12. Given the function $f(x)=9^{x}-m \cdot 3^{x}+m+1$ is always above the x-axis for $x \in(0,+\infty)$, then the range of $m$ is
A. $2-2 \sqrt{2}<m<2+2 \sqrt{2}$
B. $m<-2$
C. $m<2+2 \sqrt{2}$
D. $m \geqslant 2+2 \sqrt{2}$
Solve:
$f(x)>0 \Rightarrow m<\frac{9^{x}+1}{3^{x}-1}=3^{x}+1+\frac{2}{3^{x}-1}=3^{x}-1+\frac{2}{3^{x}-1}+2$, thus $m<2+2 \sqrt{2}$, so the answer is $C$.