Hướng dẫn giải

Ta có: \[f\left( x \right).f'\left( x \right) = \left( {2x + 1} \right)\sqrt {1 + {f^2}\left( x \right)} \Leftrightarrow \frac{{f\left( x \right).f'\left( x \right)}}{{\sqrt {1 + {f^2}\left( x \right)} }} = 2x + 1\].

Suy ra \[\int {\frac{{f\left( x \right).f'\left( x \right)}}{{\sqrt {1 + {f^2}\left( x \right)} }}dx} = \int {\left( {2x + 1} \right)dx} \Leftrightarrow \int {\frac{{d\left( {1 + {f^2}\left( x \right)} \right)}}{{2\sqrt {1 + {f^2}\left( x \right)} }}} = \int {\left( {2x + 1} \right)dx} \Leftrightarrow \sqrt {1 + {f^2}\left( x \right)} = {x^2} + x + C\]

Theo giả thiết \[f\left( 0 \right) = 2\sqrt 2 \], suy ra \[\sqrt {1 + {{\left( {2\sqrt 2 } \right)}^2}} = C \Leftrightarrow C = 3\]

Với \[C = 3\] thì \[\sqrt {1 + {f^2}\left( x \right)} = {x^2} + x + 3 \Rightarrow f\left( x \right) = \sqrt {{{\left( {{x^2} + x + 3} \right)}^2} - 1} \]

Vậy \[f\left( 1 \right) = \sqrt {24} = 2\sqrt 6 \]

Chọn D.