(THPT Yên Lạc – Vĩnh Phúc – 2022) Cho hàm số \(y = f\left( x \right)\) liên tục trên \(\left( {0; + \infty } \right)\) thỏa mãn \(2x.f’\left( x \right) + f\left( x \right) = 3{x^2}\sqrt x ,\forall x \in \left( {0; + \infty } \right)\). Biết \(f\left( 1 \right) = \frac{1}{2}\), tính \(f\left( 4 \right)\).

A. \(14\).

B. \(4\).

C. \(24\).

D. \(16\).

Lời giải:

Chọn D

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

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

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

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

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

Thay \(x = 1\) vào \(\left( * \right)\) 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 \(f\left( 4 \right) = 16\).