If m is slope of common tangents y=x2−x+1 , y=x2−3x+1 then m is

A point of first curve is P(t1,t12−t+1) , slope of tangent at P=2t1−1 . So, the equation of tangent at P is   y−(t12−t+1)=(2t1−1)(x−t1).......(1) A point  on second curve is  Q(t2,  t22−3t2+1) , slope of tangent at  Q=2t2−3 So, the equation of tangent at Q isy−(t22−3t+1)=(2t2−3)(x−t2)..........(2) and (2) are same lines, so,2t1−1=2t2−3   ⇒t2−t1=1 …………..(4)And −t1(2t1−1)+t12−t+1=−t2(2t2−3)+t22−3t2+1 ⇒−t12+1=−t22+1 t1=±t2 ⇒t2=−t1 (∵t1≠t2) ∴From(4),   t1=−12 Desired slope =2t1−1=−2