9. Given the function $f(x)=\sin \left(2 x-\frac{\pi}{6}\right)-m$ has two zeros in $\left[0, \frac{\pi}{2}\right]$, then the range of $\mathrm{m}$ is
A. $\left(\frac{1}{2}, 1\right) \quad$ B $\left[\frac{1}{2}, 1\right]$
C. $\left[\frac{1}{2}, 1\right)$
D. $\left(\frac{1}{2}, 1\right]$
Solution: The problem is equivalent to the function $f(x)=\sin \left(2 x-\frac{\pi}{6}\right)$ intersecting the line $y=m$ at two points in the interval $\left[0, \frac{\pi}{2}\right]$, so the range of values for $\mathrm{m}$ is $\left[\frac{1}{2}, 1\right)$. The correct answer is $C$.