Question

For real values of $x$ the expression $\frac{x^2-x+1}{x^2+x+1}$ takes values in interval

• A. $[\frac{1}{3},3]$
• B. $(-\infty ,\frac{1}{3}]$
• C. $[3,\infty )$
• D. $[2,\infty )$

Answer 1

Answer: (A)

$[\frac{1}{3},3]$

$\frac{x^2+x+1-x+x}{x^2+x+1}$
$\Rightarrow 1-\left(\frac{2x}{x^2+x+1}\right)$
$\Rightarrow f\left(x\right)=\left(\frac{2x}{x^2+x+1}\right)$
$\Rightarrow f^{\prime }\left(x\right)=0$

$\Rightarrow \frac{2(x^2+x+1)-2x(2x+1)}{(x^2+x+1{)}^2}=0$
$\Rightarrow 2x^2+2x+2-4x^2-2x=0$
$\Rightarrow x^2=1$
$\Rightarrow x=\pm 1$
$f\left(max\right)=\frac{2}{3}$
$f\left(min\right)=-2$
$f\left(x\right)=(1-\frac{2x}{x^2+x+1})$
$f\left(x\right)\left(max\right)=3$
$f\left(x\right)\left(min\right)=\frac{1}{3}$
$\therefore$ Range = $[\frac{1}{3},3]$
Hence. option 'A' is correct.