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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$ thỏa mãn ${f}'\left( x \right)+xf\left( x \right)=\dfrac{2x}{{{e}^{{{x}^{2}}}}}$, $\forall x\in \mathbb{R}$ và $f\left( 0 \right)=-2$. Tính $f\left( -2 \right)$.
A. $f\left( -2 \right)=\dfrac{-2}{{{e}^{4}}}$.
B. $f\left( -2 \right)=\dfrac{2}{{{e}^{4}}}$.
C. $f\left( -2 \right)=2$.
D. $f\left( -2 \right)={{e}^{2}}$.
A. $f\left( -2 \right)=\dfrac{-2}{{{e}^{4}}}$.
B. $f\left( -2 \right)=\dfrac{2}{{{e}^{4}}}$.
C. $f\left( -2 \right)=2$.
D. $f\left( -2 \right)={{e}^{2}}$.
Ta có ${f}'\left( x \right)+xf\left( x \right)=\dfrac{2x}{{{e}^{{{x}^{2}}}}}\Leftrightarrow {{e}^{\dfrac{{{x}^{2}}}{2}}}{f}'\left( x \right)+{{e}^{\dfrac{{{x}^{2}}}{2}}}xf\left( x \right)=\dfrac{2x}{{{e}^{\dfrac{{{x}^{2}}}{2}}}}\Leftrightarrow {{\left[ {{e}^{\dfrac{{{x}^{2}}}{2}}}f\left( x \right) \right]}^{\prime }}=\dfrac{2x}{{{e}^{\dfrac{{{x}^{2}}}{2}}}}$
$\Rightarrow \int\limits_{-2}^{0}{{{\left[ {{e}^{\dfrac{{{x}^{2}}}{2}}}f\left( x \right) \right]}^{\prime }}\text{d}x}=\int\limits_{-2}^{0}{\dfrac{2x}{{{e}^{\dfrac{{{x}^{2}}}{2}}}}\text{d}x}\Leftrightarrow \left. {{e}^{\dfrac{{{x}^{2}}}{2}}}f\left( x \right) \right|_{-2}^{0}=2\int\limits_{-2}^{0}{\dfrac{\text{d}\left( \dfrac{{{x}^{2}}}{2} \right)}{{{e}^{\dfrac{{{x}^{2}}}{2}}}}\text{d}x}=\left. \dfrac{-2}{{{e}^{\dfrac{{{x}^{2}}}{2}}}} \right|_{-2}^{0}$
$\Leftrightarrow f\left( 0 \right)-{{e}^{2}}f\left( -2 \right)=-2\left( 1-\dfrac{1}{{{e}^{2}}} \right)\Leftrightarrow -{{e}^{2}}f\left( -2 \right)=\dfrac{2}{{{e}^{2}}}\Leftrightarrow f\left( -2 \right)=\dfrac{-2}{{{e}^{4}}}$.
$\Rightarrow \int\limits_{-2}^{0}{{{\left[ {{e}^{\dfrac{{{x}^{2}}}{2}}}f\left( x \right) \right]}^{\prime }}\text{d}x}=\int\limits_{-2}^{0}{\dfrac{2x}{{{e}^{\dfrac{{{x}^{2}}}{2}}}}\text{d}x}\Leftrightarrow \left. {{e}^{\dfrac{{{x}^{2}}}{2}}}f\left( x \right) \right|_{-2}^{0}=2\int\limits_{-2}^{0}{\dfrac{\text{d}\left( \dfrac{{{x}^{2}}}{2} \right)}{{{e}^{\dfrac{{{x}^{2}}}{2}}}}\text{d}x}=\left. \dfrac{-2}{{{e}^{\dfrac{{{x}^{2}}}{2}}}} \right|_{-2}^{0}$
$\Leftrightarrow f\left( 0 \right)-{{e}^{2}}f\left( -2 \right)=-2\left( 1-\dfrac{1}{{{e}^{2}}} \right)\Leftrightarrow -{{e}^{2}}f\left( -2 \right)=\dfrac{2}{{{e}^{2}}}\Leftrightarrow f\left( -2 \right)=\dfrac{-2}{{{e}^{4}}}$.
Đáp án A.