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

The correct option is A $[-4,1]$

Consider the given inequality,

$2-{\log }_2(x^2+3x)\ge 0$

$2\ge {\log }_2(x^2+3x)$

$\because {\log }_{a}X=Y\Rightarrow X=a^y$

$\therefore 2^2\ge x^2+3x$

$x^2+3x-4\le 0$

$x^2+4x-x-4\le 0$

$x(x+4)-1(x+4)\le 0$

$(x+4)(x-1)\le 0$

When,

$(x+4)\le 0$

$x\le -4$

When,

$x-1\le 0$

$x\le 1$

$x\in [-4,1]$

Hence, this is the answer.