$\because$ If the solution set of the inequality $(m-1) x<\sqrt{4} x-x^{2}$ about $x$ is $\left.|x| 0<x<2\right\}$, then the value of the real number $m$ is
(A) $\frac{1}{2}$
(B) 1
(C) 2
(D) 0
3. (C).

From the problem, we know $4 x-x^{2} \geqslant 0$, which gives $0 \leqslant x \leqslant 4$. Also, $x \neq 0$. Therefore, $0<x \leqslant 4$. Combining with the graph, we know $2(m-1)$ “ $\sqrt{4 \times 2-2^{2}}$, which gives $m=2$.