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# The domain of the function f(x)

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, $\operatorname{dom}(f)=[-1-\sqrt{3},-1+\sqrt{3}]$.