Question

The solution set of $16-x^2\ge 0$ is

• A. $[-4,4]$
• B. $(-\infty ,-4]\cup [4,\infty )$
• C. $(-4,4)$
• D. $(-16,16)$

Answer 1

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

$16-x^2\ge 0\Rightarrow x^2-16\le 0\Rightarrow (x+4)(x-4)\le 0$


$\therefore x\in [-4,4]$