1. Given the function $f(x)=\log _{a}\left(a x^{2}-x+\frac{1}{2}\right)$ is always positive on $[1,2]$, then the range of the real number $a$ is
1. $\frac{1}{2}\frac{3}{2}$ Detailed explanation: When $a>1$, $a x^{2}-x+\frac{1}{2}>1 \Leftrightarrow a>\frac{1}{2}\left(\frac{1}{x}\right)^{2}+\frac{1}{x}$, from $\frac{1}{x} \in$ $\left[\frac{1}{2}, 1\right]$ we know $a>\frac{3}{2}$; When $0<a<1$, $0<a x^{2}-x+\frac{1}{2}<1 \Leftrightarrow-\frac{1}{2}\left(\frac{1}{x}\right)^{2}+\frac{1}{x}<a<\frac{1}{2}\left(\frac{1}{x}\right)^{2}+\frac{1}{x}$, and from $\frac{1}{x} \in\left[\frac{1}{2}, 1\right]$ we can get $\frac{1}{2}<a<\frac{5}{8}$.