Câu hỏi 39

\(\begin{array}{l}f''\left( x \right)f\left( x \right) + \frac{{{{\left[ {f\left( x \right)} \right]}^2}}}{{{{\sqrt {\left( {2x + 1} \right)} }^3}}} = {\left[ {f'\left( x \right)} \right]^2}\\ \Leftrightarrow f''\left( x \right)f\left( x \right) - {\left[ {f'\left( x \right)} \right]^2} =  - \frac{{{{\left[ {f\left( x \right)} \right]}^2}}}{{{{\sqrt {\left( {2x + 1} \right)} }^3}}}\\ \Leftrightarrow \frac{{f''\left( x \right)f\left( x \right) - {{\left[ {f'\left( x \right)} \right]}^2}}}{{{{\left[ {f\left( x \right)} \right]}^2}}} =  - \frac{1}{{{{\sqrt {\left( {2x + 1} \right)} }^3}}}\\ \Leftrightarrow \left[ {\frac{{f'\left( x \right)}}{{f\left( x \right)}}} \right]' =  - \frac{1}{{{{\sqrt {\left( {2x + 1} \right)} }^3}}}\\ \Leftrightarrow \frac{{f'\left( x \right)}}{{f\left( x \right)}} =  - \int\limits_{}^{} {\frac{{dx}}{{{{\sqrt {\left( {2x + 1} \right)} }^3}}}}  =  - \int\limits_{}^{} {{{\left( {2x + 1} \right)}^{ - \frac{3}{2}}}dx} \\ \Leftrightarrow \frac{{f'\left( x \right)}}{{f\left( x \right)}} =  - \frac{{{{\left( {2x + 1} \right)}^{ - \frac{1}{2}}}}}{{ - \frac{1}{2}.2}} + C = {\left( {2x + 1} \right)^{ - \frac{1}{2}}} + C\end{array}\)

Thay \(x = 0\) ta có: \(\frac{{f'\left( 0 \right)}}{{f\left( 0 \right)}} = 1 + C \Leftrightarrow 1 = 1 + C \Leftrightarrow C = 0 \Rightarrow  \Leftrightarrow \frac{{f'\left( x \right)}}{{f\left( x \right)}} = {\left( {2x + 1} \right)^{ - \frac{1}{2}}}\).

Lấy nguyên hàm 2 vế ta có:

\(\int\limits_{}^{} {\frac{{f'\left( x \right)}}{{f\left( x \right)}}dx}  = \int\limits_{}^{} {{{\left( {2x + 1} \right)}^{ - \frac{1}{2}}}dx}  \Leftrightarrow \ln \left| {f\left( x \right)} \right| = \frac{{{{\left( {2x + 1} \right)}^{\frac{1}{2}}}}}{{\frac{1}{2}.2}} + C = {\left( {2x + 1} \right)^{\frac{1}{2}}} + C\).

Do \(f\left( x \right) > 0\,\,\forall x \in \left[ {0;4} \right] \Rightarrow \ln f\left( x \right) = {\left( {2x + 1} \right)^{\frac{1}{2}}} + C\).

Thay \(x = 0\) ta có \(\ln f\left( 0 \right) = 1 + C \Leftrightarrow \ln 1 = 1 + C \Leftrightarrow 1 + C = 0 \Leftrightarrow C =  - 1\).

\( \Rightarrow \ln f\left( x \right) = {\left( {2x + 1} \right)^{\frac{1}{2}}} - 1 \Leftrightarrow f\left( x \right) = {e^{{{\left( {2x + 1} \right)}^{\frac{1}{2}}} - 1}} \Rightarrow f\left( 4 \right) = {e^{3 - 1}} = {e^2}\). 

Chọn A.