Cho hàm số \(f(x)>0\) có đạo hàm liên tục trên...

D

- Hướng dẫn giải

Ta có: \(f''\left( x \right).f\left( x \right) + {\left[ {\frac{{f\left( x \right)}}{{{\rm{cos}}x}}} \right]^2} = {\left[ {f'\left( x \right)} \right]^2} \Leftrightarrow \frac{{f''\left( x \right).f\left( x \right) - {{\left[ {f'\left( x \right)} \right]}^2}}}{{{f^2}\left( x \right)}} =  - \frac{1}{{{\rm{co}}{{\rm{s}}^2}x}}\)

\(\left[ {\frac{{f'\left( x \right)}}{{f\left( x \right)}}} \right]' =  - \frac{1}{{{{\cos }^2}x}} \Rightarrow \frac{{f'\left( x \right)}}{{f\left( x \right)}} = \tan x + C\). Do \(\left\{ \begin{array}{l}
f'\left( 0 \right) = 0\\
f\left( 0 \right) = 1
\end{array} \right.\) nên C = 0

Do đó \(\frac{{f'\left( x \right)}}{{f\left( x \right)}} =  - \tan x\). Suy ra \(\int\limits_0^{\frac{\pi }{3}} {\frac{{df\left( x \right)}}{{f\left( x \right)}} = \int\limits_0^{\frac{\pi }{3}} {\frac{{d\left( {{\rm{cos}}x} \right)}}{{{\rm{cos}}x}}} }  \Leftrightarrow \left. {\ln f\left( x \right)} \right|\begin{array}{*{20}{c}}
{\frac{\pi }{3}}\\
0
\end{array} = \left. {\ln \cos x} \right|\begin{array}{*{20}{c}}
{\frac{\pi }{3}}\\
0
\end{array} \Leftrightarrow \ln f\left( {\frac{\pi }{3}} \right) - \ln f\left( 0 \right) = \ln \frac{1}{2} \Leftrightarrow f\left( {\frac{\pi }{3}} \right) = \frac{1}{2}\)

Vậy \(f\left( {\frac{\pi }{3}} \right) = \frac{1}{2}\)