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Câu hỏi: Cho hàm số $y=f\left( x \right)$ xác định và liên tục trên $\left[ 0;2 \right]$ thỏa mãn ${{e}^{x}}{{f}^{2}}\left( x \right)+f\left( x \right)=f'\left( x \right)-\dfrac{1}{{{e}^{x}}}$ và $f\left( 0 \right)=1.$ Tính $f\left( 2 \right)$
A. $\dfrac{1}{{{e}^{2}}}$
B. $-\dfrac{5}{3{{e}^{2}}}$
C. $-\dfrac{1}{{{e}^{2}}}$
D. $-\dfrac{2}{3{{e}^{2}}}$
A. $\dfrac{1}{{{e}^{2}}}$
B. $-\dfrac{5}{3{{e}^{2}}}$
C. $-\dfrac{1}{{{e}^{2}}}$
D. $-\dfrac{2}{3{{e}^{2}}}$
Ta có ${{e}^{x}}{{f}^{2}}\left( x \right)+f\left( x \right)=f'\left( x \right)-\dfrac{1}{{{e}^{x}}}\Leftrightarrow {{\left( {{e}^{x}}.f\left( x \right) \right)}^{2}}+{{e}^{x}}.f\left( x \right)={{e}^{x}}.f'\left( x \right)-1$
$\Leftrightarrow {{\left( {{e}^{x}}.f\left( x \right) \right)}^{2}}+2{{e}^{x}}.f\left( x \right)+1={{e}^{x}}.f'\left( x \right)+\left( {{e}^{x}} \right)'.f\left( x \right)\Leftrightarrow {{\left[ {{e}^{x}}.f\left( x \right)+1 \right]}^{2}}=\left( {{e}^{x}}.f\left( x \right) \right)'$
Đặt $g\left( x \right)={{e}^{x}}.f\left( x \right)$ suy ra ${{\left[ g\left( x \right)+1 \right]}^{2}}=g'\left( x \right)\Leftrightarrow \dfrac{g'\left( x \right)}{{{\left[ g\left( x \right)+1 \right]}^{2}}}=1\Leftrightarrow \int{\dfrac{g'\left( x \right)}{{{\left[ g\left( x \right)+1 \right]}^{2}}}}=x+C$
$\Leftrightarrow \int{\dfrac{d\left( g\left( x \right)+1 \right)}{{{\left[ g\left( x \right)+1 \right]}^{2}}}}=x+C\Leftrightarrow \dfrac{1}{g\left( x \right)+1}=x+C$ mà $f\left( 0 \right)=1\Rightarrow g\left( 0 \right)=1$ nên $C=-\dfrac{1}{2}$
Do đó $-\dfrac{1}{{{e}^{x}}.f\left( x \right)+1}=x-\dfrac{1}{2}\Leftrightarrow {{e}^{x}}.f\left( x \right)+1=\dfrac{2}{1-2\text{x}}.$ Vậy $f\left( 2 \right)=-\dfrac{5}{3{{\text{e}}^{2}}}$
$\Leftrightarrow {{\left( {{e}^{x}}.f\left( x \right) \right)}^{2}}+2{{e}^{x}}.f\left( x \right)+1={{e}^{x}}.f'\left( x \right)+\left( {{e}^{x}} \right)'.f\left( x \right)\Leftrightarrow {{\left[ {{e}^{x}}.f\left( x \right)+1 \right]}^{2}}=\left( {{e}^{x}}.f\left( x \right) \right)'$
Đặt $g\left( x \right)={{e}^{x}}.f\left( x \right)$ suy ra ${{\left[ g\left( x \right)+1 \right]}^{2}}=g'\left( x \right)\Leftrightarrow \dfrac{g'\left( x \right)}{{{\left[ g\left( x \right)+1 \right]}^{2}}}=1\Leftrightarrow \int{\dfrac{g'\left( x \right)}{{{\left[ g\left( x \right)+1 \right]}^{2}}}}=x+C$
$\Leftrightarrow \int{\dfrac{d\left( g\left( x \right)+1 \right)}{{{\left[ g\left( x \right)+1 \right]}^{2}}}}=x+C\Leftrightarrow \dfrac{1}{g\left( x \right)+1}=x+C$ mà $f\left( 0 \right)=1\Rightarrow g\left( 0 \right)=1$ nên $C=-\dfrac{1}{2}$
Do đó $-\dfrac{1}{{{e}^{x}}.f\left( x \right)+1}=x-\dfrac{1}{2}\Leftrightarrow {{e}^{x}}.f\left( x \right)+1=\dfrac{2}{1-2\text{x}}.$ Vậy $f\left( 2 \right)=-\dfrac{5}{3{{\text{e}}^{2}}}$
Đáp án B.