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