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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\left( 0;+\infty \right)$ thỏa mãn $2x\cdot {f}'\left( x \right)+f\left( x \right)=4x\sqrt{x}$. Biết $f\left( 1 \right)=2$. Giá trị của $f\left( 4 \right)$ bằng:
A. $\dfrac{15}{4}$.
B. $\dfrac{17}{4}$.
C. $\dfrac{15}{2}$.
D. $\dfrac{17}{2}$.
A. $\dfrac{15}{4}$.
B. $\dfrac{17}{4}$.
C. $\dfrac{15}{2}$.
D. $\dfrac{17}{2}$.
Ta có: $2x\cdot {f}'\left( x \right)+f\left( x \right)=4x\sqrt{x}$ $\Leftrightarrow \sqrt{x}\cdot {f}'\left( x \right)+\dfrac{1}{2\sqrt{x}}\cdot f\left( x \right)=2x$ $\Leftrightarrow {{\left( \sqrt{x}\cdot f\left( x \right) \right)}^{\prime }}=2x$
$\Rightarrow \sqrt{x}\cdot f\left( x \right)={{x}^{2}}+C$
Mà $f\left( 1 \right)=2$ nên $\sqrt{1}\cdot 2={{1}^{2}}+C$ $\Leftrightarrow C=1$
$\Rightarrow \sqrt{x}\cdot f\left( x \right)={{x}^{2}}+1$ $\Leftrightarrow f\left( x \right)=x\sqrt{x}+\dfrac{1}{\sqrt{x}}$
$\Rightarrow f\left( 4 \right)=4\sqrt{4}+\dfrac{1}{\sqrt{4}}=\dfrac{17}{2}$
$\Rightarrow \sqrt{x}\cdot f\left( x \right)={{x}^{2}}+C$
Mà $f\left( 1 \right)=2$ nên $\sqrt{1}\cdot 2={{1}^{2}}+C$ $\Leftrightarrow C=1$
$\Rightarrow \sqrt{x}\cdot f\left( x \right)={{x}^{2}}+1$ $\Leftrightarrow f\left( x \right)=x\sqrt{x}+\dfrac{1}{\sqrt{x}}$
$\Rightarrow f\left( 4 \right)=4\sqrt{4}+\dfrac{1}{\sqrt{4}}=\dfrac{17}{2}$
Đáp án D.