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 $y=\frac{x^2-x+1}{x^2+x+1}$

$\Rightarrow y{x}^2+yx+y=x^2-x+1$

$\Rightarrow (x-1)x^2+(y+1)x+(y-1)=0$

If $x\in R$, then

Discriminant $\ge 0$

$\Rightarrow {(y+1)}^2-4{(y-1)}^2\ge 0$

$\Rightarrow -3y^2+10y-3\ge 0$

$\Rightarrow 3y^2-10y+3\le 0$

$\Rightarrow (3y-1)(y-3)\le 0$

$\Rightarrow \frac{1}{3}\le y\le 3$

$\therefore$ Range $=[\frac{1}{3},3]$