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 $-1-\sqrt{3}\le x\le -1+\sqrt{3}$

$f\left(x\right)=\sqrt{2-2x-x^2}$

$2-2x-x^2\ge 0$

$2-2x-x^2=0$

$x^2+2x-2=0$

$x=\frac{-2\pm \sqrt{4+8}}{2}$

$x=\frac{-2\pm \sqrt{12}}{2}$

$x=-1\pm \sqrt{3}$

$\therefore -1-\sqrt{3}\le x\le -1+\sqrt{3}$

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