int\limits_{1}^{2}{2{{x}^{2}}{{e}^{{{x}^{2}}-1}}\text{dx}}+\int\limits_{1}^{2}{{{e}^{{{x}^{2}}-1}}\text{dx}}.$
Đặt ${{I}_{1}}=\int\limits_{1}^{2}{{{e}^{{{x}^{2}}-1}}}\text{dx} \text{;} {{I}_{2}}=\int\limits_{1}^{2}{2{{x}^{2}}{{e}^{{{x}^{2}}-1}}}\text{dx}\Rightarrow $ $\left. \left( \dfrac{f\left( x \right)}{{{e}^{3x}}} \right) \right|_{1}^{2}={{I}_{1}}+{{I}_{2}}.$
Xét ${{I}_{1}}=\int\limits_{1}^{2}{{{e}^{{{x}^{2}}-1}}}\text{dx}$ đặt $\left\{ \begin{matrix}
u={{e}^{{{x}^{2}}-1}} \\
dv=dx \\
\end{matrix} \right.\Leftrightarrow \left\{ \begin{matrix}
du=2x{{e}^{{{x}^{2}}-1}}dx \\
v=x \\
\end{matrix} \right.$.
${{I}_{1}}=\int\limits_{1}^{2}{{{e}^{{{x}^{2}}-1}}}\text{dx} \text{=} \left. x{{e}^{{{x}^{2}}-1}} \right|_{1}^{2}-\int\limits_{1}^{2}{2{{x}^{2}}{{e}^{{{x}^{2}}-1}}}\text{dx}\Leftrightarrow {{I}_{1}}\text{=} \left. x{{e}^{{{x}^{2}}-1}} \right|_{1}^{2}-{{I}_{2}}$ $\Leftrightarrow {{I}_{1}}+{{I}_{2}}=\left. x{{e}^{{{x}^{2}}-1}} \right|_{1}^{2}=2{{e}^{3}}-1.$
Do đó:
$\left. \left( \dfrac{f\left( x \right)}{{{e}^{3x}}} \right) \right|_{1}^{2}=\int\limits_{1}^{2}{2{{x}^{2}}{{e}^{{{x}^{2}}-1}}\text{dx}}+\int\limits_{1}^{2}{{{e}^{{{x}^{2}}-1}}\text{dx}}\Leftrightarrow \dfrac{f\left( 2 \right)}{{{e}^{6}}}-\dfrac{f\left( 1 \right)}{{{e}^{3}}}=2{{e}^{3}}-1$
$\Leftrightarrow \dfrac{2{{e}^{9}}}{{{e}^{6}}}-\dfrac{a{{e}^{b}}}{{{e}^{3}}}=2{{e}^{3}}-1\Leftrightarrow a{{e}^{b}}={{e}^{3}}\Rightarrow \left\{ \begin{matrix}
a=1 \\
b=3 \\
\end{matrix} \right..$
Vậy $a+3b=10.$
${f}'\left( x \right)-3f\left( x \right)=\left( 2{{x}^{2}}+1 \right){{e}^{{{x}^{2}}+3x-1}}\Leftrightarrow \dfrac{{f}'\left( x \right)-3f\left( x \right)}{{{e}^{3x}}}=\left( 2{{x}^{2}}+1 \right){{e}^{{{x}^{2}}-1}}$
$\Leftrightarrow {{\left( \dfrac{f\left( x \right)}{{{e}^{3x}}} \right)}^{\prime }}=\left( 2{{x}^{2}}+1 \right){{e}^{{{x}^{2}}-1}}\Leftrightarrow \int\limits_{1}^{2}{{{\left( \dfrac{f\left( x \right)}{{{e}^{3x}}} \right)}^{\prime }}\text{dx}}=\int\limits_{1}^{2}{\left( 2{{x}^{2}}+1 \right){{e}^{{{x}^{2}}-1}}\text{dx}}$
$\Leftrightarrow \left. \left( \dfrac{f\left( x \right)}{{{e}^{3x}}} \right) \right|_{1}^{2}=\int\limits_{1}^{2}{2{{x}^{2}}{{e}^{{{x}^{2}}-1}}\text{dx}}+\int\limits_{1}^{2}{{{e}^{{{x}^{2}}-1}}\text{dx}}.$
Đặt ${{I}_{1}}=\int\limits_{1}^{2}{{{e}^{{{x}^{2}}-1}}}\text{dx} \text{;} {{I}_{2}}=\int\limits_{1}^{2}{2{{x}^{2}}{{e}^{{{x}^{2}}-1}}}\text{dx}\Rightarrow $ $\left. \left( \dfrac{f\left( x \right)}{{{e}^{3x}}} \right) \right|_{1}^{2}={{I}_{1}}+{{I}_{2}}.$
Xét ${{I}_{1}}=\int\limits_{1}^{2}{{{e}^{{{x}^{2}}-1}}}\text{dx}$ đặt $\left\{ \begin{matrix}
u={{e}^{{{x}^{2}}-1}} \\
dv=dx \\
\end{matrix} \right.\Leftrightarrow \left\{ \begin{matrix}
du=2x{{e}^{{{x}^{2}}-1}}dx \\
v=x \\
\end{matrix} \right.$.
${{I}_{1}}=\int\limits_{1}^{2}{{{e}^{{{x}^{2}}-1}}}\text{dx} \text{=} \left. x{{e}^{{{x}^{2}}-1}} \right|_{1}^{2}-\int\limits_{1}^{2}{2{{x}^{2}}{{e}^{{{x}^{2}}-1}}}\text{dx}\Leftrightarrow {{I}_{1}}\text{=} \left. x{{e}^{{{x}^{2}}-1}} \right|_{1}^{2}-{{I}_{2}}$ $\Leftrightarrow {{I}_{1}}+{{I}_{2}}=\left. x{{e}^{{{x}^{2}}-1}} \right|_{1}^{2}=2{{e}^{3}}-1.$
Do đó:
$\left. \left( \dfrac{f\left( x \right)}{{{e}^{3x}}} \right) \right|_{1}^{2}=\int\limits_{1}^{2}{2{{x}^{2}}{{e}^{{{x}^{2}}-1}}\text{dx}}+\int\limits_{1}^{2}{{{e}^{{{x}^{2}}-1}}\text{dx}}\Leftrightarrow \dfrac{f\left( 2 \right)}{{{e}^{6}}}-\dfrac{f\left( 1 \right)}{{{e}^{3}}}=2{{e}^{3}}-1$
$\Leftrightarrow \dfrac{2{{e}^{9}}}{{{e}^{6}}}-\dfrac{a{{e}^{b}}}{{{e}^{3}}}=2{{e}^{3}}-1\Leftrightarrow a{{e}^{b}}={{e}^{3}}\Rightarrow \left\{ \begin{matrix}
a=1 \\
b=3 \\
\end{matrix} \right..$
Vậy $a+3b=10.$
Đáp án C.