Test of Mathematics Solution Subjective 76 - Range of a Rational Polynomial

This is a Test of Mathematics Solution Subjective 76 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Problem

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

Solution

${y} = {\frac{x^2-x}{1-mx}}$
${\Leftrightarrow}$ ${\displaystyle{y - myx = x^2 - x}}$
${\Leftrightarrow}$ ${x^2 + x(my - 1) - y = 0}$
${\Leftrightarrow}$ ${\displaystyle{x} = {\frac{1 - my {\pm} {\sqrt{(my-1)^2+4y}}}{2}}}$
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.
${\Rightarrow}$ ${\displaystyle{(4 - 2m)^2 - 4m^2} {\le {0}}}$
${\Rightarrow}$ ${{-16m + 16} {\le {0}}}$
${\Rightarrow}$ ${-m + 1 {\le{0}}}$
${\Rightarrow}$ ${1 {\le{m}}}$

Conclusion:

If ${1 {\le{m}}}$ , then ${\displaystyle{y = {\frac{x^2 - x}{1 - mx}}}}$ can take all the values as ${x}$ varies over |R.

Problem

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

Solution

${y} = {\frac{x^2-x}{1-mx}}$
${\Leftrightarrow}$ ${\displaystyle{y - myx = x^2 - x}}$
${\Leftrightarrow}$ ${x^2 + x(my - 1) - y = 0}$
${\Leftrightarrow}$ ${\displaystyle{x} = {\frac{1 - my {\pm} {\sqrt{(my-1)^2+4y}}}{2}}}$
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.
${\Rightarrow}$ ${\displaystyle{(4 - 2m)^2 - 4m^2} {\le {0}}}$
${\Rightarrow}$ ${{-16m + 16} {\le {0}}}$
${\Rightarrow}$ ${-m + 1 {\le{0}}}$
${\Rightarrow}$ ${1 {\le{m}}}$