The number of values of 'a' for which the equation $(x-1{)}^2=|x-a|$ has exactly three solutions is

The correct option is C

3

$|x-a|=(x-1{)}^{2}Iffa=x\pm (x-1{)}^2$ No of solutions = no of intersection its between y = a ; f(x)=$x^2$ -x+1 and g(x) = $-x^2$ + 3x - 1. clearly the graphs of f(x), g(x) are tangents to each other at A(1,1) . The line y = a intersects the two graphs at three points Iff it passes through one of the three pts A,B, C. Here $B=(\frac{1}{2},\frac{3}{4})$ vertex of f and $C=(\frac{3}{2},\frac{5}{4})$ vertex of ‘g’ i.e if a $ϵ\{\frac{3}{4},\frac{5}{4},1\}$