Question

Domain of the function $f\left(x\right)=\sqrt{2-2x-x^2}$ is

• A. $-\sqrt{3}\le x\le \sqrt{3}$
• B. $-1-\sqrt{3}\le x\le -1+\sqrt{3}$
• C. $-2\le x\le 2$
• D. $-2-\sqrt{3}\le x\le -2+\sqrt{3}$

Answer 1

The correct option is B $-1-\sqrt{3}\le x\le -1+\sqrt{3}$

Given $f\left(x\right)=\sqrt{2-2x-x^2}$
for function to be defined
$2-2x-x^2\ge 0$
$\Rightarrow x^2+2x+1\le 3$
$\Rightarrow (x+1{)}^2-(\sqrt{3}{)}^2\le 0\Rightarrow (x+1+\sqrt{3})(x+1-\sqrt{3})\le 0$
$-1-\sqrt{3}\le x\le -1+\sqrt{3}$

$\therefore$ Domain of $f\left(x\right)$ is :
$-1-\sqrt{3}\le x\le -1+\sqrt{3}$