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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $-x{f}'\left( x \right).\ln x+f\left( x \right)=2{{x}^{2}}{{f}^{2}}\left( x \right), \forall x\in \left( 1;+\infty \right)$, $f\left( x \right)>0,\forall x\in \left( 1;+\infty \right)$ và $f\left( \text{e} \right)=\dfrac{1}{{{\text{e}}^{2}}}$. Tính diện tích $S$ hình phẳng giới hạn bởi đồ thị $y=xf\left( x \right),y=0,x=e,x={{e}^{2}}$.
A. $S=\dfrac{3}{2}$.
B. $S=\dfrac{1}{2}$.
C. $S=\dfrac{5}{3}$.
D. $S=2$.
A. $S=\dfrac{3}{2}$.
B. $S=\dfrac{1}{2}$.
C. $S=\dfrac{5}{3}$.
D. $S=2$.
Ta có: $-xf'\left( x \right)\ln x+f\left( x \right)=2{{x}^{2}}{{f}^{2}}\left( x \right)\Leftrightarrow -x\dfrac{f'\left( x \right)}{{{f}^{2}}\left( x \right)}\ln x+\dfrac{1}{f\left( x \right)}=2{{x}^{2}}$, $\forall x\in \left( 1;+\infty \right)$.
$\Leftrightarrow x{g}'\left( x \right).\ln x+g\left( x \right)=2{{x}^{2}}, \forall x\in \left( 1;+\infty \right)$ với $g\left( x \right)=\dfrac{1}{f\left( x \right)}$
$\Leftrightarrow {g}'\left( x \right)\ln x+\dfrac{g\left( x \right)}{x}=2x$, $\forall x\in \left( 1;+\infty \right)$ $\Rightarrow \int{{g}'\left( x \right)\ln x\text{d}x+\int{\dfrac{g\left( x \right)}{x}\text{d}x=\int{2x\text{d}x}}}$
$\Leftrightarrow g\left( x \right)\ln x-\int{\dfrac{g\left( x \right)}{x}\text{d}x+\int{\dfrac{g\left( x \right)}{x}\text{d}x={{x}^{2}}+C}}$ $\Leftrightarrow g\left( x \right)\ln x={{x}^{2}}+C$, $\forall x\in \left( 1;+\infty \right)$.
Do $f\left( \text{e} \right)=\dfrac{1}{{{\text{e}}^{2}}}\Leftrightarrow g\left( e \right)={{e}^{2}}\Leftrightarrow C=0$.
Suy ra $g\left( x \right)\ln x={{x}^{2}}$, $\forall x\in \left( 1;+\infty \right)$
$\Rightarrow g\left( x \right)=\dfrac{{{x}^{2}}}{\ln x}>0, \forall x\in \left( 1;+\infty \right)$
$\Rightarrow y=xf\left( x \right)=\dfrac{x}{g\left( x \right)}=\dfrac{\ln x}{x}$, $\forall x\in \left( 1;+\infty \right)$.
Ta có $S=\int_{\text{e}}^{{{\text{e}}^{2}}}{xf\left( x \right)\text{d}x}=\int_{\text{e}}^{{{\text{e}}^{2}}}{\dfrac{\ln x}{x}\text{d}x}=\dfrac{1}{2}{{\ln }^{2}}x\left| \begin{aligned}
& {{\text{e}}^{2}} \\
& \text{e} \\
\end{aligned} \right.=\dfrac{3}{2}$.
$\Leftrightarrow x{g}'\left( x \right).\ln x+g\left( x \right)=2{{x}^{2}}, \forall x\in \left( 1;+\infty \right)$ với $g\left( x \right)=\dfrac{1}{f\left( x \right)}$
$\Leftrightarrow {g}'\left( x \right)\ln x+\dfrac{g\left( x \right)}{x}=2x$, $\forall x\in \left( 1;+\infty \right)$ $\Rightarrow \int{{g}'\left( x \right)\ln x\text{d}x+\int{\dfrac{g\left( x \right)}{x}\text{d}x=\int{2x\text{d}x}}}$
$\Leftrightarrow g\left( x \right)\ln x-\int{\dfrac{g\left( x \right)}{x}\text{d}x+\int{\dfrac{g\left( x \right)}{x}\text{d}x={{x}^{2}}+C}}$ $\Leftrightarrow g\left( x \right)\ln x={{x}^{2}}+C$, $\forall x\in \left( 1;+\infty \right)$.
Do $f\left( \text{e} \right)=\dfrac{1}{{{\text{e}}^{2}}}\Leftrightarrow g\left( e \right)={{e}^{2}}\Leftrightarrow C=0$.
Suy ra $g\left( x \right)\ln x={{x}^{2}}$, $\forall x\in \left( 1;+\infty \right)$
$\Rightarrow g\left( x \right)=\dfrac{{{x}^{2}}}{\ln x}>0, \forall x\in \left( 1;+\infty \right)$
$\Rightarrow y=xf\left( x \right)=\dfrac{x}{g\left( x \right)}=\dfrac{\ln x}{x}$, $\forall x\in \left( 1;+\infty \right)$.
Ta có $S=\int_{\text{e}}^{{{\text{e}}^{2}}}{xf\left( x \right)\text{d}x}=\int_{\text{e}}^{{{\text{e}}^{2}}}{\dfrac{\ln x}{x}\text{d}x}=\dfrac{1}{2}{{\ln }^{2}}x\left| \begin{aligned}
& {{\text{e}}^{2}} \\
& \text{e} \\
\end{aligned} \right.=\dfrac{3}{2}$.
Đáp án A.