Here, f1(x)=x2−x+1 and f2(x)=x3−x2−2x+1
or f1′(x1)=2x1−1 and f2′(x2)=3x22−2x2−2
Let the tangents drawn to the curves y=f1(x) and y=f2(x)
at (x1,f1(x1)) and (x2,f2(x2)) be parallel. Then 2x1−1=3x22−2x2−2
or 2x1=(3x22−2x2−1)
which is possible for infinite numbers of ordered pairs.
So, there are infinite solutions.