Question

The range of $\frac{x^2-x+1}{x^2+x+1}$ is

• A. $[\frac{1}{3},3]$
• B. $[\frac{1}{3},1]$
• C. $[1,3]$
• D. $(-\infty ,\frac{1}{3}]\cup [3,\infty )$

Answer 1

The correct option is A $[\frac{1}{3},3]$

Let $y=\frac{x^2-x+1}{x^2+x+1}\Rightarrow y{x}^2+yx+y=x^2-x+1\Rightarrow (y-1)x^2+(y+1)x+(y-1)=0$
$xϵR\Rightarrow Discriminant\ge 0\Rightarrow (y+1{)}^2-4(y-1{)}^2\ge 0\Rightarrow -3y^2+10y-3\ge 0$
$\Rightarrow 3y^2-10y+3\le 0\Rightarrow (3y-1)(y-3)\le 0\Rightarrow \frac{1}{3}\le y\le 3$
Range= $[\frac{1}{3},3]$