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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\left( 0;+\infty \right)$ thỏa mãn $2xf'\left( x \right)+f\left( x \right)=3{{x}^{2}}\sqrt{x}, \forall x\in \left( 0;+\infty \right)$. Biết $f\left( 1 \right)=\dfrac{1}{2}$, tính $f\left( 4 \right)$.
A. $14$.
B. $4$.
C. $24$.
D. $16$.
A. $14$.
B. $4$.
C. $24$.
D. $16$.
$2xf'\left( x \right)+f\left( x \right)=3{{x}^{2}}\sqrt{x}\Leftrightarrow \sqrt{x}f'\left( x \right)+\dfrac{1}{2\sqrt{x}}f\left( x \right)=\dfrac{3}{2}{{x}^{2}}\Leftrightarrow \left( \sqrt{x}f\left( x \right) \right)'=\dfrac{3}{2}{{x}^{2}}$
Suy ra $\left. \left( \sqrt{x}f\left( x \right) \right) \right|_{1}^{4}=\int\limits_{1}^{4}{\dfrac{3}{2}{{x}^{2}}dx}$ hay $2f\left( 4 \right)-f\left( 1 \right)=\dfrac{63}{2}\Leftrightarrow f\left( 4 \right)=16$
Suy ra $\left. \left( \sqrt{x}f\left( x \right) \right) \right|_{1}^{4}=\int\limits_{1}^{4}{\dfrac{3}{2}{{x}^{2}}dx}$ hay $2f\left( 4 \right)-f\left( 1 \right)=\dfrac{63}{2}\Leftrightarrow f\left( 4 \right)=16$
Đáp án D.