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

Question

The range of the function $f(x)=\frac{x^2-x+1}{x^2+x+1}$ where $\mathrm{x} \in \mathrm{R}$, is, $(-\infty, 3]$, $(-\infty, \infty)$, $[3, \infty)$, $\left[\frac{1}{3}, 3\right]$

MARKS App · Accepted Answer

$\begin{aligned} & \text {Let } y=\frac{x^2-x+1}{x^2+x+1} \ & \Rightarrow \mathrm{x}^2(\mathrm{y}-1)+\mathrm{x}(\mathrm{y}+1)+(\mathrm{y}-1)=0 \ & \Rightarrow x=\frac{-(y+1) \pm \sqrt{(y+1)^2-4(y-1)^2}}{2(y-1)} \ & =\frac{-(y+1) \pm \sqrt{-3 y^2+10 y-3}}{2(y-1)} \text { is real iff } \ & \mathrm{y}-1 \neq 0 \Rightarrow \mathrm{y} \neq 1 \ & \text { If } y=1 \text { then original equation gives } \mathrm{x}=0 \text {, } \ & \text { so taking } \mathrm{y}=1 \ & \text { Also } 3 y^2-10 y+3 \leq 0 \ & \Rightarrow(3 y-1)(y-3) \leq 0 \ & \Rightarrow \mathrm{y} \in\left[\frac{1}{3}, 3\right] \quad \therefore \text { Range is }\left[\frac{1}{3}, 3\right] \ & \end{aligned}$

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