Cho hàm số \(f\left( x \right)\) 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 \(...

* Hướng dẫn giải

Ta có: \({x^2}f'\left( x \right) = f\left( x \right) + f'\left( x \right),\,\forall x \ne  \pm 1\,\, \Rightarrow f'\left( x \right).\left( {{x^2} - 1} \right) = f\left( x \right) \Leftrightarrow \dfrac{{f'\left( x \right)}}{{f\left( x \right)}} = \dfrac{1}{{{x^2} - 1}}\)

\( \Rightarrow \int\limits_0^{\frac{1}{2}} {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}} dx = \int\limits_0^{\frac{1}{2}} {\dfrac{1}{{{x^2} - 1}}dx}  \Leftrightarrow \left. {\ln \left| {f\left( x \right)} \right|} \right|_0^{\frac{1}{2}} = \frac{1}{2}\left. {\ln \left| {\dfrac{{x - 1}}{{x + 1}}} \right|} \right|_0^{\frac{1}{2}} \Leftrightarrow \ln \left| {f\left( {\frac{1}{2}} \right)} \right| - \ln \left| e \right| = \dfrac{1}{2}\left( {\ln \dfrac{1}{3} - \ln 1} \right)\)\( \Leftrightarrow \ln \left| {f\left( {\frac{1}{2}} \right)} \right| - 1 =  - \dfrac{1}{2}\ln 3 \Leftrightarrow \ln \left| {f\left( {\dfrac{1}{2}} \right)} \right| = \ln \dfrac{e}{{\sqrt 3 }} \Leftrightarrow \left| {f\left( {\dfrac{1}{2}} \right)} \right| = \dfrac{e}{{\sqrt 3 }} \Rightarrow f\left( {\dfrac{1}{2}} \right) = \dfrac{e}{{\sqrt 3 }}\) (do hàm số \(f\left( x \right)\) dương)

Chọn: D