Question

The solution set of $\frac{x^2-4}{x^2-16}\le 0$ is

• A. $(-\infty ,-4)\cup (4,\infty )$
• B. $[-4,4]$
• C. $(-4,-2]\cup [2,4)$
• D. $[-4,-2]\cup [2,4]$

Answer 1

The correct option is C $(-4,-2]\cup [2,4)$

$\frac{x^2-4}{x^2-16}\le 0\Rightarrow \frac{(x+2)(x-2)}{(x+4)(x-4)}\le 0,x\ne \pm 4$


$\therefore x\in (-4,-2]\cup [2,4)$