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Câu hỏi: Cho hàm số $y=f\left( x \right)$ xác định và liên tục trên $\left[ 1;e \right]$ thỏa mãn $x{f}'\left( x \right)=x{{\left[ f\left( x \right) \right]}^{2}}+3f\left( x \right)+\dfrac{4}{x}$ và $f\left( 1 \right)=-3$. Tính $f\left( {{e}^{2}} \right)$.
A. $\dfrac{5}{3{{e}^{2}}}$
B. $\dfrac{7}{3{{e}^{2}}}$
C. $-\dfrac{7}{3{{e}^{2}}}$
D. $-\dfrac{5}{3{{e}^{2}}}$
A. $\dfrac{5}{3{{e}^{2}}}$
B. $\dfrac{7}{3{{e}^{2}}}$
C. $-\dfrac{7}{3{{e}^{2}}}$
D. $-\dfrac{5}{3{{e}^{2}}}$
Ta có: $x{f}'\left( x \right)=x{{\left[ f\left( x \right) \right]}^{2}}+3f\left( x \right)+\dfrac{4}{x}\Leftrightarrow f\left( x \right)+x{f}'\left( x \right)=x{{\left[ f\left( x \right) \right]}^{2}}+4f\left( x \right)+\dfrac{4}{x}$
$\Leftrightarrow {{\left[ xf\left( x \right) \right]}^{\prime }}=\dfrac{1}{x}{{\left[ xf\left( x \right)+2 \right]}^{2}}\Leftrightarrow \dfrac{{{\left[ xf\left( x \right) \right]}^{\prime }}}{{{\left[ xf\left( x \right)+2 \right]}^{2}}}=\dfrac{1}{x}$
Đặt $g\left( x \right)=xf\left( x \right)$ ta có: $\dfrac{{g}'\left( x \right)}{{{\left[ g\left( x \right)+2 \right]}^{2}}}=\dfrac{1}{x}$ suy ra $\int{\dfrac{{g}'\left( x \right)dx}{{{\left[ g\left( x \right)+2 \right]}^{2}}}}=\int{\dfrac{dx}{x}}$
$\Leftrightarrow \int{\dfrac{d\left[ g\left( x \right) \right]}{{{\left[ g\left( x \right)+2 \right]}^{2}}}}=\ln \left| x \right|+C\Leftrightarrow \dfrac{-1}{g\left( x \right)+2}=\ln \left| x \right|+C\Leftrightarrow \dfrac{-1}{xf\left( x \right)+2}=\ln \left| x \right|+C$
Do $f\left( 1 \right)=-3$ nên $\dfrac{-1}{-1}=C\Leftrightarrow C=1$. Suy ra $\dfrac{-1}{{{e}^{2}}f\left( e \right)+2}=3\Leftrightarrow f\left( {{e}^{2}} \right)=\dfrac{-7}{3{{e}^{2}}}$.
$\Leftrightarrow {{\left[ xf\left( x \right) \right]}^{\prime }}=\dfrac{1}{x}{{\left[ xf\left( x \right)+2 \right]}^{2}}\Leftrightarrow \dfrac{{{\left[ xf\left( x \right) \right]}^{\prime }}}{{{\left[ xf\left( x \right)+2 \right]}^{2}}}=\dfrac{1}{x}$
Đặt $g\left( x \right)=xf\left( x \right)$ ta có: $\dfrac{{g}'\left( x \right)}{{{\left[ g\left( x \right)+2 \right]}^{2}}}=\dfrac{1}{x}$ suy ra $\int{\dfrac{{g}'\left( x \right)dx}{{{\left[ g\left( x \right)+2 \right]}^{2}}}}=\int{\dfrac{dx}{x}}$
$\Leftrightarrow \int{\dfrac{d\left[ g\left( x \right) \right]}{{{\left[ g\left( x \right)+2 \right]}^{2}}}}=\ln \left| x \right|+C\Leftrightarrow \dfrac{-1}{g\left( x \right)+2}=\ln \left| x \right|+C\Leftrightarrow \dfrac{-1}{xf\left( x \right)+2}=\ln \left| x \right|+C$
Do $f\left( 1 \right)=-3$ nên $\dfrac{-1}{-1}=C\Leftrightarrow C=1$. Suy ra $\dfrac{-1}{{{e}^{2}}f\left( e \right)+2}=3\Leftrightarrow f\left( {{e}^{2}} \right)=\dfrac{-7}{3{{e}^{2}}}$.
Đáp án C.