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Câu hỏi: Cho hàm số $f\left( x \right)$ nhận giá trị dương và có đạo hàm ${f}'\left( x \right)$ liên tục trên $\left[ 0;1 \right]$ thỏa mãn $f\left( 1 \right)=e.f\left( 0 \right)$ và $\int\limits_{0}^{1}{\dfrac{dx}{{{f}^{2}}\left( x \right)}}+\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx}\le 2$. Mệnh đề nào sau đây đúng?
A. $f\left( 1 \right)=\sqrt{\dfrac{2e}{e-1}}$.
B. $f\left( 1 \right)=\dfrac{2\left( e-2 \right)}{e-1}$.
C. $f\left( 1 \right)=\sqrt{\dfrac{2{{e}^{2}}}{{{e}^{2}}-1}}$.
D. $f\left( 1 \right)=\sqrt{\dfrac{2\left( e-2 \right)}{e-1}}$.
A. $f\left( 1 \right)=\sqrt{\dfrac{2e}{e-1}}$.
B. $f\left( 1 \right)=\dfrac{2\left( e-2 \right)}{e-1}$.
C. $f\left( 1 \right)=\sqrt{\dfrac{2{{e}^{2}}}{{{e}^{2}}-1}}$.
D. $f\left( 1 \right)=\sqrt{\dfrac{2\left( e-2 \right)}{e-1}}$.
Ta có $\int\limits_{0}^{1}{\dfrac{dx}{{{f}^{2}}\left( x \right)}}+\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx}=\int\limits_{0}^{1}{\left[ \dfrac{1}{{{f}^{2}}\left( x \right)}+{{\left[ {f}'\left( x \right) \right]}^{2}} \right]dx\overset{AM-GM}{\mathop{\ge }} 2\int\limits_{0}^{1}{\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx}}$.
$=2\ln \left| f\left( x \right) \right|\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=2\ln \left| f\left( 1 \right) \right|-2\ln \left| f\left( 0 \right) \right|=2\ln \left| \dfrac{f\left( 1 \right)}{f\left( 0 \right)} \right|=2\ln e=2$.
Mà $\int\limits_{0}^{1}{\dfrac{dx}{{{f}^{2}}\left( x \right)}}+\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx}\le 2$ nên dấu "=" xảy ra, tức là ${f}'\left( x \right)=\dfrac{1}{f\left( x \right)}\Leftrightarrow f\left( x \right).{f}'\left( x \right)=1$.
$\Rightarrow \int\limits_{{}}^{{}}{f\left( x \right).{f}'\left( x \right)dx}=\int\limits_{{}}^{{}}{xdx}\Rightarrow \dfrac{{{f}^{2}}\left( x \right)}{2}=x+C\Rightarrow f\left( x \right)=\sqrt{2x+2C}$
Theo giả thiết $f\left( 1 \right)=e.f\left( 0 \right)$ nên ta có $\sqrt{2+2C}=e\sqrt{2C}\Leftrightarrow 2+2C={{e}^{2}}.2C\Leftrightarrow C=\dfrac{1}{{{e}^{2}}-1}$
$\Rightarrow f\left( x \right)=\sqrt{2x+\dfrac{2}{{{e}^{2}}-1}}\Rightarrow f\left( 1 \right)=\sqrt{2+\dfrac{2}{{{e}^{2}}-1}}=\sqrt{\dfrac{2{{e}^{2}}}{{{e}^{2}}-1}}$.
$=2\ln \left| f\left( x \right) \right|\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=2\ln \left| f\left( 1 \right) \right|-2\ln \left| f\left( 0 \right) \right|=2\ln \left| \dfrac{f\left( 1 \right)}{f\left( 0 \right)} \right|=2\ln e=2$.
Mà $\int\limits_{0}^{1}{\dfrac{dx}{{{f}^{2}}\left( x \right)}}+\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx}\le 2$ nên dấu "=" xảy ra, tức là ${f}'\left( x \right)=\dfrac{1}{f\left( x \right)}\Leftrightarrow f\left( x \right).{f}'\left( x \right)=1$.
$\Rightarrow \int\limits_{{}}^{{}}{f\left( x \right).{f}'\left( x \right)dx}=\int\limits_{{}}^{{}}{xdx}\Rightarrow \dfrac{{{f}^{2}}\left( x \right)}{2}=x+C\Rightarrow f\left( x \right)=\sqrt{2x+2C}$
Theo giả thiết $f\left( 1 \right)=e.f\left( 0 \right)$ nên ta có $\sqrt{2+2C}=e\sqrt{2C}\Leftrightarrow 2+2C={{e}^{2}}.2C\Leftrightarrow C=\dfrac{1}{{{e}^{2}}-1}$
$\Rightarrow f\left( x \right)=\sqrt{2x+\dfrac{2}{{{e}^{2}}-1}}\Rightarrow f\left( 1 \right)=\sqrt{2+\dfrac{2}{{{e}^{2}}-1}}=\sqrt{\dfrac{2{{e}^{2}}}{{{e}^{2}}-1}}$.
Đáp án C.