Đáp án:
\(\begin{array}{l}
48,\,\,\,A\\
49,\,\,\,C
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
48;\\
- 1 \le \sin \left( {\dfrac{x}{2} - \dfrac{{3\pi }}{2}} \right) \le 1\\
\Leftrightarrow - 2 \le 2\sin \left( {\dfrac{x}{2} - \dfrac{{3\pi }}{2}} \right) \le 2\\
\Leftrightarrow - 2 \le - 2\sin \left( {\dfrac{x}{2} - \dfrac{{3\pi }}{2}} \right) \le 2\\
\Leftrightarrow 1 + \left( { - 2} \right) \le 1 + \left( { - 2\sin \left( {\dfrac{x}{2} - \dfrac{{3\pi }}{2}} \right)} \right) \le 1 + 2\\
\Leftrightarrow - 1 \le 1 - 2\sin \left( {\dfrac{x}{2} - \dfrac{{3\pi }}{2}} \right) \le 3\\
\Leftrightarrow - 1 \le y \le 3\\
\Rightarrow {y_{\max }} = 3 \Leftrightarrow \sin \left( {\dfrac{x}{2} - \dfrac{{3\pi }}{2}} \right) = - 1\\
\Leftrightarrow \dfrac{x}{2} - \dfrac{{3\pi }}{2} = - \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow \dfrac{x}{2} = \pi + k2\pi \\
\Leftrightarrow x = 2\pi + k4\pi \,\,\,\,\,\left( {k \in Z} \right)\\
Vậy\,\,\,{y_{\max }} = 3 \Leftrightarrow x = 2\pi + k4\pi \,\,\,\,\left( {k \in Z} \right)\\
49\\
- 1 \le \cos 2x \le 1\\
\Leftrightarrow - 5 \le 5\cos 2x \le 5\\
\Leftrightarrow - 5 - 1 \le 5\cos 2x - 1 \le 5 - 1\\
\Leftrightarrow - 6 \le 5\cos 2x - 1 \le 4\\
\Leftrightarrow \dfrac{{ - 6}}{2} \le \dfrac{{5\cos 2x - 1}}{2} \le \dfrac{4}{2}\\
\Leftrightarrow - 3 \le \dfrac{{5\cos 2x - 1}}{2} \le 2\\
\Rightarrow - 3 \le y \le 2\\
\Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = - 3 \Leftrightarrow \cos 2x = - 1 \Leftrightarrow 2x = \pi + k2\pi \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \\
{y_{\max }} = 2 \Leftrightarrow \cos 2x = 1 \Leftrightarrow 2x = k2\pi \Leftrightarrow x = k\pi
\end{array} \right.\\
\Rightarrow {y_{\max }} + {y_{\min }} = - 3 + 2 = - 1
\end{array}\)