SAT1000 - P321

The period $\frac{2\pi}{\omega}$ must be larger than the interval $\frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$ , otherwise the interval will include a maximum value, so $\frac{2\pi}{\omega} > \frac{\pi}{6}$ , which solves to $\omega < 12$ .

By symmetry the minimum will be exactly between $\frac{\pi}{6}$ and $\frac{\pi}{3}$ , which is at $\frac{1}{2}(\frac{\pi}{6} + \frac{\pi}{3}) = \frac{\pi}{4}$ , so $\sin (\omega frac{\pi}{4} + \frac{\pi}{3}) = -1$ , which solves to $\omega = \frac{14}{3}$ for $0 < \omega < 12$ .

Therefore, $a = 14$ , $b = 3$ , and $a + b = \boxed{17}$ .

The sine function is symmetric about it's minimum between two successive zeros. So $\omega \frac{π}{6}+\frac{π}{3}+\omega \frac{π}{3}+\frac{π}{3}=2\times \frac{3π}{2}\implies \omega =\frac{14}{3}$

So, $a=14,b=3$ and $a+b=\boxed {17}$ .