Question

The solution set of $\frac{1}{x^2-9}\ge 0$ is

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

Answer 1

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

$\frac{1}{x^2-9}\ge 0\Rightarrow x^2-9>0\Rightarrow (x+3)(x-3)>0$


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