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Câu hỏi: Cho hàm số $f\left( x \right)$ đồng biến trên [0; 2] thỏa mãn điều kiện ${{f}^{2}}\left( x \right)-f\left( x \right).f''\left( x \right)+{{\left[ f'\left( x \right) \right]}^{2}}$ và $f\left( 0 \right)=1,f\left( 2 \right)={{e}^{6}}$. Giá trị của $\ln f\left( 4 \right)$ bằng
A. 16.
B. 7.
C. 14.
D. 12.
A. 16.
B. 7.
C. 14.
D. 12.
${{f}^{2}}\left( x \right)-f\left( x \right)f''\left( x \right)+{{\left[ f'\left( x \right) \right]}^{2}}=0\Leftrightarrow 1=\dfrac{f\left( x \right)f''\left( x \right)-{{\left[ f'\left( x \right) \right]}^{2}}}{{{f}^{2}}\left( x \right)}$
$\Leftrightarrow 1=\dfrac{\left[ f'\left( x \right) \right]'.f\left( x \right)-f'\left( x \right)\left[ f\left( x \right) \right]'}{{{\left[ f\left( x \right) \right]}^{2}}}=\dfrac{u'v-uv'}{{{v}^{2}}}\Leftrightarrow 1=\left[ \dfrac{f'\left( x \right)}{f\left( x \right)} \right]$ '
$\Rightarrow \int{1dx}=\int{\left( \dfrac{f'\left( x \right)}{f\left( x \right)} \right)'dx}\Leftrightarrow \dfrac{f'\left( x \right)}{f\left( x \right)}=x+{{C}_{1}}$
$\Rightarrow \int{\dfrac{f'\left( x \right)}{f\left( x \right)}dx}=\int{\left( x+{{C}_{1}} \right)dx}\Leftrightarrow \int{\dfrac{d\left( f\left( x \right) \right)}{f\left( x \right)}=\int{\left( x+{{C}_{1}} \right)dx}}$
$\Leftrightarrow \ln \left| f\left( x \right) \right|=\dfrac{{{x}^{2}}}{2}+{{C}_{1}}x+{{C}_{2}}.$
Ta có $\left\{ \begin{aligned}
& f\left( 0 \right)=1 \\
& f\left( 2 \right)={{e}^{6}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \ln \left| f\left( 0 \right) \right|=0 \\
& \ln \left| f\left( 2 \right) \right|=6 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{C}_{2}}=0 \\
& 2+2{{C}_{1}}+{{C}_{2}}=6 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{C}_{1}}=2 \\
& {{C}_{2}}=0 \\
\end{aligned} \right.$
Do hàm $f\left( x \right)$ đồng biến và $f\left( 0 \right)=1\Rightarrow f\left( x \right)>0,\forall x>0$
$\Rightarrow \ln \left[ f\left( x \right) \right]=\dfrac{{{x}^{2}}}{2}+2x\Rightarrow \ln \left[ f\left( 4 \right) \right]=\dfrac{{{4}^{2}}}{2}+2.4=16$.
$\Leftrightarrow 1=\dfrac{\left[ f'\left( x \right) \right]'.f\left( x \right)-f'\left( x \right)\left[ f\left( x \right) \right]'}{{{\left[ f\left( x \right) \right]}^{2}}}=\dfrac{u'v-uv'}{{{v}^{2}}}\Leftrightarrow 1=\left[ \dfrac{f'\left( x \right)}{f\left( x \right)} \right]$ '
$\Rightarrow \int{1dx}=\int{\left( \dfrac{f'\left( x \right)}{f\left( x \right)} \right)'dx}\Leftrightarrow \dfrac{f'\left( x \right)}{f\left( x \right)}=x+{{C}_{1}}$
$\Rightarrow \int{\dfrac{f'\left( x \right)}{f\left( x \right)}dx}=\int{\left( x+{{C}_{1}} \right)dx}\Leftrightarrow \int{\dfrac{d\left( f\left( x \right) \right)}{f\left( x \right)}=\int{\left( x+{{C}_{1}} \right)dx}}$
$\Leftrightarrow \ln \left| f\left( x \right) \right|=\dfrac{{{x}^{2}}}{2}+{{C}_{1}}x+{{C}_{2}}.$
Ta có $\left\{ \begin{aligned}
& f\left( 0 \right)=1 \\
& f\left( 2 \right)={{e}^{6}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \ln \left| f\left( 0 \right) \right|=0 \\
& \ln \left| f\left( 2 \right) \right|=6 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{C}_{2}}=0 \\
& 2+2{{C}_{1}}+{{C}_{2}}=6 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{C}_{1}}=2 \\
& {{C}_{2}}=0 \\
\end{aligned} \right.$
Do hàm $f\left( x \right)$ đồng biến và $f\left( 0 \right)=1\Rightarrow f\left( x \right)>0,\forall x>0$
$\Rightarrow \ln \left[ f\left( x \right) \right]=\dfrac{{{x}^{2}}}{2}+2x\Rightarrow \ln \left[ f\left( 4 \right) \right]=\dfrac{{{4}^{2}}}{2}+2.4=16$.
Đáp án A.