Question

Range of rational expression $y=\frac{x^2-x+4}{x^2+x+4},x\in R$ is

• A. $[\frac{2}{3},\frac{5}{3}]$
• B. $[-\frac{5}{3},\frac{5}{3}]$
• C. $[-\frac{3}{5},\frac{3}{5}]$
• D. $[\frac{3}{5},\frac{5}{3}]$

Answer 1

The correct option is D $[\frac{3}{5},\frac{5}{3}]$

Let $y=\frac{x^2-x+4}{x^2+x+4}$
$\Rightarrow (y-1)x^2+(y+1)x+(4y-4)=0$

$\because x\in R\Rightarrow \Delta \ge 0$
$\Rightarrow (y+1{)}^2-4(y-1\left)\right(4y-4)\ge 0$
$\Rightarrow (y+1{)}^2-16(y-1{)}^2\ge 0$
$\Rightarrow -15y^2+34y-15\ge 0$
$\Rightarrow 15y^2-34y+15\le 0$
$\Rightarrow (5y-3)(3y-5)\le 0$
$\Rightarrow y\in [\frac{3}{5},\frac{5}{3}]$