`~rai~`
\(y=\sin2x+\sqrt{3}\cos2x+1\\\quad=\dfrac{1}{2}\left(\dfrac{1}{2}\sin2x+\dfrac{\sqrt{3}}{2}\cos2x\right)+1\\\quad=\dfrac{1}{2}\left(\cos\dfrac{\pi}{3}\sin2x+\sin\dfrac{\pi}{3}\cos2x\right)+1\\\quad=\dfrac{1}{2}\sin\left(2x+\dfrac{\pi}{3}\right)+1.\\\text{Ta có:}-1\le\sin\left(2x+\dfrac{\pi}{3}\right)\le 1\\\Leftrightarrow -\dfrac{1}{2}\le\dfrac{1}{2}\sin\left(2x+\dfrac{\pi}{3}\right)\le\dfrac{1}{2}\\\Leftrightarrow \dfrac{1}{2}\le\dfrac{1}{2}\sin\left(2x+\dfrac{\pi}{3}\right)+1\le\dfrac{3}{2}\\\Leftrightarrow \dfrac{1}{2}\le y\le\dfrac{3}{2}.\\\Rightarrow\text{Tập giá trị là:}\left[\dfrac{1}{2};\dfrac{3}{2}\right]\\\Rightarrow \begin{cases}a=\dfrac{1}{2}\\b=\dfrac{3}{2}\end{cases}\\\Rightarrow T=a+b=\dfrac{1}{2}+\dfrac{3}{2}=2.\\\to\text{Chọn ý B.}\)
