Question

Solution set of the inequality, $2-{\log }_2(x^2+3x)\ge 0$ is

• A. $[-4,1]$
• B. $[-4,-3)\cup (0,1]$
• C. $(-\infty ,-3)\cup (1,\infty )$
• D. $(-\infty ,-4)\cup (1,\infty )$

Answer 1

Answer: (B)

$[-4,-3)\cup (0,1]$

$2-{\log }_2(x^2+3x)\ge 0(x^2+3x)>0\Rightarrow x(x+3)>0\Rightarrow x\in (-\infty ,-3)\cup (0,\infty )$
Now,
$2\ge {\log }_2(x^2+3x)\Rightarrow 4\ge (x^2+3x)\Rightarrow x^2+3x-4\le 0\Rightarrow (x+4)(x-1)\le 0\Rightarrow x\in [-4,1]\therefore x\in [-4,-3)\cup (0,1]$