2015/032) If x is real,show that $\dfrac{x}{x^2-5x+9}$ always lie in the interval $[\frac{-1}{11},1]$

let $y = \dfrac{x}{x^2-5x+9}$
so $yx^2 - 5yx + 9y = x$

or $yx^2 - (5y+ 1) x + 9y = 0$

for it to have real root x discriminant > = 0 or
$(5y+1) ^2 - 36y^2 \ge 0$
=> $(5y + 1 + 6y)(5y+1 - 6y) \ge 0$

or $(11y + 1)(1-y) \ge 0$ or $y \ge \dfrac{-1}{11}$ and $y \le 1$

so $\dfrac{x}{x^2-5x+9}$ always lie in the interval $[\frac{-1}{11},1]$

I have solved it at https://in.answers.yahoo.com/question/index?qid=20130502090041AAY63AN where you can find some more solution