1. Given $\omega$ is a positive real number, $f(x)=4 \sin \frac{\omega x}{2} \cos \frac{\omega x}{2}$ is increasing on $\left[-\frac{\pi}{3}, \frac{\pi}{4}\right]$. Then the range of $\omega$ is ( ).
(A) $\left(0, \frac{2}{3}\right]$
(B) $\left(0, \frac{3}{2}\right]$
(C) $\left[\frac{3}{2},+\infty\right)$
(D) $[1,+\infty)$
$-1 . B$.
$f(x)=2 \sin \omega x, f(x)$ has a monotonically increasing interval $\left[-\frac{\pi}{2 \omega}, \frac{\pi}{2 \omega}\right]$. Given that $\left[-\frac{\pi}{3}, \frac{\pi}{4}\right] \subseteq\left[-\frac{\pi}{2 \omega}, \frac{\pi}{2 \omega}\right]$. Therefore, $\frac{\pi}{2 \omega} \geqslant \frac{\pi}{3}$. Hence, $0<\omega \leqslant \frac{3}{2}$.