Đáp án:
C
Giải thích các bước giải:
\(\begin{array}{l}
y' = 2\cos x\\
y' = 0\\
\to 2\cos x = 0\\
\to x = \dfrac{\pi }{2} + k\pi \\
Do:x \in \left[ { - \dfrac{\pi }{3};\dfrac{{2\pi }}{3}} \right]\\
\to x = \dfrac{\pi }{2}\\
Có:y\left( {\dfrac{\pi }{2}} \right) = 2\sin \dfrac{\pi }{2} + 3 = 5\\
y\left( { - \dfrac{\pi }{3}} \right) = 2\sin \left( { - \dfrac{\pi }{3}} \right) + 3 = 3 - \sqrt 3 \\
y\left( {\dfrac{{2\pi }}{3}} \right) = 2\sin \left( {\dfrac{{2\pi }}{3}} \right) + 3 = 3 + \sqrt 3 \\
\to Min = y\left( { - \dfrac{\pi }{3}} \right) = 3 - \sqrt 3 \\
\to a = - 1;b = 3\\
\to T = a + b = 3 - 1 = 2
\end{array}\)
