Đáp án:
\(\begin{array}{l}
d)Min = - \dfrac{2}{7}\\
Max = \dfrac{{26}}{7}\\
e)Min = - 7\\
Max = - 5\\
f)Min = 0\\
Max = \dfrac{1}{6}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
d)Do:0 \le {\cos ^2}2x \le 1\\
\to 0 \le 4{\cos ^2}2x \le 4\\
\to - \dfrac{2}{7} \le 4{\cos ^2}2x - \dfrac{2}{7} \le \dfrac{{26}}{7}\\
\to Min = - \dfrac{2}{7} \Leftrightarrow {\cos ^2}2x = 0 \to 2x = \dfrac{\pi }{2} + k\pi \to x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\\
Max = \dfrac{{26}}{7} \Leftrightarrow {\cos ^2}2x = 1 \to \left[ \begin{array}{l}
2x = k2\pi \\
2x = - \pi + k2\pi
\end{array} \right. \to \left[ \begin{array}{l}
x = k\pi \\
x = - \dfrac{\pi }{2} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
e)Do: - 1 \le \sin 2x \le 1\\
\to - 7 \le \sin 2x - 6 \le - 5\\
\to Min = - 7 \Leftrightarrow \sin 2x = - 1 \to 2x = - \dfrac{\pi }{2} + k2\pi \to x = - \dfrac{\pi }{4} + k\pi \\
Max = - 5 \to \sin 2x = 1 \to 2x = \dfrac{\pi }{2} + k2\pi \to x = \dfrac{\pi }{4} + k\pi \\
f)Do:0 \le \left| {\sin 3x} \right| \le 1\\
\to 0 \ge - \left| {\sin 3x} \right| \ge - 1\\
\to 1 \ge 1 - \left| {\sin 3x} \right| \ge 0\\
\to \dfrac{1}{6} \ge \dfrac{{1 - \left| {\sin 3x} \right|}}{6} \ge 0\\
\to Min = 0 \to \left| {\sin 3x} \right| = 1 \to \left[ \begin{array}{l}
3x = \dfrac{\pi }{2} + k2\pi \\
3x = - \dfrac{\pi }{2} + k2\pi
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k\pi \\
x = - \dfrac{\pi }{6} + k\pi
\end{array} \right.\\
Max = \dfrac{1}{6} \to \left| {\sin 3x} \right| = 0 \to 3x = k\pi \to x = \dfrac{{k\pi }}{3}
\end{array}\)
