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Câu hỏi: Cho hàm số $f\left( x \right)$ nhận giá trị dương và có đạo hàm liên tục trên $\left[ 0;3 \right]$ và thoả mãn $f\left( 0 \right)=3,f\left( 3 \right)=8$ và $\int\limits_{0}^{3}{\dfrac{{{\left( {f}'\left( x \right) \right)}^{2}}}{f\left( x \right)+1}dx=\dfrac{4}{3}}$. Giá trị của $f\left( 2 \right)$ bằng
A. $\dfrac{64}{9}$.
B. $\dfrac{55}{9}$.
C. $\dfrac{16}{3}$.
D. $\dfrac{19}{3}$.
Ta có $\int\limits_{0}^{3}{{{1}^{2}}dx.\int\limits_{0}^{3}{\dfrac{{{\left( {f}'\left( x \right) \right)}^{2}}}{f\left( x \right)+1}}dx\ge {{\left( \int\limits_{0}^{3}{\dfrac{{f}'\left( x \right)}{\sqrt{f\left( x \right)+1}}dx} \right)}^{2}}}$.
Do đó: ${{\int\limits_{0}^{3}{\dfrac{{{\left( {f}'\left( x \right) \right)}^{2}}}{f\left( x \right)+1}dx\ge \dfrac{1}{3}{{\left( \int\limits_{0}^{3}{\dfrac{{f}'\left( x \right)}{\sqrt{f\left( x \right)+1}}} \right)}^{2}}=\dfrac{1}{3}\left( \mathop{\left. 2\sqrt{f\left( x \right)+1} \right|}_{0}^{3} \right)}}^{2}}=\dfrac{4}{3}{{\left( \sqrt{f\left( 3 \right)+1}-\sqrt{f\left( 0 \right)+1} \right)}^{2}}=\dfrac{4}{3}$.
Vì vậy dấu $''=''$ phải xảy ra tức là $\dfrac{{f}'\left( x \right)}{\sqrt{f\left( x \right)+1}}=k\Rightarrow 2\sqrt{f\left( x \right)+1}=kx+C$
Vì $\left\{ \begin{aligned}
& f\left( 0 \right)=3 \\
& f\left( 3 \right)=8 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& C=4 \\
& 3k+C=6 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& k=\dfrac{2}{3} \\
& C=4 \\
\end{aligned} \right.\Rightarrow 2\sqrt{f\left( x \right)+1}=\dfrac{2}{3}x+4\Rightarrow f\left( x \right)$
$=\dfrac{1}{4}{{\left( \dfrac{2}{3}x+4 \right)}^{2}}-1\Rightarrow f\left( x \right)=\dfrac{55}{9}$
A. $\dfrac{64}{9}$.
B. $\dfrac{55}{9}$.
C. $\dfrac{16}{3}$.
D. $\dfrac{19}{3}$.
Ta có $\int\limits_{0}^{3}{{{1}^{2}}dx.\int\limits_{0}^{3}{\dfrac{{{\left( {f}'\left( x \right) \right)}^{2}}}{f\left( x \right)+1}}dx\ge {{\left( \int\limits_{0}^{3}{\dfrac{{f}'\left( x \right)}{\sqrt{f\left( x \right)+1}}dx} \right)}^{2}}}$.
Do đó: ${{\int\limits_{0}^{3}{\dfrac{{{\left( {f}'\left( x \right) \right)}^{2}}}{f\left( x \right)+1}dx\ge \dfrac{1}{3}{{\left( \int\limits_{0}^{3}{\dfrac{{f}'\left( x \right)}{\sqrt{f\left( x \right)+1}}} \right)}^{2}}=\dfrac{1}{3}\left( \mathop{\left. 2\sqrt{f\left( x \right)+1} \right|}_{0}^{3} \right)}}^{2}}=\dfrac{4}{3}{{\left( \sqrt{f\left( 3 \right)+1}-\sqrt{f\left( 0 \right)+1} \right)}^{2}}=\dfrac{4}{3}$.
Vì vậy dấu $''=''$ phải xảy ra tức là $\dfrac{{f}'\left( x \right)}{\sqrt{f\left( x \right)+1}}=k\Rightarrow 2\sqrt{f\left( x \right)+1}=kx+C$
Vì $\left\{ \begin{aligned}
& f\left( 0 \right)=3 \\
& f\left( 3 \right)=8 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& C=4 \\
& 3k+C=6 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& k=\dfrac{2}{3} \\
& C=4 \\
\end{aligned} \right.\Rightarrow 2\sqrt{f\left( x \right)+1}=\dfrac{2}{3}x+4\Rightarrow f\left( x \right)$
$=\dfrac{1}{4}{{\left( \dfrac{2}{3}x+4 \right)}^{2}}-1\Rightarrow f\left( x \right)=\dfrac{55}{9}$
Đáp án B.