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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( 1 \right)=3$ và $x\left( 4-f'\left( x \right) \right)=f\left( x \right)-1$ với mọi $x>0$. Tính $f\left( 2 \right)$.
A. $5$.
B. $2$.
C. $3$.
D. $6$.
A. $5$.
B. $2$.
C. $3$.
D. $6$.
Từ giả thiết $x\left( 4-f'\left( x \right) \right)=f\left( x \right)-1\Rightarrow x.{f}'\left( x \right)+f\left( x \right)=4x+1\Leftrightarrow {{\left[ xf\left( x \right) \right]}^{\prime }}=4x+1$.
$\Rightarrow \int\limits_{1}^{2}{{{\left[ xf\left( x \right) \right]}^{\prime }}\text{d}x}=\int\limits_{1}^{2}{\left( 4x+1 \right)\text{d}x}\Leftrightarrow \left. xf\left( x \right) \right|_{1}^{2}=\left. \left( 2{{x}^{2}}+x \right) \right|_{1}^{2}$.
$\Leftrightarrow 2f\left( 2 \right)-f\left( 1 \right)=7\Rightarrow f\left( 2 \right)=\dfrac{7+f\left( 1 \right)}{2}=\dfrac{7+3}{2}=5$.
$\Rightarrow \int\limits_{1}^{2}{{{\left[ xf\left( x \right) \right]}^{\prime }}\text{d}x}=\int\limits_{1}^{2}{\left( 4x+1 \right)\text{d}x}\Leftrightarrow \left. xf\left( x \right) \right|_{1}^{2}=\left. \left( 2{{x}^{2}}+x \right) \right|_{1}^{2}$.
$\Leftrightarrow 2f\left( 2 \right)-f\left( 1 \right)=7\Rightarrow f\left( 2 \right)=\dfrac{7+f\left( 1 \right)}{2}=\dfrac{7+3}{2}=5$.
Đáp án A.