# Range of a rational polynomial (Tomato subjective – 76)

problem: Find the set of all values of $${m}$$ such that $${\displaystyle {y} = {\frac{x^2-x}{1-mx}}}$$ can take all real values.

solution: $${\displaystyle {y} = {\frac{x^2-x}{1-mx}}}$$
$${\Leftrightarrow}$$ $${\displaystyle{y – myx = x^2 – x}}$$
$${\Leftrightarrow}$$ $${x^2 + x(my – 1) – y = 0}$$
$${\Leftrightarrow}$$ $${\displaystyle{x} = {\frac{1 – my {\pm} {\sqrt{(my-1)^2+4y}}}{2}}}$$
Now $${y}$$ takes all real values if discriminant $${(my-1)^2 + 4y}$$ is allways $${> 0}$$.
So now we have to find the all values of $${m}$$ such that $${(my-1)^2 + 4y}$$ $${> 0}$$ for all $${y}$$ $${\in}$$ |R.
$${\Leftrightarrow}$$ $${m^2 y^2 – 2my + 1 + 4y > 0}$$
$${\Leftrightarrow}$$ $${m^2 y^2 + y(4 – 2m) + 1 > 0}$$ …(i)
Now this is a equation of upside open parabola. If the discriminant is $${\le{0}}$$ of equation (i) then $${(my – 1)^2 + 4y}$$ will always positive.
$${\Rightarrow}$$ $${\displaystyle{(4 – 2m)^2 – 4m^2} {\le {0}}}$$
$${\Rightarrow}$$ $${{16m + 16} {\le {0}}}$$
$${\Rightarrow}$$ $${m + 1 {le{0}}}$$
$${\Rightarrow}$$ $${m {\le{- 1}}}$$
Conclusion: If $${m {\le{1}}}$$, then $${\displaystyle{y = {\frac{x^2 – x}{1 – mx}}}}$$ can take all the values as $${x}$$ varies over |R.