Cho hàm số \(y = f\left( x \right)\) liên tục trên \(\left( {0\,;\,\, + \infty } \right)\) thỏa mãn \(2x \cdot f'\left( x \right) + f\left( x \right) = 3{x^2}\sqrt x ,\,\,\forall x \in \left( {0\,;\,\, + \infty } \right).\) Biết \(f\left( 1 \right) = \frac{1}{2}\). Giá trị \(f\left( 4 \right)\) là

Ta có \(2x \cdot f'\left( x \right) + f\left( x \right) = 3{x^2}\sqrt x ,\,\,\forall x \in \left( {0\,;\,\, + \infty } \right).\)

\( \Leftrightarrow \sqrt x  \cdot f'\left( x \right) + \frac{1}{{2\sqrt x }}f\left( x \right) = \frac{{3{x^2}}}{2},\,\,\forall x \in \left( {0\,;\,\, + \infty } \right)\)\( \Leftrightarrow \sqrt x  \cdot f'\left( x \right) + {\left( {\sqrt x } \right)^\prime }.f\left( x \right) = \frac{{3{x^2}}}{2},\,\,\forall x \in \left( {0\,;\,\, + \infty } \right)\)

\( \Leftrightarrow {\left[ {\sqrt x  \cdot f\left( x \right)} \right]^\prime } = \frac{{3{x^2}}}{2},\,\,\forall x \in \left( {0\,;\,\, + \infty } \right)\)

\( \Leftrightarrow \sqrt x  \cdot f\left( x \right) = \int {\frac{{3{x^2}}}{2}\,} dx = \frac{{{x^3}}}{2} + C\)

\( \Leftrightarrow \sqrt x  \cdot f\left( x \right) = \frac{{{x^3}}}{2} + C\,\,\,(*)\)

Thay \(x = 1\) vào \((*)\) ta được: \(f\left( 1 \right) = \frac{1}{2} + C\)\( \Rightarrow C = f\left( 1 \right) - \frac{1}{2} = \frac{1}{2} - \frac{1}{2} = 0 \Rightarrow f\left( x \right) = \frac{{{x^3}}}{{2\sqrt x }}.\)

Vậy \(f\left( 4 \right) = 16.\) Chọn D.