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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\left[ 0;4 \right]$ thỏa mãn ${f}''\left( x \right)f\left( x \right)+\dfrac{{{f}^{2}}\left( x \right)}{\sqrt{{{\left( 2x+1 \right)}^{3}}}}={{\left( {f}'\left( x \right) \right)}^{2}}$ 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. $2e$.
C. ${{e}^{3}}$.
D. ${{e}^{2}}+1$.
A. ${{e}^{2}}$.
B. $2e$.
C. ${{e}^{3}}$.
D. ${{e}^{2}}+1$.
Ta có: ${f}''\left( x \right).f\left( x \right)+\dfrac{{{f}^{2}}\left( x \right)}{\sqrt{{{\left( 2x+1 \right)}^{3}}}}={{\left( {f}'\left( x \right) \right)}^{2}}\Leftrightarrow {f}''\left( x \right).f\left( x \right)-{{\left( {f}'\left( x \right) \right)}^{2}}=-\dfrac{{{f}^{2}}\left( x \right)}{\sqrt{{{\left( 2x+1 \right)}^{3}}}}$
$\Leftrightarrow \dfrac{{f}''\left( x \right).f\left( x \right)-{{\left( {f}'\left( x \right) \right)}^{2}}}{{{f}^{2}}\left( x \right)}=-\dfrac{1}{\sqrt{{{\left( 2x+1 \right)}^{3}}}}\Leftrightarrow {{\left( \dfrac{{f}'\left( x \right)}{f\left( x \right)} \right)}^{\prime }}=-\dfrac{1}{\sqrt{{{\left( 2x+1 \right)}^{3}}}}$
$\Leftrightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=-\int{\dfrac{dx}{\sqrt{{{\left( 2x+1 \right)}^{3}}}}\Leftrightarrow }\dfrac{{f}'\left( x \right)}{f\left( x \right)}=-\int{{{\left( 2x+1 \right)}^{-\dfrac{3}{2}}}dx}\Leftrightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=\dfrac{1}{\sqrt{2x+1}}+{{C}_{1}}$
Thay $x=0$ ta được: ${{C}_{1}}=0\Rightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=\dfrac{1}{\sqrt{2x+1}}\Rightarrow \int{\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx}=\int{\dfrac{dx}{\sqrt{2x+1}}}$
$\Leftrightarrow \ln \left( f\left( x \right) \right)=\sqrt{2x+1}+{{C}_{2}}$.
Thay $x=0$ ta được: ${{C}_{2}}=-1\Rightarrow \ln \left( f\left( x \right) \right)=\sqrt{2x+1}-1$.
Thay $x=4$ ta được $\ln \left( f\left( 4 \right) \right)=2\Rightarrow f\left( 4 \right)={{e}^{2}}$.
$\Leftrightarrow \dfrac{{f}''\left( x \right).f\left( x \right)-{{\left( {f}'\left( x \right) \right)}^{2}}}{{{f}^{2}}\left( x \right)}=-\dfrac{1}{\sqrt{{{\left( 2x+1 \right)}^{3}}}}\Leftrightarrow {{\left( \dfrac{{f}'\left( x \right)}{f\left( x \right)} \right)}^{\prime }}=-\dfrac{1}{\sqrt{{{\left( 2x+1 \right)}^{3}}}}$
$\Leftrightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=-\int{\dfrac{dx}{\sqrt{{{\left( 2x+1 \right)}^{3}}}}\Leftrightarrow }\dfrac{{f}'\left( x \right)}{f\left( x \right)}=-\int{{{\left( 2x+1 \right)}^{-\dfrac{3}{2}}}dx}\Leftrightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=\dfrac{1}{\sqrt{2x+1}}+{{C}_{1}}$
Thay $x=0$ ta được: ${{C}_{1}}=0\Rightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=\dfrac{1}{\sqrt{2x+1}}\Rightarrow \int{\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx}=\int{\dfrac{dx}{\sqrt{2x+1}}}$
$\Leftrightarrow \ln \left( f\left( x \right) \right)=\sqrt{2x+1}+{{C}_{2}}$.
Thay $x=0$ ta được: ${{C}_{2}}=-1\Rightarrow \ln \left( f\left( x \right) \right)=\sqrt{2x+1}-1$.
Thay $x=4$ ta được $\ln \left( f\left( 4 \right) \right)=2\Rightarrow f\left( 4 \right)={{e}^{2}}$.
Đáp án A.