If x is real show that x / (x2-5x+9) always lies in the interval [ -1/ 11, 1] - Maths - Complex Numbers and Quadratic Equations

Question

If x is real show that x / (x2-5x+9) always lies in the interval
[ -1/ 11, 1]

Santosh Kumar · Accepted Answer

$let \frac{x}{x^{2} - 5 x + 9} = y \Rightarrow x^{2} y - 5 x y + 9 y = x \Rightarrow y x^{2} - (5 y + 1) x + 9 y = 0$

As x is real, so D > 0

$[- (5 y + 1)]^{2} - 4 y (9 y) > = 0 \Rightarrow (5 y + 1)^{2} - 36 y^{2} > 0 \Rightarrow 25 y^{2} + 10 y + 1 - 36 y^{2} > 0 \Rightarrow - 11 y^{2} + 10 y + 1 > 0 \Rightarrow 11 y^{2} - 10 y - 1 < 0 \Rightarrow 11 y^{2} - 11 y + y - 1 < 0 \Rightarrow 11 y (y - 1) + 1 (y - 1) < 0 \Rightarrow (11 y + 1) (y - 1) < 0 \Rightarrow \frac{- 1}{11} < y < 1$