Question

Find the interval of $x$ such that $\left|\frac{x^2-3x-1}{x^2+x+1}\right|<3$ is satisfied is

• A. $(-\infty ,-2)\cup (-1,\infty )$
• B. $[\frac{8}{5},2]\cup [\frac{5}{2},\infty ]$
• C. $(-5,-2)\cup (2,5)$
• D. None of these

Answer 1

The correct option is A

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

As $x^2+x+1>0\forall x$, we get $|x^2-3x-1|<3(x^2+x+1)$

$\iff -3(x^2+x+1)<x^2-3x-1<3(x^2+x+1)$
$\iff -2<4x^2$ and $2x^2+6x+4>0$
$\iff x^2+3x+2>0$
$\iff$ x < -2 or x > -1