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Range of a Polynomial (Tomato subjective 66)

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.

June 15, 2015
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