Question

The range of the function $f\left(x\right)=\frac{x^{2}-x+1}{x^{2}+x+1}$ where $x \in R$

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

Answer 1

The Correct Option is D

We have, $y=\frac{x^{2}-x+1}{x^{2}+x+1}$ $\Rightarrow y\,x^{2}+yx+y-x^{2}+x-1=0$ $\Rightarrow x^{2}\left(y-1\right)+x\left(y+1\right)+\left(y-1\right)=0$ $\therefore x=\frac{-\left(y+1\right) \pm\sqrt{\left(y+1\right)^{2}-4\left(y-1\right)^{2}}}{2\left(y-1\right)}$ $=\frac{-\left(y+1\right) \pm\sqrt{-3y^{2}+10y-3}}{2\left(y-1\right)}$ If $y = 1$ then original equation gives $x = 0$. Also, $-3y^{2}+10y-3 \ge 0$ $\Rightarrow 3y^{2}-10y+3 \le 0$ $\Rightarrow 3y^{2}-9y-y+3 \le 0$ $\Rightarrow \left(3y-1\right)\left(y-3\right) \le 0$ $\Rightarrow y \in\left[\frac{1}{3}, 3\right]$ $\therefore$ Range is $\left[\frac{1}{3}, 3\right]$.