problem: If c is a real number with 0 < c < 1, then show that the values taken by the function y = , as x varies over real numbers, range over all real numbers.
solution: y =
or + 4yx + 3cy = + 2x + c
or (y-1) + (4y-2)x + 3cy – c = 0
Now if we can show that the discriminant for all y then for all real y there exist a real x.
Now, discriminant is – 4(y-1)(3cy-c)
We need to show – 4(y-1)(3cy-c) > 0 for any 0 < c < 1. (16-12c) – (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
– 4 (16-12c)(4-4c) < 0.
or – (4-3c)(1-c) < 0
or -C < 0
Now given 1>c>0 so <c
or -c < 0
Conclusion: So for 1>c>0 y =
Range over all real number when x varies over all real number.