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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên $\left[ 1;2 \right]$ thỏa mãn $f\left( 1 \right)=4$ và $f\left( x \right)=x{f}'\left( x \right)-2{{x}^{3}}-3{{x}^{2}}.$ Tính giá trị $f\left( 2 \right)$
A. 5.
B. 20.
C. 10.
D. 15.
A. 5.
B. 20.
C. 10.
D. 15.
Ta có $f\left( x \right)=x{f}'\left( x \right)-2{{x}^{3}}-3{{x}^{2}}\Leftrightarrow \dfrac{x{f}'\left( x \right)-f\left( x \right)}{{{x}^{2}}}=2x+3$
$\Leftrightarrow {{\left[ \dfrac{f\left( x \right)}{x} \right]}^{\prime }}=2x+3\Leftrightarrow \int\limits_{1}^{2}{{{\left[ \dfrac{f\left( x \right)}{x} \right]}^{\prime }}}dx=\int\limits_{1}^{2}{\left( 2x+3 \right)dx}\Leftrightarrow \left. \dfrac{f\left( x \right)}{x} \right|_{1}^{2}=6$
$\Leftrightarrow \dfrac{f\left( 2 \right)}{2}-\dfrac{f\left( 1 \right)}{1}=6\Rightarrow f\left( 2 \right)=20.$
$\Leftrightarrow {{\left[ \dfrac{f\left( x \right)}{x} \right]}^{\prime }}=2x+3\Leftrightarrow \int\limits_{1}^{2}{{{\left[ \dfrac{f\left( x \right)}{x} \right]}^{\prime }}}dx=\int\limits_{1}^{2}{\left( 2x+3 \right)dx}\Leftrightarrow \left. \dfrac{f\left( x \right)}{x} \right|_{1}^{2}=6$
$\Leftrightarrow \dfrac{f\left( 2 \right)}{2}-\dfrac{f\left( 1 \right)}{1}=6\Rightarrow f\left( 2 \right)=20.$
Đáp án B.