1. Let real numbers $a$, $b$, $c$ satisfy $c+b=3a^2-4a+6$, $c-b=a^2-4a+4$. Then the relationship between $a$, $b$, $c$ is ( ).
(A) $a<b \leqslant c$
(B) $b<a \leqslant c$
(C) $b<c \leqslant a$
(D) $c<a \leqslant b$
$-1 . A$.
Since $b=\frac{(c+b)-(c-b)}{2}=a^{2}+1$, then $b-a=a^{2}-a+1=\left(a-\frac{1}{2}\right)^{2}+\frac{3}{4}>0$. Also, $c-b=a^{2}-4 a+4=(a-2)^{2} \geqslant 0$, thus $a<b \leqslant c$.