Question:
The domain of the function $f(x)=\sqrt{2-2 x-x^{2}}$ is
(a) $[-\sqrt{3}, \sqrt{3}]$
(b) $[-1-\sqrt{3},-1+\sqrt{3}]$
(c) $[-2,2]$
(d) $[-2-\sqrt{3},-2+\sqrt{3}]$
Solution:
(b) $[-1-\sqrt{3},-1+\sqrt{3}]$
$f(x)=\sqrt{2-2 x-x^{2}}$
Since, $2-2 x-x^{2} \geq 0$
$x^{2}+2 x-2 \leq 0$
$\Rightarrow x^{2}-2 x-2+1-1 \leq 0$
$\Rightarrow(x-1)^{2}-(\sqrt{3})^{2} \leq 0$
$\Rightarrow[x-(-1-\sqrt{3})][x-(-1+\sqrt{3})] \leq 0$
$\Rightarrow(-1-\sqrt{3}) \leq x \leq(-1+\sqrt{3})$
Thus, $\operatorname{dom}(f)=[-1-\sqrt{3},-1+\sqrt{3}]$.