Câu hỏi 5

Ta có: \(y = \sqrt 3 \sin 2x - \cos 2x + 2019\)

\(\begin{array}{l} \Rightarrow y = 2\left( {\frac{{\sqrt 3 }}{2}\sin 2x - \frac{1}{2}\cos 2x} \right) + 2019\\\,\,\,\,\,\,\,\,\,\,\,\, = 2\left( {\sin 2x.\cos \frac{\pi }{6} - \cos 2x.\sin \frac{\pi }{6}} \right) + 2019\\\,\,\,\,\,\,\,\,\,\,\,\, = 2\sin \left( {2x - \frac{\pi }{6}} \right) + 2019.\end{array}\)

Ta có: \( - 1 \le \sin \left( {2x + \frac{\pi }{6}} \right) \le 1 \Rightarrow  - 2 \le 2\sin \left( {2x + \frac{\pi }{6}} \right) \le 2\)

\(\begin{array}{l} \Rightarrow  - 2 + 2019 \le 2\sin \left( {2x + \frac{\pi }{6}} \right) + 2019 \le 2 + 2019\\ \Rightarrow 2017 \le 2\sin \left( {2x + \frac{\pi }{6}} \right) + 2019 \le 2021.\\ \Rightarrow G = \left[ {2017;\,\,2021} \right].\end{array}\)

Chọn  C.