The value range of the quadratic function $f(x)=x^{2}-2 a x+2 a+4$ is $[1,+\infty)$, then the range of the real number $a$ is
(A) $[-1,3]$
(B) $(-\infty,-1) \cup(3, \infty)$
(C) $\mid-1,3\}$
(D) None of the above
1. (C).
$f(x)=(x-a)^{2}-a^{2}+2a+4$ has a minimum value of $-a^{2}-2a+4$. According to the problem, we should have $-a^{2}+2a+4=1$. Therefore, $a=-1$ or 3.