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Câu hỏi: Cho hàm số $f\left( x \right)$ nhận giá trị dương trên khoảng $\left( 0;+\infty \right)$ có đạo hàm trên khoảng đó và thỏa mãn $f\left( x \right)\ln f\left( x \right)=x\left( 2f\left( x \right)-f'\left( x \right) \right), \forall x\in \left( 0;+\infty \right)$. Biết $f\left( 1 \right)=f\left( 3 \right)$, giá trị $f\left( 2 \right)$ thuộc khoảng nào dưới đây?
A. $\left( 40;42 \right)$.
B. $\left( 3;5 \right)$.
C. $\left( 32;34 \right)$.
D. $\left( 1;3 \right)$.
A. $\left( 40;42 \right)$.
B. $\left( 3;5 \right)$.
C. $\left( 32;34 \right)$.
D. $\left( 1;3 \right)$.
Ta có: $\forall x\in \left( 0;+\infty \right)$
$f\left( x \right)\ln f\left( x \right)=x\left( 2f\left( x \right)-f'\left( x \right) \right)$
$\Rightarrow f\left( x \right)\ln f\left( x \right)=2xf\left( x \right)-xf'\left( x \right)$
$\Rightarrow f\left( x \right)\ln f\left( x \right)+xf'\left( x \right)=2xf\left( x \right)$
$\Rightarrow \ln f\left( x \right)+\dfrac{xf'\left( x \right)}{f\left( x \right)}=2x$
$\Rightarrow \left( x\ln f\left( x \right) \right)'=2x$
$\Rightarrow x\ln f\left( x \right)={{x}^{2}}+C$.
Có:
$\left\{ \begin{aligned}
& 1\ln f\left( 1 \right)=1+C \\
& 3\ln f\left( 3 \right)=9+C \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 3\ln f\left( 1 \right)=3+3C \\
& 3\ln f\left( 3 \right)=9+C \\
\end{aligned} \right.$$\Rightarrow 0=-6+2C\Rightarrow C=3$.
Vậy: $x\ln f\left( x \right)={{x}^{2}}+3\Rightarrow \ln f\left( x \right)=x+\dfrac{3}{x}$
$\Rightarrow f\left( x \right)={{e}^{x+\dfrac{3}{x}}}$
$f\left( 2 \right)={{e}^{2+\dfrac{3}{2}}}\approx 33,12$.
$f\left( x \right)\ln f\left( x \right)=x\left( 2f\left( x \right)-f'\left( x \right) \right)$
$\Rightarrow f\left( x \right)\ln f\left( x \right)=2xf\left( x \right)-xf'\left( x \right)$
$\Rightarrow f\left( x \right)\ln f\left( x \right)+xf'\left( x \right)=2xf\left( x \right)$
$\Rightarrow \ln f\left( x \right)+\dfrac{xf'\left( x \right)}{f\left( x \right)}=2x$
$\Rightarrow \left( x\ln f\left( x \right) \right)'=2x$
$\Rightarrow x\ln f\left( x \right)={{x}^{2}}+C$.
Có:
$\left\{ \begin{aligned}
& 1\ln f\left( 1 \right)=1+C \\
& 3\ln f\left( 3 \right)=9+C \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 3\ln f\left( 1 \right)=3+3C \\
& 3\ln f\left( 3 \right)=9+C \\
\end{aligned} \right.$$\Rightarrow 0=-6+2C\Rightarrow C=3$.
Vậy: $x\ln f\left( x \right)={{x}^{2}}+3\Rightarrow \ln f\left( x \right)=x+\dfrac{3}{x}$
$\Rightarrow f\left( x \right)={{e}^{x+\dfrac{3}{x}}}$
$f\left( 2 \right)={{e}^{2+\dfrac{3}{2}}}\approx 33,12$.
Đáp án C.