13. (10th "Hope Cup" Invitational Competition Question) Given the function $f(x)=\lg \left(a x^{2}+2 x+1\right)$ has a range of all real numbers, then the range of real number $a$ is

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13. (10th "Hope Cup" Invitational Competition Question) Given the function $f(x)=\lg \left(a x^{2}+2 x+1\right)$ has a range of all real numbers, then the range of real number $a$ is
13. Fill $0 \leqslant a \leqslant 1$. Reason: When $a=0$, it only requires $x>-\frac{1}{2}$, and the function $\lg \left(a x^{2}+2 x+1\right)$ can take all real numbers. When $a>0$, it only requires the discriminant of $a x^{2}+2 x+1$, $\Delta_{x} \geqslant 0$, i.e., $a \leqslant 1$, for the function $\lg \left(a x^{2}+2 x+1\right)$ to take all real numbers. When $a<0$, since $a x^{2}+2 x+1 \leqslant 1-\frac{4}{4 a}=1-\frac{1}{a}$, the function $\lg \left(a x^{2}+2 x+1\right)$ cannot take all real numbers.