* problem*: 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.

* solution:* 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.