函数 $y=2 \sin \left(\frac{\pi}{3}-x\right)-\cos \left(\frac{\pi}{6}+x\right)(x \in \mathbf{R})$ 的最小值等于 $(\quad)$
A -3
B. -2
$\mathrm{C} \cdot-1$
D. $-\sqrt{5}$

解析: $\because\left(\frac{\pi}{3}-x\right)+\left(\frac{\pi}{6}+x\right)=\frac{\pi}{2}$,

$\therefore y=2 \sin \left[\frac{\pi}{2}-\left(\frac{\pi}{6}+x\right)\right]-\cos \left(x+\frac{\pi}{6}\right)$

$=2 \cos \left(x+\frac{\pi}{6}\right)-\cos \left(x+\frac{\pi}{6}\right)$

$=\cos \left(x+\frac{\pi}{6}\right), \quad \therefore y_{\min }=-1$.

答案: C

11.函数 $y=2 \sin \left(\omega x+\frac{\pi}{4}\right)(\omega>0)$ 的周期为 $\pi$, 则其单调递增区间为 $(\quad)$
周期 $T=\pi, \therefore \frac{2 \pi}{\omega}=\pi, \therefore \omega=2, \therefore y=2 \sin \left(2 x+\frac{\pi}{4}\right)$.由 $-\frac{\pi}{2}+2 k \pi \leqslant 2 x+\frac{\pi}{4} \leqslant 2 k \pi+\frac{\pi}{2}, k \in \mathbf{Z}$, 得 $k \pi-\frac{3}{8}$ $\pi \leqslant x \leqslant k \pi+\frac{\pi}{8}, \quad k \in \mathbf{Z}$.

答案: C