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Câu hỏi: Cho hàm số f(x) thỏa mãn ${f}'\left( x \right)=\left( 2x+1 \right).{{\left[ f\left( x \right) \right]}^{2}}$ và $f\left( 2 \right)=-\dfrac{1}{3}.$ Giá trị của $f\left( 1 \right)$ bằng
A. $\dfrac{11}{3}.$
B. $\dfrac{13}{3}.$
C. $-1.$
D. 1.
A. $\dfrac{11}{3}.$
B. $\dfrac{13}{3}.$
C. $-1.$
D. 1.
Ta có $\dfrac{{f}'\left( x \right)}{{{\left[ f\left( x \right) \right]}^{2}}}=2x+1\Rightarrow \int\limits_{1}^{2}{\dfrac{{f}'\left( x \right)}{{{\left[ f\left( x \right) \right]}^{2}}}dx}=\int\limits_{1}^{2}{\left( 2x+1 \right)dx}$
$\Rightarrow \int\limits_{1}^{2}{\dfrac{1}{{{\left[ f\left( x \right) \right]}^{2}}}d\left[ f\left( x \right) \right]}=\left. \left( {{x}^{2}}+x \right) \right|_{1}^{2}\Rightarrow \left. -\dfrac{1}{f\left( x \right)} \right|_{1}^{2}=4\Rightarrow -\dfrac{1}{f\left( 2 \right)}+\dfrac{1}{f\left( 1 \right)}=4$.
Mà $f\left( 2 \right)=-\dfrac{1}{3}\Rightarrow f\left( 1 \right)=1$.
$\Rightarrow \int\limits_{1}^{2}{\dfrac{1}{{{\left[ f\left( x \right) \right]}^{2}}}d\left[ f\left( x \right) \right]}=\left. \left( {{x}^{2}}+x \right) \right|_{1}^{2}\Rightarrow \left. -\dfrac{1}{f\left( x \right)} \right|_{1}^{2}=4\Rightarrow -\dfrac{1}{f\left( 2 \right)}+\dfrac{1}{f\left( 1 \right)}=4$.
Mà $f\left( 2 \right)=-\dfrac{1}{3}\Rightarrow f\left( 1 \right)=1$.
Đáp án D.