Question

Solve the irrational inequality $\sqrt{4-x^2}+\frac{\sqrt{x^2}}{x}\ge 0$

• A. $[-\sqrt{3},0)\cup (0,2]$
• B. $[-2,2]$
• C. $[2,4]\cup [5,13]$
• D. $\varphi$

Answer 1

The correct option is A $[-\sqrt{3},0)\cup (0,2]$

Given $\sqrt{4-x^2}+\frac{\sqrt{x^2}}{x}\ge 0$ will be true
if $4-x^2\ge 0\Rightarrow (x-2)(x+2)\le 0,x\ne 0$
$x\in [-2,2]-\left\{0\right\}\cdots \left(1\right)$

Case $1:$ when $x\ge 0$
$\sqrt{4-x^2}\ge -\frac{\sqrt{x^2}}{x}\Rightarrow -\frac{\sqrt{x^2}}{x}<0\Rightarrow \frac{\sqrt{x^2}}{x}>0\Rightarrow x>0\cdots \left(1\right)$

Case $2:$ when $x<0$
$\sqrt{4-x^2}\ge -\frac{\sqrt{x^2}}{x}\Rightarrow 4-x^2\ge 1,x\ne 0\Rightarrow x^2-3\le 0,x\ne 0\Rightarrow (x-\sqrt{3})(x+\sqrt{3})\le 0,x\ne 0$
$\therefore x\in [-\sqrt{3},0)\cdots \left(3\right)$

Hence from $\left(1\right),\left(2\right)$ and $\left(3\right)$
$x\in [-\sqrt{3},2]-\left\{0\right\}$