This is a Test of Mathematics Solution Subjective 66 (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
If c is a real number with 0 < c < 1, then show that the values taken by the function y = $ {frac {x^2+2x+c}{x^2+4x+3c}} $ , as x varies over real numbers, range over all real numbers.
y = $ {\frac {x^2+2x+c}{x^2+4x+3c}} $
or $ {yx^2} $ + 4yx + 3cy = $ {x^2} $ + 2x + c
or $ {x^2} $ (y-1) + (4y-2)x + 3cy - c = 0
Now if we can show that the discriminant $ {\ge{0}} $ for all y then for all real y there exist a real x.
Now, discriminant is $ {(4y-2)^2} $ - 4(y-1)(3cy-c)
We need to show $ {(4y-2)^2} $ - 4(y-1)(3cy-c) > 0 for any 0 < c < 1. (16-12c) $ {4^2} $ - (16-16c)y + (4-4c) > 0.
This is parabola opening upword.
Now if its discriminant < 0 then this equation is always > 0.
So this is equivalent to prove
$ {(16-16c)^2} $ - 4 (16-12c)(4-4c) < 0.
or $ {(2-2c)^2} $ - (4-3c)(1-c) < 0
or $ {c^2} $ -C < 0
Now given 1>c>0 so $ {c^2} $ <c
or $ {c^2} $ -c < 0
Conclusion: So for 1>c>0 y = $ {\frac{x^2+2x+c}{x^2+4x+3c}} $
Range over all real number when x varies over all real number.
This is a Test of Mathematics Solution Subjective 66 (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
If c is a real number with 0 < c < 1, then show that the values taken by the function y = $ {frac {x^2+2x+c}{x^2+4x+3c}} $ , as x varies over real numbers, range over all real numbers.
y = $ {\frac {x^2+2x+c}{x^2+4x+3c}} $
or $ {yx^2} $ + 4yx + 3cy = $ {x^2} $ + 2x + c
or $ {x^2} $ (y-1) + (4y-2)x + 3cy - c = 0
Now if we can show that the discriminant $ {\ge{0}} $ for all y then for all real y there exist a real x.
Now, discriminant is $ {(4y-2)^2} $ - 4(y-1)(3cy-c)
We need to show $ {(4y-2)^2} $ - 4(y-1)(3cy-c) > 0 for any 0 < c < 1. (16-12c) $ {4^2} $ - (16-16c)y + (4-4c) > 0.
This is parabola opening upword.
Now if its discriminant < 0 then this equation is always > 0.
So this is equivalent to prove
$ {(16-16c)^2} $ - 4 (16-12c)(4-4c) < 0.
or $ {(2-2c)^2} $ - (4-3c)(1-c) < 0
or $ {c^2} $ -C < 0
Now given 1>c>0 so $ {c^2} $ <c
or $ {c^2} $ -c < 0
Conclusion: So for 1>c>0 y = $ {\frac{x^2+2x+c}{x^2+4x+3c}} $
Range over all real number when x varies over all real number.
