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