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Câu hỏi: Cho hàm số có đạo hàm liên tục trên khoảng $\left( 1;+\infty \right)$ và thỏa mãn $\left( x{f}'\left( x \right)-2f\left( x \right) \right)\ln x={{x}^{3}}-f\left( x \right),\forall x\in \left( 1;+\infty \right)$ ; biết $f\left( \sqrt[3]{e} \right)=3e$. Giá trị $f\left( 2 \right)$ thuộc khoảng nào dưới đây?
A. $\left( 12;\dfrac{25}{2} \right)$.
B. $\left( 13;\dfrac{27}{2} \right)$.
C. $\left( \dfrac{23}{2};12 \right)$.
D. $\left( 14;\dfrac{29}{2} \right)$.
A. $\left( 12;\dfrac{25}{2} \right)$.
B. $\left( 13;\dfrac{27}{2} \right)$.
C. $\left( \dfrac{23}{2};12 \right)$.
D. $\left( 14;\dfrac{29}{2} \right)$.
Vì $x\in \left( 1;+\infty \right)$ nên ta có $\left( {{x}^{2}}{f}'\left( x \right)-2xf\left( x \right) \right)\ln x={{x}^{4}}-xf\left( x \right)$
$\Leftrightarrow \left( \dfrac{{{x}^{2}}{f}'\left( x \right)-2xf\left( x \right)}{{{x}^{4}}} \right)\ln x=1-\dfrac{f\left( x \right)}{{{x}^{3}}}$
$\Rightarrow {{\left( \dfrac{f\left( x \right)}{{{x}^{2}}} \right)}^{\prime }}\ln x=1-\dfrac{f\left( x \right)}{{{x}^{3}}}\Leftrightarrow \int{{{\left( \dfrac{f\left( x \right)}{{{x}^{2}}} \right)}^{\prime }}\ln xdx}=\int{\left( 1-\dfrac{f\left( x \right)}{{{x}^{3}}} \right)dx}$
$\Leftrightarrow \dfrac{f\left( x \right)\ln x}{{{x}^{2}}}-\int{\dfrac{f\left( x \right)}{{{x}^{3}}}dx}=x-\int{\dfrac{f\left( x \right)}{{{x}^{3}}}dx}+C$
$\Leftrightarrow \dfrac{f\left( x \right)\ln x}{{{x}^{2}}}=x+C\Leftrightarrow \dfrac{f\left( x \right)\ln x}{{{x}^{2}}}=x+C\Leftrightarrow f\left( x \right)=\dfrac{{{x}^{2}}\left( x+C \right)}{\ln x}$.
Theo đề bài $f\left( \sqrt[3]{e} \right)=3e\Rightarrow C=0\Rightarrow f\left( x \right)=\dfrac{{{x}^{3}}}{\ln x}f\left( 2 \right)=\dfrac{8}{\ln 2}\in \left( \dfrac{23}{2};12 \right)$.
$\Leftrightarrow \left( \dfrac{{{x}^{2}}{f}'\left( x \right)-2xf\left( x \right)}{{{x}^{4}}} \right)\ln x=1-\dfrac{f\left( x \right)}{{{x}^{3}}}$
$\Rightarrow {{\left( \dfrac{f\left( x \right)}{{{x}^{2}}} \right)}^{\prime }}\ln x=1-\dfrac{f\left( x \right)}{{{x}^{3}}}\Leftrightarrow \int{{{\left( \dfrac{f\left( x \right)}{{{x}^{2}}} \right)}^{\prime }}\ln xdx}=\int{\left( 1-\dfrac{f\left( x \right)}{{{x}^{3}}} \right)dx}$
$\Leftrightarrow \dfrac{f\left( x \right)\ln x}{{{x}^{2}}}-\int{\dfrac{f\left( x \right)}{{{x}^{3}}}dx}=x-\int{\dfrac{f\left( x \right)}{{{x}^{3}}}dx}+C$
$\Leftrightarrow \dfrac{f\left( x \right)\ln x}{{{x}^{2}}}=x+C\Leftrightarrow \dfrac{f\left( x \right)\ln x}{{{x}^{2}}}=x+C\Leftrightarrow f\left( x \right)=\dfrac{{{x}^{2}}\left( x+C \right)}{\ln x}$.
Theo đề bài $f\left( \sqrt[3]{e} \right)=3e\Rightarrow C=0\Rightarrow f\left( x \right)=\dfrac{{{x}^{3}}}{\ln x}f\left( 2 \right)=\dfrac{8}{\ln 2}\in \left( \dfrac{23}{2};12 \right)$.
Đáp án C.