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:

Chỉ \[\left( I \right)\] đúng

Chỉ \[\left( {II} \right)\] đúng

Cả hai đều đúng

Cả hai đều sai

\[\frac{1}{{\cos x}}\]

\[ - \frac{1}{{\cos x}}\]

\(\cot x\)

\(tanx\)

\[3\]

\[6\]

\(12\)

\(24\)

\[\left[ { - 1;2} \right]\]

\[\left( { - \infty ;0} \right]\]

\(\left\{ { - 1} \right\}\)

\(\emptyset \)

\[y'''\left( 1 \right) = \frac{3}{8}\]

\[y'''\left( 1 \right) = \frac{1}{8}\]

\[y'''\left( 1 \right) = - \frac{3}{8}\]

\[y'''\left( 1 \right) = - \frac{1}{4}\]

\[64\]

\[ - 64\]

\(64\sqrt 3 \)

\( - 64\sqrt 3 \)

\(y'' = - \sin 2x\)

\(y'' = - 4\sin x\)

\(y'' = \sin 2x\)

\(y'' = - 4\sin 2x\)

\({y^{(n)}} = {2^n}\sin (2x + n\frac{\pi }{3})\)

\({y^{(n)}} = {2^n}\sin (2x + \frac{\pi }{2})\)

\({y^{(n)}} = {2^n}\sin (x + \frac{\pi }{2})\)

\({y^{(n)}} = {2^n}\sin (2x + n\frac{\pi }{2})\)

\({y^{(n)}} = \frac{{{{(2)}^n}.{a^n}.n!}}{{{{(ax + b)}^{n + 1}}}}\)

\({y^{(n)}} = \frac{{{{( - 1)}^n}.{a^n}.n!}}{{{{(x + 1)}^{n + 1}}}}\)

\({y^{(n)}} = \frac{{{{( - 1)}^n}.n!}}{{{{(ax + b)}^{n + 1}}}}\)

\({y^{(n)}} = \frac{{{{( - 1)}^n}.{a^n}.n!}}{{{{(ax + b)}^{n + 1}}}}\)

\({y^{(n)}} = {\left( { - 1} \right)^n}\cos \left( {2x + n\frac{\pi }{2}} \right)\)

\({y^{(n)}} = {2^n}\cos \left( {2x + \frac{\pi }{2}} \right)\)

\({y^{(n)}} = {2^{n + 1}}\cos \left( {2x + n\frac{\pi }{2}} \right)\)

\({y^{(n)}} = {2^n}\cos \left( {2x + n\frac{\pi }{2}} \right)\)