The shortest distance of the point from the curve y = |x -1| + |x + 1| is

Correct Option: A

The given curve is:
$y=|x-1|+|x+1|$

Let us analyze the behavior of this function by dividing it into intervals:
1) If $x<-1$, then $y=-(x-1)-(x+1)=-2x$
2) If $-1\le x\le 1$, then $y=-(x-1)+(x+1)=2$
3) If $x>1$, then $y=(x-1)+(x+1)=2x$

Thus, the minimum y-value of the curve is 2, which occurs for any x in the interval [-1, 1]. For all other x, $y>2$.

Assuming the source question specifies the point as (0, 1), we can find its shortest distance to the curve:
Since the bottom-most boundary of the curve is the horizontal line segment $y=2$ for $x\in [-1,1]$, the closest point on the curve to the point (0, 1) is (0, 2).
The perpendicular distance from (0, 1) to the line segment $y=2$ is:
$d=|2-1|=1$.

Hence, the shortest distance is 1.