Đáp án:
$T = 2$
Giải thích các bước giải:
$\begin{array}{l}y = \sin2x + \sqrt3\cos2x + 1\\ = 2.\left(\dfrac{1}{2}\sin2x + \dfrac{\sqrt3}{2}\cos2x\right) + 1\\ = 2.\left(\sin2x.\cos\dfrac{\pi}{3} + \cos2x.\sin\dfrac{\pi}{3}\right)+1\\ = 2.\sin\left(2x + \dfrac{\pi}{3}\right) + 1\\ Ta\,\,có:\\ -1 \leq \sin\left(2x + \dfrac{\pi}{3}\right) \leq 1\\ \Leftrightarrow -2 \leq 2\sin\left(2x + \dfrac{\pi}{3}\right) \leq 2\\ \Leftrightarrow -1 \leq 2\sin\left(2x + \dfrac{\pi}{3}\right) + 1 \leq 3\\ Hay -1 \leq y \leq 3\\ Vậy\,\,T = [-1;3]\\\Rightarrow \begin{cases}a = -1\\b = 3\end{cases}\Rightarrow a + b = 2\end{array}$
