Complete set of values of ‘a’ such that x2−x1−ax attains all real values is

Let y=(x2−x)(1−ax);  x∈R ⇒x2−x=y−ayx   ⇒ x2+(ay−1)x−y=0 Since y attains all values in R for real x.Discriminant : D=(ay−1)2+4y≥0 ;∀  y∈R ⇒     a2y2+(4−2a)y+1≥0  ;∀  y∈R ⇒     4(2−a)2−4a2≤0 ⇒      4−4a+a2−a2≤0 ⇒    a≥1 Hence a∈[1,  ∞)