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) \[-2+\sqrt{3}\le x\le -2-\sqrt{3}\]
Correct Answer: B
Solution :
\[f(x)=\sqrt{2-2x-{{x}^{2}}}\]is defined, if \[2-2x-{{x}^{2}}\ge 0\] \[\Rightarrow \] \[{{x}^{2}}+2x-2\le 0\] Now, \[x=\frac{-2\pm \sqrt{4+8}}{2}\] \[\Rightarrow \] \[x=\frac{-2\pm 2\sqrt{3}}{2}\] \[\Rightarrow \] \[x=-1\pm \sqrt{3}\] \[\therefore \] \[-1-\sqrt{3}\le x\le -1+\sqrt{3}\]
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