Question 30, Let the function $\mathrm{f}(\mathrm{x})=\mathrm{x}-\ln (\mathrm{ax}+2 \mathrm{a}+1)+2$, if for any $\mathrm{x} \geq-2$, it always holds that $\mathrm{f}(\mathrm{x}) \geq 0$, try to find the range of the real number $\mathrm{a}$.
Question 30, Solution: First, according to the domain of the function, for any $x \geq-2$, we have $a x+2 a+1>0 \Rightarrow$ $a \geq 0$.

Furthermore, according to the condition, for any $\varepsilon \in \mathrm{R}^{+}$, we have $\mathrm{f}(-2+\varepsilon)=\varepsilon-\ln (\mathrm{a} \varepsilon+1) \geq 0$, hence $1 \geq$ $\lim _{\varepsilon \rightarrow 0^{+}} \frac{\ln (a \varepsilon+1)}{\varepsilon}=\lim _{\varepsilon \rightarrow 0^{+}} \frac{a}{a \varepsilon+1}=a$.

On the other hand, when $a \in[0,1]$, for any $x \geq-2$, we have $f(x)=x-\ln (a x+2 a+1)+2=(x+$ 2) $-\ln (a(x+2)+1) \geq(x+2)-a(x+2) \geq 0$.
In summary, the required $\mathrm{a} \in[0,1]$.