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