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Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( 1 \right)=1$ và $2x.{f}'\left( x \right)-f\left( x \right)=2\left( {{x}^{3}}+{{x}^{2}} \right)\sqrt{x}$, $\forall x>0$. Giá trị của $f\left( 4 \right)$ bằng
A. $59$.
B. $58$.
C. $56$.
D. $57$.
A. $59$.
B. $58$.
C. $56$.
D. $57$.
Với mọi $x>0$, ta có:
$2x{f}'\left( x \right)-f\left( x \right)=2\left( {{x}^{3}}+{{x}^{2}} \right)\sqrt{x}$ $\Leftrightarrow \dfrac{\sqrt{x}{f}'\left( x \right)-\dfrac{f\left( x \right)}{2\sqrt{x}}}{x}={{x}^{2}}+x$ $\Leftrightarrow {{\left[ \dfrac{f\left( x \right)}{\sqrt{x}} \right]}^{\prime }}={{x}^{2}}+x$
$\Rightarrow \int\limits_{1}^{4}{{{\left[ \dfrac{f\left( x \right)}{\sqrt{x}} \right]}^{\prime }}\text{d}x}=\int\limits_{1}^{4}{\left( {{x}^{2}}+x \right)\text{d}x}$ $\Leftrightarrow \left. \dfrac{f\left( x \right)}{\sqrt{x}} \right|_{1}^{4}=\left. \left( \dfrac{1}{3}{{x}^{3}}+\dfrac{1}{2}{{x}^{2}} \right) \right|_{1}^{4}$ $\Leftrightarrow \dfrac{f\left( 4 \right)}{2}-f\left( 1 \right)=\dfrac{57}{2}$
$\Leftrightarrow f\left( 4 \right)=57+2f\left( 1 \right)$ $=59$.
$2x{f}'\left( x \right)-f\left( x \right)=2\left( {{x}^{3}}+{{x}^{2}} \right)\sqrt{x}$ $\Leftrightarrow \dfrac{\sqrt{x}{f}'\left( x \right)-\dfrac{f\left( x \right)}{2\sqrt{x}}}{x}={{x}^{2}}+x$ $\Leftrightarrow {{\left[ \dfrac{f\left( x \right)}{\sqrt{x}} \right]}^{\prime }}={{x}^{2}}+x$
$\Rightarrow \int\limits_{1}^{4}{{{\left[ \dfrac{f\left( x \right)}{\sqrt{x}} \right]}^{\prime }}\text{d}x}=\int\limits_{1}^{4}{\left( {{x}^{2}}+x \right)\text{d}x}$ $\Leftrightarrow \left. \dfrac{f\left( x \right)}{\sqrt{x}} \right|_{1}^{4}=\left. \left( \dfrac{1}{3}{{x}^{3}}+\dfrac{1}{2}{{x}^{2}} \right) \right|_{1}^{4}$ $\Leftrightarrow \dfrac{f\left( 4 \right)}{2}-f\left( 1 \right)=\dfrac{57}{2}$
$\Leftrightarrow f\left( 4 \right)=57+2f\left( 1 \right)$ $=59$.
Đáp án A.