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
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1. $0 \leqslant a \leqslant 4$ Detailed explanation: To make the range of $f(x)$ be $\mathbf{R}$, the argument $a x^{2}-4 x+a-3$ must be able to take all positive values. Therefore, either $a=0$; or $a>0$ and $4^{2}-4 a(a-3) \geqslant 0$, solving this yields $0 \leqslant a \leqslant 4$.