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Câu hỏi: . Cho hàm số f(x) dương thỏa mãn $f\left( 0 \right)=e$ và ${{x}^{2}}{f}'\left( x \right)=f\left( x \right)+{f}'\left( x \right),\forall x\ne \pm 1.$ Giá trị $f\left( \dfrac{1}{2} \right)$ là
A. ${{e}^{\sqrt{3}}}.$
B. $e\sqrt{3}.$
C. ${{e}^{2}}.$
D. $\dfrac{e}{\sqrt{3}}.$
A. ${{e}^{\sqrt{3}}}.$
B. $e\sqrt{3}.$
C. ${{e}^{2}}.$
D. $\dfrac{e}{\sqrt{3}}.$
Với $f\left( x \right)>0,\forall x\ne \pm 1$, ta có ${{x}^{2}}{f}'\left( x \right)=f\left( x \right)+{f}'\left( x \right)\Leftrightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=\dfrac{1}{{{x}^{2}}-1}$.
Suy ra $\int{\dfrac{{f}'\left( x \right)}{f\left( x \right)}d\text{x}}=\int{\dfrac{d\text{x}}{{{x}^{2}}-1}}\Leftrightarrow \ln f\left( x \right)=\dfrac{1}{2}\ln \left| \dfrac{x-1}{x+1} \right|+C$.
Xét trên khoảng $\left( -1;1 \right)$, ta có $\ln f\left( x \right)=\dfrac{1}{2}\ln \dfrac{1-x}{x+1}+C$.
Do $f\left( 0 \right)=e\Rightarrow C=1$. Do đó $\ln f\left( x \right)=\dfrac{1}{2}\ln \dfrac{1-x}{x+1}+1\Leftrightarrow f\left( x \right)=e\sqrt{\dfrac{1-x}{x+1}}\Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{e}{\sqrt{3}}$.
Suy ra $\int{\dfrac{{f}'\left( x \right)}{f\left( x \right)}d\text{x}}=\int{\dfrac{d\text{x}}{{{x}^{2}}-1}}\Leftrightarrow \ln f\left( x \right)=\dfrac{1}{2}\ln \left| \dfrac{x-1}{x+1} \right|+C$.
Xét trên khoảng $\left( -1;1 \right)$, ta có $\ln f\left( x \right)=\dfrac{1}{2}\ln \dfrac{1-x}{x+1}+C$.
Do $f\left( 0 \right)=e\Rightarrow C=1$. Do đó $\ln f\left( x \right)=\dfrac{1}{2}\ln \dfrac{1-x}{x+1}+1\Leftrightarrow f\left( x \right)=e\sqrt{\dfrac{1-x}{x+1}}\Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{e}{\sqrt{3}}$.
Đáp án D.