Find the complete set of values of a such that $\frac{x^2-x}{1-ax}$ all real values.

[1, ∞) (-∞, ∞) (-∞, 0) [1, 0)

[1, ∞)

$y=\frac{x^2-x}{1-ax}$ or $x^2- x=y-axy$ or $x^2+x (ay-1)-y=0$ Since x is real we get $(ay - 1)^2+ 4y ≥0$ or $a^2y^2+2y(2-a) +1≥0\, ∀\,y∈ R$ So, as $a^2 > 0$, $4(2-a)^2-4a^2≤0$ or $4-4a≤0$ or $a∈ [1,∞)$