$2 \cdot 83$ If the real number $x$ satisfies the equation $|1-x|=1+|x|$, then $\sqrt{(x-1)^{2}}$ equals
(A) $x-1$.
(B) $1-x$.
(C) $\pm(x-1)$.
(D) 1 .
(E) -1 .
(Chongqing, Sichuan Province, China Junior High School Mathematics Competition, 1983)
[Solution] When $x \leqslant 0$, $|1-x|=1-x$, and $1+|x|=1-x$, so the equation always holds.

When $0<x<1$, the original equation becomes $1-x=1+x$, which simplifies to $x=0$, but this contradicts $0<x<1$.
Thus, the original equation has no solution in this interval.
When $x \geqslant 1$, the original equation becomes $x-1=1+x$, which also has no solution.
$\therefore |1-x|=1+|x|$ has the solution $x \leqslant 0$.
At this time, $\sqrt{(x-1)^{2}}=1-x$.
Therefore, the answer is $(B)$.