Question

Range of the expression $f\left(x\right)=\frac{x^2+x+1}{x^2-x+1},x\in R$ is

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

Answer 1

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

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

$\because x\in R\Rightarrow \Delta \ge 0$
$\Rightarrow (y+1{)}^2-4(y-1{)}^2\ge 0$
$\Rightarrow -3y^2+10y-3\ge 0$
$\Rightarrow 3y^2-10y+3\le 0$
$\Rightarrow (3y-1)(y-3)\le 0$
$\Rightarrow y\in [\frac{1}{3},3]$