Question

Find the range of $x$ for the following expression:

$\left|\frac{x^2-3x+2}{x^2+3x+2}\right|\ge 1$

• A. (0,81)
• B. $(-\infty ,1)$
• C. $(-\infty ,-2)\cup (-2,-1)\cup (-1,0)$
• D. None of these

Answer 1

The correct option is C

$(-\infty ,-2)\cup (-2,-1)\cup (-1,0)$

$x^2-3x+2=(x-1)(x-2)and{x}^2+3x+2=(x+1)(x+2)$

For $x<-2$:
$x^2-3x+2\ge x^2+3x+2⟹x\le 0$
So the interval is $(-\infty ,-2)$

For $-2<x<-1$, we obtain:
$x^2-3x+2\ge -(x^2+3x+2)$
$⟹2(x^2+2)\ge 0$
$⟹x\in R$
So, the interval is $(-2,-1)$

For $-1<x<1$, we obtain:
$x^2-3x+2\ge x^2+3x+2$
$⟹-3x\ge 3x$
$⟹x\le 0$
So the interval is $(-1,0]$

For $1\le x\le 2$, we obtain:
$-(x^2-3x+2)\ge x^2+3x+2$
$⟹2(x^2+2)\le 0$
So, no value is possible in this interval.

For $2<x$:
$x^2-3x+2\ge x^2+3x+2$
$⟹-3x\ge 3x$
$⟹x\le 0$
So, no value is possible in this interval.