This is a Test of Mathematics Solution Subjective 71 (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.
Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta
Consider the following simultaneous equations in x and y :
$ {\displaystyle{4x + y + axy = a}} $
$ {\displaystyle{x - 2y -xy^2 = 0}} $
where $ {\displaystyle{a}} $ is a real constant. Show that these equations admit real solutions in x and y.
$ {\displaystyle{4x + y + axy = a}} $ ... (i)
$ {\displaystyle{x - 2y -xy^2 = 0}} $ ... (ii)
$ {\displaystyle{x - 2y -xy^2 = 0}} $
$ {\Rightarrow} $ $ {\displaystyle{x = {\frac{2y}{1-y^2}}}} $
$ {\displaystyle{x + y + axy = a}} $
$ {\Rightarrow} $ $ {\displaystyle{\frac{2y}{1-y^2} + 4 + a{\frac{2y^2}{1-y^2}}}} $ $ {\displaystyle{= a}} $ [ replacing x in terms of y. ]
$ {\Rightarrow} $ $ {\displaystyle{3y - y^3 + 3ay^2 -a = 0}} $
This is a 3rd degree polynomial over y.
Now as we know any odd degree polynomial has at least one real root.That is why y has also at least one real root.Now for that particular root we get $ {\displaystyle{x = {\frac{2y}{1-y^2}}}} $ also real.
Conclusion: $ {\displaystyle{x + y + axy = a}} $ & $ {\displaystyle{x - 2y -xy^2 = 0}} $ admits real solutions in x and y.
This is a Test of Mathematics Solution Subjective 71 (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.
Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta
Consider the following simultaneous equations in x and y :
$ {\displaystyle{4x + y + axy = a}} $
$ {\displaystyle{x - 2y -xy^2 = 0}} $
where $ {\displaystyle{a}} $ is a real constant. Show that these equations admit real solutions in x and y.
$ {\displaystyle{4x + y + axy = a}} $ ... (i)
$ {\displaystyle{x - 2y -xy^2 = 0}} $ ... (ii)
$ {\displaystyle{x - 2y -xy^2 = 0}} $
$ {\Rightarrow} $ $ {\displaystyle{x = {\frac{2y}{1-y^2}}}} $
$ {\displaystyle{x + y + axy = a}} $
$ {\Rightarrow} $ $ {\displaystyle{\frac{2y}{1-y^2} + 4 + a{\frac{2y^2}{1-y^2}}}} $ $ {\displaystyle{= a}} $ [ replacing x in terms of y. ]
$ {\Rightarrow} $ $ {\displaystyle{3y - y^3 + 3ay^2 -a = 0}} $
This is a 3rd degree polynomial over y.
Now as we know any odd degree polynomial has at least one real root.That is why y has also at least one real root.Now for that particular root we get $ {\displaystyle{x = {\frac{2y}{1-y^2}}}} $ also real.
Conclusion: $ {\displaystyle{x + y + axy = a}} $ & $ {\displaystyle{x - 2y -xy^2 = 0}} $ admits real solutions in x and y.
