Đáp án:
\(\left\{ \begin{array}{l}
Min\,y = \frac{5}{2}\\
Max\,y = \frac{{161}}{{32}}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
y = {\cos ^2}2x - \sin x\cos x + 4 = 1 - 2{\sin ^2}2x - \frac{1}{2}\sin 2x + 4\\
= - 2{\sin ^2}2x - \frac{1}{2}\sin 2x + 5.\\
Dat\,\,\sin 2x = t\,\,\,\left( { - 1 \le t \le 1} \right).\\
\Rightarrow y = - 2{t^2} - \frac{1}{2}t + 5\\
\Rightarrow y' = - 4t - \frac{1}{2}\\
\Rightarrow y' = 0 \Leftrightarrow - 4t - \frac{1}{2} = 0 \Leftrightarrow t = - \frac{1}{8}\\
Ta\,\,co:\,\,\,\left\{ \begin{array}{l}
y\left( { - 1} \right) = \frac{7}{2}\\
y\left( { - \frac{1}{8}} \right) = \frac{{161}}{{32}}\\
y\left( 1 \right) = \frac{5}{2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
Min\,y = \frac{5}{2}\\
Max\,y = \frac{{161}}{{32}}
\end{array} \right..
\end{array}\)
