Cho hàm số \(f(x)\) thỏa mãn \(f\left( x \right) + 2\sqrt x f'\left( x \right) = 3x{e^{ - \sqrt x }},\forall x \in \left[ {0; + \infty } \right).\) Giá trị \(f(1)\) bằng

* Hướng dẫn giải

Ta có: \(f\left( x \right) + 2\sqrt x f'\left( x \right) = 3x{e^{ - \sqrt x }},\forall x \in \left[ {0; + \infty } \right)\,\,\,\left( * \right)\)

\( \Rightarrow {e^{\sqrt x }}f\left( x \right) + 2\sqrt x {e^{\sqrt x }}f'\left( x \right) = 3x\) \( \Rightarrow \dfrac{{{e^{\sqrt x }}f\left( x \right)}}{{2\sqrt x }} + {e^{\sqrt x }}f'\left( x \right) = \dfrac{{3x}}{{2\sqrt x }}\) (với \(x > 0\))

\( \Rightarrow \left[ {{e^{\sqrt x }}f\left( x \right)} \right]' = \dfrac{{3\sqrt x }}{2} \Rightarrow \int\limits_0^1 {\left[ {{e^{\sqrt x }}f\left( x \right)} \right]'dx}  = \int\limits_0^1 {\dfrac{{3\sqrt x }}{2}dx} \)

\( \Rightarrow \left. {\left[ {{e^{\sqrt x }}f\left( x \right)} \right]} \right|_0^1 = \left. {\left( {{{\sqrt x }^3}} \right)} \right|_0^1 \Rightarrow e.f\left( 1 \right) - f\left( 0 \right) = 1\)

Mà từ \(\left( * \right)\) ta có: \(f\left( 0 \right) = 0\) nên \(e.f\left( 1 \right) = 1 \Rightarrow f\left( 1 \right) = \dfrac{1}{e}\).

Chọn C.