Question

The solution set of $x^2-16\le 0$ and $x^2-9\ge 0$ is

• A. $[4,\infty )$
• B. $(-\infty ,3]$
• C. $(-\infty ,-3]\cup [3,\infty )$
• D. $[-4,-3]\cup [3,4]$

Answer 1

The correct option is D $[-4,-3]\cup [3,4]$

$x^2-16\le 0\Rightarrow (x+4)(x-4)\le 0\Rightarrow x\in [-4,4]\cdots \left(1\right)$
$x^2-9\ge 0\Rightarrow (x+3)(x-3)\ge 0\Rightarrow x\in (-\infty ,-3]\cup [3,\infty )\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,
$x\in [-4,-3]\cup [3,4]$