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Question

# The domain of the function $f\left(x\right)=\sqrt{2-2x-{x}^{2}}$ is (a) $\left[-\sqrt{3},\sqrt{3}\right]$ (b) $\left[-1-\sqrt{3},-1+\sqrt{3}\right]$ (c) [−2, 2] (d) $\left[-2-\sqrt{3},-2+\sqrt{3}\right]$

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Solution

## (b) $\left[-1-\sqrt{3},-1+\sqrt{3}\right]$ $f\left(x\right)=\sqrt{2-2x-{x}^{2}}$ $\mathrm{Since},2-2x-{x}^{2}\ge 0,\phantom{\rule{0ex}{0ex}}{x}^{2}+2x-2\le 0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2x-2+1-1\le 0\phantom{\rule{0ex}{0ex}}⇒{\left(x-1\right)}^{2}-{\left(\sqrt{3}\right)}^{2}\le 0\phantom{\rule{0ex}{0ex}}⇒\left[x-\left(-1-\sqrt{3}\right)\right]\left[x-\left(-1+\sqrt{3}\right)\right]\le 0\phantom{\rule{0ex}{0ex}}⇒\left(-1-\sqrt{3}\right)\le x\le \left(-1+\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{dom}\left(f\right)=\left[-1-\sqrt{3},-1+\sqrt{3}\right].\phantom{\rule{0ex}{0ex}}$

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