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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R}$ thỏa mãn ${f}'\left( x \right)+2x.f\left( x \right)={{e}^{-{{x}^{2}}}} \forall x\in \mathbb{R}$ và $f\left( 0 \right)=0$. Tính $f\left( -1 \right)$
A. $f\left( -1 \right)={{e}^{2}}$
B. $f\left( -1 \right)=\dfrac{-1}{e}$
C. $f\left( -1 \right)=\dfrac{1}{{{e}^{2}}}$
D. $f\left( -1 \right)=\dfrac{1}{e}$
A. $f\left( -1 \right)={{e}^{2}}$
B. $f\left( -1 \right)=\dfrac{-1}{e}$
C. $f\left( -1 \right)=\dfrac{1}{{{e}^{2}}}$
D. $f\left( -1 \right)=\dfrac{1}{e}$
Ta có ${f}'\left( x \right)+2x.f\left( x \right)={{e}^{-{{x}^{2}}}}\Leftrightarrow {{e}^{{{x}^{2}}}}.f\left( \left( x \right)+2x.f\left( x \right) \right)=1$
$\Leftrightarrow {f}'\left( x \right).{{e}^{{{x}^{2}}}}+2x.{{e}^{{{x}^{2}}}}.f\left( x \right)=1\Leftrightarrow \left( f\left( x \right).{{e}^{{{x}^{2}}}} \right)=1\Leftrightarrow \int{\left( f\left( x \right).{{e}^{{{x}^{2}}}} \right)dx}=\int{dx}$
$\Leftrightarrow f\left( x \right).{{e}^{{{x}^{2}}}}=x+C$
$f\left( 0 \right)\Leftrightarrow C=0\Rightarrow f\left( x \right)=\dfrac{x}{{{e}^{{{x}^{2}}}}}\Rightarrow f\left( -1 \right)=-\dfrac{1}{e}$
$\Leftrightarrow {f}'\left( x \right).{{e}^{{{x}^{2}}}}+2x.{{e}^{{{x}^{2}}}}.f\left( x \right)=1\Leftrightarrow \left( f\left( x \right).{{e}^{{{x}^{2}}}} \right)=1\Leftrightarrow \int{\left( f\left( x \right).{{e}^{{{x}^{2}}}} \right)dx}=\int{dx}$
$\Leftrightarrow f\left( x \right).{{e}^{{{x}^{2}}}}=x+C$
$f\left( 0 \right)\Leftrightarrow C=0\Rightarrow f\left( x \right)=\dfrac{x}{{{e}^{{{x}^{2}}}}}\Rightarrow f\left( -1 \right)=-\dfrac{1}{e}$
Đáp án B.