Question

The solution set of $3^{x^2-4}\ge 243$ is

• A. $(-2,2)$
• B. $[-2,2]$
• C. $(-\infty ,-2]\cup [2,\infty )$
• D. $(-\infty ,-3]\cup [3,\infty )$

Answer 1

The correct option is D $(-\infty ,-3]\cup [3,\infty )$

$3^{x^2-4}\ge 243\Rightarrow 3^{x^2-4}\ge 3^5\Rightarrow x^2-4\ge 5\Rightarrow x^2-9\ge 0\Rightarrow (x+3)(x-3)\ge 0$


$\therefore x\in (-\infty ,-3]\cup [3,\infty )$