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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn ${{e}^{3x}}\left( 4f\left( x \right)+f'\left( x \right) \right)=2\sqrt{f\left( x \right)},f\left( x \right)>0 \forall x\ge 0$ và $f\left( 0 \right)=1$. Tính
$\int\limits_{0}^{\ln 2}{f\left( x \right)} dx$
A. $\dfrac{201}{640}$.
B. $\dfrac{11}{24}$.
C. $\dfrac{209}{640}$.
D. $-\dfrac{1}{12}$.
$\int\limits_{0}^{\ln 2}{f\left( x \right)} dx$
A. $\dfrac{201}{640}$.
B. $\dfrac{11}{24}$.
C. $\dfrac{209}{640}$.
D. $-\dfrac{1}{12}$.
Ta có: ${{e}^{3x}}\left( 4f\left( x \right)+f'\left( x \right) \right)=2\sqrt{f\left( x \right)}$ ${{e}^{3x}}\left( 4f\left( x \right)+f'\left( x \right) \right)=2\sqrt{f\left( x \right)}$ $\begin{aligned}
& \Leftrightarrow {{e}^{3x}}\left( 4\dfrac{f\left( x \right)}{2\sqrt{f\left( x \right)}}+\dfrac{f'\left( x \right)}{2\sqrt{f\left( x \right)}} \right)=1 \\
& \Leftrightarrow {{e}^{3x}}\left( 2\sqrt{f\left( x \right)}+\dfrac{f'\left( x \right)}{2\sqrt{f\left( x \right)}} \right)=1 \\
& \Leftrightarrow 2{{e}^{2x}}\sqrt{f\left( x \right)}+\dfrac{{{e}^{2x}}.f'\left( x \right)}{2\sqrt{f\left( x \right)}}={{e}^{-x}} \\
& \Leftrightarrow \left( {{e}^{2x}}\sqrt{f\left( x \right)} \right)'={{e}^{-x}} \\
& \Leftrightarrow {{e}^{2x}}\sqrt{f\left( x \right)}=\int{{{e}^{-x}}}dx \\
& \Leftrightarrow {{e}^{2x}}\sqrt{f\left( x \right)}=-{{e}^{-x}}+C \\
\end{aligned}$
Vì $f\left( 0 \right)=1$ nên $\Leftrightarrow {{e}^{0}}\sqrt{f\left( 0 \right)}=-{{e}^{0}}+C\Leftrightarrow C=2$
Suy ra $\sqrt{f\left( x \right)}=-{{e}^{-3x}}+2{{e}^{-2x}}\Leftrightarrow f\left( x \right)={{\left( -{{e}^{-3x}}+2{{e}^{-2x}} \right)}^{2}}$
$\int\limits_{0}^{\ln 2}{f\left( x \right)} dx=\int\limits_{0}^{\ln 2}{{{\left( -{{e}^{-3x}}+2{{e}^{-2x}} \right)}^{2}}}dx=\dfrac{209}{640}$.
& \Leftrightarrow {{e}^{3x}}\left( 4\dfrac{f\left( x \right)}{2\sqrt{f\left( x \right)}}+\dfrac{f'\left( x \right)}{2\sqrt{f\left( x \right)}} \right)=1 \\
& \Leftrightarrow {{e}^{3x}}\left( 2\sqrt{f\left( x \right)}+\dfrac{f'\left( x \right)}{2\sqrt{f\left( x \right)}} \right)=1 \\
& \Leftrightarrow 2{{e}^{2x}}\sqrt{f\left( x \right)}+\dfrac{{{e}^{2x}}.f'\left( x \right)}{2\sqrt{f\left( x \right)}}={{e}^{-x}} \\
& \Leftrightarrow \left( {{e}^{2x}}\sqrt{f\left( x \right)} \right)'={{e}^{-x}} \\
& \Leftrightarrow {{e}^{2x}}\sqrt{f\left( x \right)}=\int{{{e}^{-x}}}dx \\
& \Leftrightarrow {{e}^{2x}}\sqrt{f\left( x \right)}=-{{e}^{-x}}+C \\
\end{aligned}$
Vì $f\left( 0 \right)=1$ nên $\Leftrightarrow {{e}^{0}}\sqrt{f\left( 0 \right)}=-{{e}^{0}}+C\Leftrightarrow C=2$
Suy ra $\sqrt{f\left( x \right)}=-{{e}^{-3x}}+2{{e}^{-2x}}\Leftrightarrow f\left( x \right)={{\left( -{{e}^{-3x}}+2{{e}^{-2x}} \right)}^{2}}$
$\int\limits_{0}^{\ln 2}{f\left( x \right)} dx=\int\limits_{0}^{\ln 2}{{{\left( -{{e}^{-3x}}+2{{e}^{-2x}} \right)}^{2}}}dx=\dfrac{209}{640}$.
Đáp án C.