Câu hỏi 6

\(\begin{array}{l}\,\,\,\,\,z = \cos 2\alpha  + \left( {\sin \alpha  - \cos \alpha } \right)i\\ \Rightarrow \left| z \right| = \sqrt {{{\cos }^2}2\alpha  + {{\left( {\sin \alpha  - \cos \alpha } \right)}^2}} \\\,\,\,\,\,\,\left| z \right| = \sqrt {{{\cos }^2}2\alpha  + 1 - \sin 2\alpha } \\\,\,\,\,\,\,\left| z \right| = \sqrt {1 - {{\sin }^2}2\alpha  + 1 - \sin 2\alpha } \\\,\,\,\,\,\,\left| z \right| = \sqrt { - {{\sin }^2}2\alpha  - \sin 2\alpha  + 2} \\ \Rightarrow \left| z \right| = \sqrt { - {{\left( {\sin 2\alpha  + \frac{1}{2}} \right)}^2} + \frac{9}{4}}  \le \sqrt {\frac{9}{4}}  = \frac{3}{2}\end{array}\)

Dấu bằng xảy ra \( \Leftrightarrow \sin 2\alpha  =  - \frac{1}{2} \Leftrightarrow \left[ \begin{array}{l}2\alpha  =  - \frac{\pi }{6} + k2\pi \\2\alpha  = \frac{{7\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\alpha  =  - \frac{\pi }{{12}} + k\pi \\\alpha  = \frac{{7\pi }}{{12}} + k\pi \end{array} \right.\,\,\,\left( {k \in Z} \right)\).

Vậy \({\left| z \right|_{\max }} = \frac{3}{2}\).

Chọn D.