Đáp án: A
Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
f'\left( x \right) \ge 0\\
f\left( x \right) > 0
\end{array} \right.\\
{\left[ {f'\left( x \right)} \right]^2} = \left( {x + 1} \right)f\left( x \right)\\
\Rightarrow \dfrac{{f'\left( x \right)}}{{\sqrt {f\left( x \right)} }} = \sqrt {x + 1} \\
\Rightarrow \int {\dfrac{{f'\left( x \right)}}{{\sqrt {f\left( x \right)} }}dx} = \int {\sqrt {x + 1} dx} \\
\Rightarrow \int {\dfrac{1}{{\sqrt {f\left( x \right)} }}d\left( {f\left( x \right)} \right)} = \int {\sqrt {x + 1} dx} \\
\Rightarrow 2\int {\dfrac{1}{{2\sqrt {f\left( x \right)} }}d\left( {f\left( x \right)} \right)} = \dfrac{2}{3}\left( {x + 1} \right)\sqrt {x + 1} + C\\
\Rightarrow 2.\sqrt {f\left( x \right)} = \dfrac{2}{3}\left( {x + 1} \right)\sqrt {x + 1} + C\\
Do:f\left( 3 \right) = \dfrac{3}{2}\\
\Rightarrow 2.\sqrt {\dfrac{3}{2}} = \dfrac{2}{3}.\left( {3 + 1} \right).\sqrt {3 + 1} + C\\
\Rightarrow C = \dfrac{{3\sqrt 6 - 16}}{3}\\
\Rightarrow 2.\sqrt {f\left( x \right)} = \dfrac{2}{3}\left( {x + 1} \right)\sqrt {x + 1} + \dfrac{{3\sqrt 6 - 16}}{3}\\
\Rightarrow 2.\sqrt {f\left( 8 \right)} = \dfrac{2}{3}.\left( {8 + 1} \right).\sqrt {8 + 1} + \dfrac{{3\sqrt 6 - 16}}{3}\\
\Rightarrow {f^2}\left( 8 \right) = 3263,21
\end{array}$
