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Câu hỏi: Cho hàm số $y=f\left( x \right)$ xác định và liên tục trên $\mathbb{R}\backslash \!\!\{0\}$ thỏa mãn $f\left( 1 \right)=-2$ $f\left( x \right)\ne -\dfrac{1}{x}$ và ${{x}^{2}}{{f}^{2}}\left( x \right)+\left( 2x-1 \right)f\left( x \right)=x{f}'\left( x \right)-1$ với $\forall x\in \mathbb{R}\backslash \!\!\{0\}$. Tính diện tích hình phẳng giới hạn bởi $y=f\left( x \right),y=0,x=1,x=3$.
A. $2\ln 3+\dfrac{1}{3}$.
B. $\ln 3+\dfrac{1}{4}$.
C. $2\ln 3+\dfrac{3}{4}$.
D. $\ln 3+\dfrac{2}{3}$.
A. $2\ln 3+\dfrac{1}{3}$.
B. $\ln 3+\dfrac{1}{4}$.
C. $2\ln 3+\dfrac{3}{4}$.
D. $\ln 3+\dfrac{2}{3}$.
Từ giả thiết
$\begin{aligned}
& {{x}^{2}}{{f}^{2}}\left( x \right)+\left( 2x-1 \right)f\left( x \right)=x{f}'\left( x \right)-1\Leftrightarrow {{x}^{2}}{{f}^{2}}\left( x \right)+2xf\left( x \right)+1=x{f}'\left( x \right)+f\left( x \right) \\
& \Leftrightarrow {{\left[ xf\left( x \right)+1 \right]}^{2}}={{\left( xf\left( x \right) \right)}^{\prime }}\Leftrightarrow {{\left[ xf\left( x \right)+1 \right]}^{2}}={{\left( xf\left( x \right)+1 \right)}^{\prime }} \\
& \Leftrightarrow \dfrac{{{\left( xf\left( x \right)+1 \right)}^{\prime }}}{{{\left[ xf\left( x \right)+1 \right]}^{2}}}=1 \left( \text{do} f\left( x \right)\ne -\dfrac{1}{x},\forall x \right) \\
\end{aligned}$
Do đó $\int{\dfrac{{{\left( xf\left( x \right)+1 \right)}^{\prime }}}{{{\left[ xf\left( x \right)+1 \right]}^{2}}}}dx=\int{1}\text{d}x\Leftrightarrow -\dfrac{1}{xf\left( x \right)+1}=x+C$
Mà $f\left( 1 \right)=-2\Rightarrow -\dfrac{1}{f\left( 1 \right)+1}=1+C\Rightarrow C=0$.
Suy ra $-\dfrac{1}{xf\left( x \right)+1}=x\Leftrightarrow -1={{x}^{2}}f\left( x \right)+x\Leftrightarrow f\left( x \right)=-\dfrac{1}{{{x}^{2}}}-\dfrac{1}{x}$
Vậy $\int\limits_{1}^{3}{\left| f\left( x \right) \right|}dx=\int\limits_{1}^{3}{\left| \dfrac{-1}{{{x}^{2}}}-\dfrac{1}{x} \right|}dx=\int\limits_{1}^{3}{\left( \dfrac{1}{{{x}^{2}}}+\dfrac{1}{x} \right)}dx=\left. \left( \dfrac{-1}{x}+\ln x \right) \right|_{1}^{3}=\ln 3+\dfrac{2}{3}$.
$\begin{aligned}
& {{x}^{2}}{{f}^{2}}\left( x \right)+\left( 2x-1 \right)f\left( x \right)=x{f}'\left( x \right)-1\Leftrightarrow {{x}^{2}}{{f}^{2}}\left( x \right)+2xf\left( x \right)+1=x{f}'\left( x \right)+f\left( x \right) \\
& \Leftrightarrow {{\left[ xf\left( x \right)+1 \right]}^{2}}={{\left( xf\left( x \right) \right)}^{\prime }}\Leftrightarrow {{\left[ xf\left( x \right)+1 \right]}^{2}}={{\left( xf\left( x \right)+1 \right)}^{\prime }} \\
& \Leftrightarrow \dfrac{{{\left( xf\left( x \right)+1 \right)}^{\prime }}}{{{\left[ xf\left( x \right)+1 \right]}^{2}}}=1 \left( \text{do} f\left( x \right)\ne -\dfrac{1}{x},\forall x \right) \\
\end{aligned}$
Do đó $\int{\dfrac{{{\left( xf\left( x \right)+1 \right)}^{\prime }}}{{{\left[ xf\left( x \right)+1 \right]}^{2}}}}dx=\int{1}\text{d}x\Leftrightarrow -\dfrac{1}{xf\left( x \right)+1}=x+C$
Mà $f\left( 1 \right)=-2\Rightarrow -\dfrac{1}{f\left( 1 \right)+1}=1+C\Rightarrow C=0$.
Suy ra $-\dfrac{1}{xf\left( x \right)+1}=x\Leftrightarrow -1={{x}^{2}}f\left( x \right)+x\Leftrightarrow f\left( x \right)=-\dfrac{1}{{{x}^{2}}}-\dfrac{1}{x}$
Vậy $\int\limits_{1}^{3}{\left| f\left( x \right) \right|}dx=\int\limits_{1}^{3}{\left| \dfrac{-1}{{{x}^{2}}}-\dfrac{1}{x} \right|}dx=\int\limits_{1}^{3}{\left( \dfrac{1}{{{x}^{2}}}+\dfrac{1}{x} \right)}dx=\left. \left( \dfrac{-1}{x}+\ln x \right) \right|_{1}^{3}=\ln 3+\dfrac{2}{3}$.
Đáp án D.