⇒ x∈R−{1,2} k(2x−4−x+1)=2(x2−3x+2) k(x−3)=2(x2−3x+2) 2x2−(6+k)x+3k+4=0 For no real roots b2−4ac<0 ∴(k+6)2−8.(3k+4)<0 ⇒k2−12k−4<0 ⇒(k−6)2−32<0 ⇒(k−6)2<32 ⇒−4√2<k−6<4√2 6−4√2<k<6+4√2 Integral k∈{1,2,3,4,....11} Sum = 66