Hàm số \(y = f\left( x \right) = \cos \left( {2x - \frac{\pi }{3}} \right)\) . Phương trình \({f^{\left( 4 \right)}}\left( x \right) = - 8\) có nghiệm \(x \in \left[ {0;\frac{\pi }{2}} \right]\) là:

Hướng dẫn giải:

Chọn A.

Ta có: \(y' = - 2{\rm{sin}}\left( {2x - \frac{\pi }{3}} \right)\). \[y'' = - 4{\rm{cos}}\left( {2x - \frac{\pi }{3}} \right)\]. \[y''' = 8{\rm{sin}}\left( {2x - \frac{\pi }{3}} \right)\]. \[{y^{\left( 4 \right)}} = 16{\rm{cos}}\left( {2x - \frac{\pi }{3}} \right)\]

Khi đó : \({f^{\left( 4 \right)}}\left( x \right) = - 8\) \[ \Leftrightarrow 16{\rm{cos}}\left( {2x - \frac{\pi }{3}} \right) = - 8\] \[ \Leftrightarrow {\rm{cos}}\left( {2x - \frac{\pi }{3}} \right) = - \frac{1}{2}\]

\( \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{3} = \frac{{2\pi }}{3} + k2\pi \\2x - \frac{\pi }{3} = - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\) \( \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k\pi \\x = - \frac{\pi }{6} + k\pi \end{array} \right.\) .