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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

## 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.