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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( 2 \right)=-\dfrac{1}{3}$ và ${f}'\left( x \right)=x{{\left[ f\left( x \right) \right]}^{2}}$ với mọi $x\in \mathbb{R}$. Giá trị của $f\left( 1 \right)$ bằng
A. $-\dfrac{11}{2}$.
B. $-\dfrac{2}{3}$.
C. $-\dfrac{2}{9}$.
D. $-\dfrac{7}{6}$.
A. $-\dfrac{11}{2}$.
B. $-\dfrac{2}{3}$.
C. $-\dfrac{2}{9}$.
D. $-\dfrac{7}{6}$.
Ta có ${f}'\left( x \right)=x{{\left[ f\left( x \right) \right]}^{2}}\Leftrightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=x$. Do đó
$\int\limits_{{}}^{{}}{\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx}=\int\limits_{{}}^{{}}{xdx}\Leftrightarrow \int\limits_{{}}^{{}}{d\left( \dfrac{1}{f\left( x \right)} \right)}=\int\limits_{{}}^{{}}{xdx}\Leftrightarrow -\dfrac{1}{f\left( x \right)}=\dfrac{1}{2}{{x}^{2}}+C\Leftrightarrow f\left( x \right)=-\dfrac{1}{\dfrac{1}{2}{{x}^{2}}+C}$
Theo giả thiết $f\left( 2 \right)=-\dfrac{1}{4}\Rightarrow C=1\Rightarrow f\left( x \right)=-\dfrac{1}{\dfrac{1}{2}{{x}^{2}}+1}$. Từ đó suy ra $f\left( 1 \right)=-\dfrac{2}{3}$.
$\int\limits_{{}}^{{}}{\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx}=\int\limits_{{}}^{{}}{xdx}\Leftrightarrow \int\limits_{{}}^{{}}{d\left( \dfrac{1}{f\left( x \right)} \right)}=\int\limits_{{}}^{{}}{xdx}\Leftrightarrow -\dfrac{1}{f\left( x \right)}=\dfrac{1}{2}{{x}^{2}}+C\Leftrightarrow f\left( x \right)=-\dfrac{1}{\dfrac{1}{2}{{x}^{2}}+C}$
Theo giả thiết $f\left( 2 \right)=-\dfrac{1}{4}\Rightarrow C=1\Rightarrow f\left( x \right)=-\dfrac{1}{\dfrac{1}{2}{{x}^{2}}+1}$. Từ đó suy ra $f\left( 1 \right)=-\dfrac{2}{3}$.
Đáp án B.