2. Solve the inequality $\sqrt{9-x}-3 \geqslant x|x-3|+\ln (1+x)$.
Answer: $(-1 ; 0]$.

Solution. Domain: $(-1 ; 9]$. Let

$$
f(x)=\sqrt{9-x}-3, \quad g(x)=x|x-3|+\ln (1+x)
$$

At $x=0$, both functions are zero.

If $-1<x<0$, then $f(x)>0$, while $g(x)<0$. In this case, $f(x)>g(x)$.

If $0<x<9$, then $f(x)<0$, while $g(x)>0$ (the first term is non-negative, and the second term is positive). In this case, $f(x)<g(x)$.

Thus, the inequality $f(x) \geqslant g(x)$ holds only for $x \in$ $(-1 ; 0]$.

Evaluation. Full solution: 12 points. If it is noted that the parts of the inequality are sign-constant on the half-axes, but the answer is incorrect due to (possibly partial) ignoring of the domain, 4 points.