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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm cấp $2$ trên $\left( -\infty ;0 \right)$ và thỏa mãn: ${{x}^{2}}f\left( x \right)\left[ {{f}'}'\left( x \right)+1 \right]=f\left( x \right)\left[ {{x}^{2}}-4f\left( x \right) \right]+{{x}^{2}}.{{\left[ {f}'\left( x \right) \right]}^{2}}$, biết $f\left( -1 \right)=1$, ${f}'\left( -1 \right)=-4$. Tính $I=\int_{-2}^{-1}{\dfrac{f\left( x \right)}{{{x}^{2}}}dx}$
A. $3$.
B. $\dfrac{3}{5}$.
C. $\dfrac{7}{3}$.
D. $\dfrac{11}{3}$
A. $3$.
B. $\dfrac{3}{5}$.
C. $\dfrac{7}{3}$.
D. $\dfrac{11}{3}$
${{x}^{2}}.f\left( x \right).{{f}'}'\left( x \right)=-4{{f}^{2}}\left( x \right)+{{x}^{2}}{{\left[ {f}'\left( x \right) \right]}^{2}}$ $\Leftrightarrow {{x}^{2}}\left( f\left( x \right).{{f}'}'\left( x \right)-{{\left[ {f}'\left( x \right) \right]}^{2}} \right)=-4{{f}^{2}}\left( x \right)$
$\Leftrightarrow \dfrac{{{f}'}'\left( x \right).f\left( x \right)-{{\left[ {f}'\left( x \right) \right]}^{2}}}{{{f}^{2}}\left( x \right)}=-\dfrac{4}{{{x}^{2}}}$ $\Rightarrow {{\left( \dfrac{{f}'\left( x \right)}{f\left( x \right)} \right)}^{\prime }}=-\dfrac{4}{{{x}^{2}}}$ $\Rightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=\int{\dfrac{-4}{{{x}^{2}}}dx}=\dfrac{4}{x}+C$
* Theo giả thiết, ta có: $\dfrac{{f}'\left( -1 \right)}{f\left( -1 \right)}=-4+C\Rightarrow C=0$
Vậy $\dfrac{{f}'\left( x \right)}{f\left( x \right)}=\dfrac{4}{x}$
$\Rightarrow \ln \left[ f\left( x \right) \right]=\int{\dfrac{4}{x}dx}=4\ln \left( -x \right)+C$
* Theo giả thiết, ta có: $\ln \left[ f\left( -1 \right) \right]=4.\ln \left( 1 \right)+C\Rightarrow C=0$
Vậy $\ln \left[ f\left( x \right) \right]=4\ln \left( -x \right)$ $\Rightarrow f\left( x \right)={{e}^{4\ln \left( -x \right)}}={{x}^{4}}$
* $I=\int_{-2}^{-1}{\dfrac{f\left( x \right)}{{{x}^{2}}}dx}=\int_{-2}^{-1}{{{x}^{2}}dx}=\dfrac{7}{3}$
$\Leftrightarrow \dfrac{{{f}'}'\left( x \right).f\left( x \right)-{{\left[ {f}'\left( x \right) \right]}^{2}}}{{{f}^{2}}\left( x \right)}=-\dfrac{4}{{{x}^{2}}}$ $\Rightarrow {{\left( \dfrac{{f}'\left( x \right)}{f\left( x \right)} \right)}^{\prime }}=-\dfrac{4}{{{x}^{2}}}$ $\Rightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=\int{\dfrac{-4}{{{x}^{2}}}dx}=\dfrac{4}{x}+C$
* Theo giả thiết, ta có: $\dfrac{{f}'\left( -1 \right)}{f\left( -1 \right)}=-4+C\Rightarrow C=0$
Vậy $\dfrac{{f}'\left( x \right)}{f\left( x \right)}=\dfrac{4}{x}$
$\Rightarrow \ln \left[ f\left( x \right) \right]=\int{\dfrac{4}{x}dx}=4\ln \left( -x \right)+C$
* Theo giả thiết, ta có: $\ln \left[ f\left( -1 \right) \right]=4.\ln \left( 1 \right)+C\Rightarrow C=0$
Vậy $\ln \left[ f\left( x \right) \right]=4\ln \left( -x \right)$ $\Rightarrow f\left( x \right)={{e}^{4\ln \left( -x \right)}}={{x}^{4}}$
* $I=\int_{-2}^{-1}{\dfrac{f\left( x \right)}{{{x}^{2}}}dx}=\int_{-2}^{-1}{{{x}^{2}}dx}=\dfrac{7}{3}$
Đáp án C.