Lời giải:Ta có: \(f\left( x \right) = f'\left( x \right)\sqrt {3x + 1} \Rightarrow \dfrac{{f'\left( x \right)}}{{f\left( x \right)}} = \dfrac{1}{{\sqrt {3x + 1} }}\)
Lấy nguyên hàm hai vế ta được:
\(\int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}dx} = \int {\dfrac{1}{{\sqrt {3x + 1} }}dx} \Rightarrow \int {\dfrac{{d\left( {f\left( x \right)} \right)}}{{f\left( x \right)}}} = \int {{{\left( {3x + 1} \right)}^{ - \dfrac{1}{2}}}dx} \)
\( \Rightarrow \ln f\left( x \right) = \dfrac{2}{3}\sqrt {3x + 1} + C\,\,\left( {Do\,\,f\left( x \right) > 0\,\,\forall x > 0} \right) \Rightarrow f\left( x \right) = {e^{\dfrac{2}{3}\sqrt {3x + 1} + C}}\)
Do \(f\left( 1 \right) = 1\) nên \({e^{\dfrac{2}{3}\sqrt {3.1 + 1} + C}} = 1 \Leftrightarrow \dfrac{4}{3} + C = 0 \Leftrightarrow C = - \dfrac{4}{3}\) hay \(f\left( x \right) = {e^{\dfrac{2}{3}\sqrt {3x + 1} - \dfrac{4}{3}}}\)
Do đó \(f\left( 5 \right) = {e^{\dfrac{2}{3}\sqrt {3.5 + 1} - \dfrac{4}{3}}} = {e^{\dfrac{4}{3}}} \approx 3,79\).
Vậy \(3 < f\left( 5 \right) < 4\).
Chọn D.
