1. Let the set $M=\left\{x \left\lvert\,-\frac{1}{2}<x<\frac{1}{2}\right.\right\}, N=\left\{x \mid x^{2} \leqslant x\right\}$, then $M \cap N=$
A. $\left[0, \frac{1}{2}\right)$
B. $\left(-\frac{1}{2}, 1\right]$
C. $\left[-1, \frac{1}{2}\right)$
D. $\left(-\frac{1}{2}, 0\right]$
$M=\left(-\frac{1}{2}, \frac{1}{2}\right), N=[0,1] \Rightarrow M \cap N=\left[0, \frac{1}{2}\right)$, so the answer is $A$.