Complete set of values of ‘a’ such that $\frac{x^2-x}{1-ax}$ attains all real values is

Let $y=\frac{\left(x^2-x\right)}{\left(1-ax\right)};x\in R$

$\Rightarrow x^2-x=y-ayx$  

$\Rightarrow$ $x^2+\left(ay-1\right)x-y=0$

Since $y$ attains all values in R for real x.

Discriminant : $D={\left(ay-1\right)}^2+4y\ge 0$ $;\forall y\in R$

$\Rightarrow$     $a^2{y}^2+\left(4-2a\right)y+1\ge 0$  $;\forall y\in R$

$\Rightarrow$     $4{\left(2-a\right)}^2-4a^2\le 0$

$\Rightarrow$      $4-4a+a^2-a^2\le 0$

$\Rightarrow$    $a\ge 1$

Hence $a\in \left[1,\infty \right)$