(1) If the function $f(x)=\lg \left(a x^{2}-4 x+a-3\right)$ has a range of $\mathbf{R}$, then the range of real number $a$ is ( ).
(A) $(4,+\infty)$
(B) $[0,4]$
(C) $(0,4)$
(D) $(-\infty,-1) \cup(4,+\infty)$
(1) To make the range of $f(x)$ be $\mathbf{R}$, we should ensure that $a x^{2}-4 x+a-3$ can take all positive values. Therefore, either $a=0$, or $a>0$ and $4^{2}-4 a(a-3) \geqslant 0$. Solving this, we get $0 \leqslant a \leqslant 4$. Hence, the correct choice is B.