3. Let $x$ be a real number, $y=|x-1|+|x-1|$. Among the following four conclusions:

I . $y$ has no minimum value;
II. There is only one $x$ that makes $y$ reach its minimum value;
III. There are a finite number of $x$ (more than one) that make $y$ reach its maximum value;
$\mathbb{N}$. There are infinitely many $x$ that make $y$ reach its minimum value.
The correct one is
(A) I ;
(B) $\mathbb{I I}$;
(C) III ;
(D) $\mathrm{N}$.
3. (D)

Since $|x-1|$ and $|x+1|$ represent the distances from point $x$ to points 1 and -1 on the number line, respectively.
Therefore, when $-1 \leqslant x \leqslant 1$, $y=|x-1|+|x+1|=2$;
when $x < -1$ or $x > 1$, $y=|x-1|+|x+1|=2+2|x-1|>2$.
And there are infinitely many real numbers $x$ between -1 and 1, so there are infinitely many $x$ that make $y$ take its minimum value.