1. If $f(x)$ is a decreasing function on $\mathbf{R}$, and the graph of $f(x)$ passes through the points $A(0,3)$ and $B(3,-1)$, then the solution set of the inequality $|f(x+1)-1|<2$ is ( ).
(A) $(-\infty, 3)$
(B) $(-\infty, 2)$
(C) $(0,3)$
(D) $(-1,2)$
-、1.(D).
It is easy to get $-1<f(x+1)<3$.
$\because f(0)=3, f(3)=-1$, and $f(x)$ is a decreasing function,
$\therefore$ the solution to the inequality $-1<f(x)<3$ is $0<x<3$.
Therefore, when $-1<f(x+1)<3$, we have $0<x+1<3$.