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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên đoạn $\left[ 0;4 \right]$ thỏa mãn $f''\left( x \right)f\left( x \right)+\dfrac{{{\left[ f\left( x \right) \right]}^{2}}}{\sqrt{{{\left( 2\text{x}+1 \right)}^{3}}}}={{\left[ f'\left( x \right) \right]}^{2}}$ và $f\left( x \right)>0$ với mọi $x\in \left[ 0;4 \right].$ Biết rằng $f'\left( 0 \right)=f\left( 0 \right)=1,$ giá trị của $f\left( 4 \right)$ bằng
A. ${{e}^{2}}$
B. $2\text{e}$
C. ${{e}^{3}}$
D. ${{e}^{2}}+1$
A. ${{e}^{2}}$
B. $2\text{e}$
C. ${{e}^{3}}$
D. ${{e}^{2}}+1$
Ta có: $f''\left( x \right)f\left( x \right)+\dfrac{{{\left[ f\left( x \right) \right]}^{2}}}{\sqrt{{{\left( 2\text{x}+1 \right)}^{3}}}}={{\left[ f'\left( x \right) \right]}^{2}}\Leftrightarrow f''\left( x \right)f\left( x \right)-{{\left[ f'\left( x \right) \right]}^{2}}=-\dfrac{{{\left[ f\left( x \right) \right]}^{2}}}{\sqrt{{{\left( 2\text{x}+1 \right)}^{3}}}}$
$\Leftrightarrow \dfrac{f''\left( x \right)f\left( x \right)-{{\left[ f'\left( x \right) \right]}^{2}}}{{{\left[ f\left( x \right) \right]}^{2}}}=-\dfrac{1}{\sqrt{{{\left( 2\text{x}+1 \right)}^{3}}}}\Leftrightarrow {{\left( \dfrac{f'\left( x \right)}{f\left( x \right)} \right)}^{'}}=-\dfrac{1}{\sqrt{{{\left( 2\text{x}+1 \right)}^{3}}}}$
$\Leftrightarrow \dfrac{f'\left( x \right)}{f\left( x \right)}=-\int{\dfrac{1}{\sqrt{{{\left( 2\text{x}+1 \right)}^{3}}}}}dx\Leftrightarrow \dfrac{f'\left( x \right)}{f\left( x \right)}=-\int{{{\left( 2\text{x}+1 \right)}^{\dfrac{-3}{2}}}}dx\Leftrightarrow \dfrac{f'\left( x \right)}{f\left( x \right)}=\dfrac{1}{\sqrt{2\text{x}+1}}+{{C}_{1}}$
Thay $x=0$ ta được ${{C}_{1}}=0$
$\Rightarrow \dfrac{f'\left( x \right)}{f\left( x \right)}=\dfrac{1}{\sqrt{2\text{x}+1}}\Rightarrow \int\limits_{{}}^{{}}{\dfrac{f'\left( x \right)}{f\left( x \right)}dx}=\int{\dfrac{dx}{\sqrt{2\text{x}+1}}}\Leftrightarrow \ln \left[ f\left( x \right) \right]=\sqrt{2\text{x}+1}+{{C}_{2}}$
Thay $x=0$ ta được ${{C}_{2}}=-1.$
$\Rightarrow \ln \left[ f\left( x \right) \right]=\sqrt{2\text{x}+1}-1$
Thay $x=4$ ta được $\ln \left[ f\left( 4 \right) \right]=2\Rightarrow f\left( 4 \right)={{e}^{2}}.$
$\Leftrightarrow \dfrac{f''\left( x \right)f\left( x \right)-{{\left[ f'\left( x \right) \right]}^{2}}}{{{\left[ f\left( x \right) \right]}^{2}}}=-\dfrac{1}{\sqrt{{{\left( 2\text{x}+1 \right)}^{3}}}}\Leftrightarrow {{\left( \dfrac{f'\left( x \right)}{f\left( x \right)} \right)}^{'}}=-\dfrac{1}{\sqrt{{{\left( 2\text{x}+1 \right)}^{3}}}}$
$\Leftrightarrow \dfrac{f'\left( x \right)}{f\left( x \right)}=-\int{\dfrac{1}{\sqrt{{{\left( 2\text{x}+1 \right)}^{3}}}}}dx\Leftrightarrow \dfrac{f'\left( x \right)}{f\left( x \right)}=-\int{{{\left( 2\text{x}+1 \right)}^{\dfrac{-3}{2}}}}dx\Leftrightarrow \dfrac{f'\left( x \right)}{f\left( x \right)}=\dfrac{1}{\sqrt{2\text{x}+1}}+{{C}_{1}}$
Thay $x=0$ ta được ${{C}_{1}}=0$
$\Rightarrow \dfrac{f'\left( x \right)}{f\left( x \right)}=\dfrac{1}{\sqrt{2\text{x}+1}}\Rightarrow \int\limits_{{}}^{{}}{\dfrac{f'\left( x \right)}{f\left( x \right)}dx}=\int{\dfrac{dx}{\sqrt{2\text{x}+1}}}\Leftrightarrow \ln \left[ f\left( x \right) \right]=\sqrt{2\text{x}+1}+{{C}_{2}}$
Thay $x=0$ ta được ${{C}_{2}}=-1.$
$\Rightarrow \ln \left[ f\left( x \right) \right]=\sqrt{2\text{x}+1}-1$
Thay $x=4$ ta được $\ln \left[ f\left( 4 \right) \right]=2\Rightarrow f\left( 4 \right)={{e}^{2}}.$
Đáp án A.