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Câu hỏi: Cho hàm số $f\left( x \right)$ nhận giá trị dương trên $[0;1]$, có đạo hàm dương và liên tục trên $[0;1]$, thỏa mãn $f\left( 0 \right)=1$ và $\int\limits_{0}^{1}{\left[ {{f}^{3}}\left( x \right)+4{{\left[ f'\left( x \right) \right]}^{3}} \right]dx}\le 3\int\limits_{0}^{1}{f'\left( x \right).{{f}^{2}}\left( x \right)dx.}$ Tính $I=\int\limits_{0}^{1}{f\left( x \right)dx.}$
A. $I=2\left( \sqrt{e}-1 \right).$
B. $I=2\left( {{e}^{2}}-1 \right).$
C. $I=\dfrac{\sqrt{e}-1}{2}.$
D. $I=\dfrac{{{e}^{2}}-1}{2}.$
A. $I=2\left( \sqrt{e}-1 \right).$
B. $I=2\left( {{e}^{2}}-1 \right).$
C. $I=\dfrac{\sqrt{e}-1}{2}.$
D. $I=\dfrac{{{e}^{2}}-1}{2}.$
Áp dụng bất đẳng thức AM-GM cho ba số dương ta có
$\begin{aligned}
& {{f}^{3}}\left( x \right)+4{{[f'\left( x \right)]}^{3}}=4{{[f'\left( x \right)]}^{3}}+\dfrac{{{f}^{3}}\left( x \right)}{2}+\dfrac{{{f}^{3}}\left( x \right)}{2}\ge 3\sqrt[3]{4{{[f'\left( x \right)]}^{3}}.\dfrac{{{f}^{3}}\left( x \right)}{2}.\dfrac{{{f}^{3}}\left( x \right)}{2}} \\
& =3f'\left( x \right).{{f}^{2}}\left( x \right) \\
\end{aligned}$
Suy ra $\int\limits_{0}^{1}{\left[ {{f}^{3}}\left( x \right)+4{{[f'\left( x \right)]}^{3}} \right]dx}\ge 3\int\limits_{0}^{1}{f'\left( x \right).{{f}^{2}}\left( x \right)dx}$.
Mà $\int\limits_{0}^{1}{\left\{ {{f}^{3}}\left( x \right)+4{{[f'\left( x \right)]}^{3}} \right\}dx}\le 3\int\limits_{0}^{1}{f'\left( x \right).{{f}^{2}}\left( x \right)dx}$ nên dấu "=" xảy ra, tức là
$\begin{aligned}
& 4{{[f'\left( x \right)]}^{3}}=\dfrac{{{f}^{3}}\left( x \right)}{2}=\dfrac{{{f}^{3}}\left( x \right)}{2}\Leftrightarrow f'\left( x \right)=\dfrac{1}{2}f\left( x \right) \\
& \dfrac{f'\left( x \right)}{f\left( x \right)}=\dfrac{1}{2}\Rightarrow \int{\dfrac{f'\left( x \right)}{f\left( x \right)}dx}=\dfrac{1}{2}\int{dx}\Rightarrow \ln \left| f\left( x \right) \right|=\dfrac{1}{2}x+C\Rightarrow f\left( x \right)={{e}^{\dfrac{1}{2}x+C}} \\
\end{aligned}$
Theo giả thiết $f\left( 0 \right)=1\Rightarrow C=0\Rightarrow f\left( x \right)={{e}^{\dfrac{1}{2}x}}\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=2\left( \sqrt{e}-1 \right)$
$\begin{aligned}
& {{f}^{3}}\left( x \right)+4{{[f'\left( x \right)]}^{3}}=4{{[f'\left( x \right)]}^{3}}+\dfrac{{{f}^{3}}\left( x \right)}{2}+\dfrac{{{f}^{3}}\left( x \right)}{2}\ge 3\sqrt[3]{4{{[f'\left( x \right)]}^{3}}.\dfrac{{{f}^{3}}\left( x \right)}{2}.\dfrac{{{f}^{3}}\left( x \right)}{2}} \\
& =3f'\left( x \right).{{f}^{2}}\left( x \right) \\
\end{aligned}$
Suy ra $\int\limits_{0}^{1}{\left[ {{f}^{3}}\left( x \right)+4{{[f'\left( x \right)]}^{3}} \right]dx}\ge 3\int\limits_{0}^{1}{f'\left( x \right).{{f}^{2}}\left( x \right)dx}$.
Mà $\int\limits_{0}^{1}{\left\{ {{f}^{3}}\left( x \right)+4{{[f'\left( x \right)]}^{3}} \right\}dx}\le 3\int\limits_{0}^{1}{f'\left( x \right).{{f}^{2}}\left( x \right)dx}$ nên dấu "=" xảy ra, tức là
$\begin{aligned}
& 4{{[f'\left( x \right)]}^{3}}=\dfrac{{{f}^{3}}\left( x \right)}{2}=\dfrac{{{f}^{3}}\left( x \right)}{2}\Leftrightarrow f'\left( x \right)=\dfrac{1}{2}f\left( x \right) \\
& \dfrac{f'\left( x \right)}{f\left( x \right)}=\dfrac{1}{2}\Rightarrow \int{\dfrac{f'\left( x \right)}{f\left( x \right)}dx}=\dfrac{1}{2}\int{dx}\Rightarrow \ln \left| f\left( x \right) \right|=\dfrac{1}{2}x+C\Rightarrow f\left( x \right)={{e}^{\dfrac{1}{2}x+C}} \\
\end{aligned}$
Theo giả thiết $f\left( 0 \right)=1\Rightarrow C=0\Rightarrow f\left( x \right)={{e}^{\dfrac{1}{2}x}}\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=2\left( \sqrt{e}-1 \right)$
Đáp án A.