Question

Exhaustive set of value of $x$ satisfying ${\log }_{|x|}(x^2+x+1)\ge 0$ is

• A. $(-1,0)$
• B. $(-\infty ,1)\cup (1,\infty )$
• C. $(-\infty ,\infty )-(-1,0,1)$
• D. $(-\infty ,-1)\cup (-1,0)\cup (1,\infty )$

Answer 1

The correct option is D $(-\infty ,-1)\cup (-1,0)\cup (1,\infty )$

${\log }_{|x|}(x^2+x+1)\ge 0$
${\log }_x(x^2+x+1)\ge 0$ (since, base of logarithm is to be positive)
Here, the base $x$ can be either $>1$ or $0<x<1$
Case I: When $x>1$
$f\left(x\right)\ge 1$
$\Rightarrow x^2+x+1\ge 1$
$\Rightarrow x(x+1)\ge 0$
$\Rightarrow x\in (-\infty ,-1)\cup (0,\infty )$ ....(1)
Case II: When $0<|x|<1$
$f\left(x\right)\le 1$
$\Rightarrow x^2+x+1\le 1$
$\Rightarrow x(x+1)\le 0$
$\Rightarrow x\in (-1,0)$ ....(2)
From (1) and (2), it follows that
$x\in (-\infty ,-1)\cup (-1,0)\cup (1,\infty )$