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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm trên hoảng $\left( 0 ; +\infty \right)$ và thỏa mãn: $2{{x}^{2}}+f\left( x \right)=2x{f}'\left( x \right)$. Cho biết $f\left( 1 \right)=\dfrac{5}{3}$, giá trị $f\left( 4 \right)$ bằng
A. $1$.
B. $\dfrac{38}{3}$.
C. $53$.
D. $\dfrac{187}{2}$.
A. $1$.
B. $\dfrac{38}{3}$.
C. $53$.
D. $\dfrac{187}{2}$.
Với $x>0$ ta có
$2{{x}^{2}}+f\left( x \right)=2x{f}'\left( x \right)$ $\Leftrightarrow 2x{f}'\left( x \right)-f\left( x \right)=2{{x}^{2}}$ $\Leftrightarrow {f}'\left( x \right)-\dfrac{1}{2x}f\left( x \right)=x$
$\Leftrightarrow \dfrac{1}{\sqrt{x}}{f}'\left( x \right)-\dfrac{1}{2x\sqrt{x}}f\left( x \right)=\sqrt{x}$ $\Leftrightarrow {{\left( \dfrac{1}{\sqrt{x}}.f\left( x \right) \right)}^{\prime }}=\sqrt{x}$
Nguyên hàm hai vế ta được:
$\int{{{\left( \dfrac{1}{\sqrt{x}}.f\left( x \right) \right)}^{\prime }}dx}=\int{\sqrt{x}}dx\Leftrightarrow \dfrac{1}{\sqrt{x}}f\left( x \right)=\dfrac{2}{3}x\sqrt{x}+C$
$f\left( 1 \right)=\dfrac{5}{3}\Rightarrow \dfrac{1}{\sqrt{1}}f\left( 1 \right)=\dfrac{2}{3}+C\Leftrightarrow C=\dfrac{5}{3}-\dfrac{2}{3}=1$.
Suy ra: $\dfrac{1}{\sqrt{4}}f\left( 4 \right)=\dfrac{2}{3}.4\sqrt{4}+1\Leftrightarrow f\left( 4 \right)=\dfrac{38}{3}$.
$2{{x}^{2}}+f\left( x \right)=2x{f}'\left( x \right)$ $\Leftrightarrow 2x{f}'\left( x \right)-f\left( x \right)=2{{x}^{2}}$ $\Leftrightarrow {f}'\left( x \right)-\dfrac{1}{2x}f\left( x \right)=x$
$\Leftrightarrow \dfrac{1}{\sqrt{x}}{f}'\left( x \right)-\dfrac{1}{2x\sqrt{x}}f\left( x \right)=\sqrt{x}$ $\Leftrightarrow {{\left( \dfrac{1}{\sqrt{x}}.f\left( x \right) \right)}^{\prime }}=\sqrt{x}$
Nguyên hàm hai vế ta được:
$\int{{{\left( \dfrac{1}{\sqrt{x}}.f\left( x \right) \right)}^{\prime }}dx}=\int{\sqrt{x}}dx\Leftrightarrow \dfrac{1}{\sqrt{x}}f\left( x \right)=\dfrac{2}{3}x\sqrt{x}+C$
$f\left( 1 \right)=\dfrac{5}{3}\Rightarrow \dfrac{1}{\sqrt{1}}f\left( 1 \right)=\dfrac{2}{3}+C\Leftrightarrow C=\dfrac{5}{3}-\dfrac{2}{3}=1$.
Suy ra: $\dfrac{1}{\sqrt{4}}f\left( 4 \right)=\dfrac{2}{3}.4\sqrt{4}+1\Leftrightarrow f\left( 4 \right)=\dfrac{38}{3}$.
Đáp án B.