Question

The range of $f\left(x\right)=\frac{x^2-x+1}{x^2+x+1}$ is

• A. $\left[\frac{1}{3},3\right]$
• B. $\left[\frac{1}{2},2\right]$
• C. $[0,1]$
• D. $[-1,1]$

Answer 1

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

Given, $f\left(x\right)=\frac{x^2-x+1}{x^2+x+1}=y$ (say)
$\Rightarrow (y-1)x^2+(y+1)x+(y-1)=0$
Here $x$ is real, Thus discriminant of above quadratic $D\ge 0$
$\Rightarrow (y+1{)}^2-4(y-1{)}^2\ge 0$
$\Rightarrow [y+1+2(y-1\left)\right][y+1-2(y-1\left)\right]\ge 0$
$\Rightarrow (3y-1)(3-y)\ge 0\Rightarrow (3y-1)(y-3)\le 0$
$\Rightarrow y\in [\frac{1}{3},3]$, which is range of $f\left(x\right)$