A) \[[-1,\,\,2]-\{0\}\]
B) \[[-2,\,\,2]-(-1,\,\,1)\]
C) \[[-2,\,\,2]-\{0\}\]
D) \[[1,\,\,2]\]
Correct Answer: B
Solution :
Key Idea The domain of\[{{\sin }^{-1}}x\]is\[[-1,\,\,1]\]. Since, domain of \[{{\sin }^{-1}}x\]is\[[-1,\,\,1]\]. \[\therefore \] \[{{\sin }^{-1}}\left[ {{\log }_{2}}\frac{{{x}^{2}}}{2} \right]\] \[\Rightarrow \] \[-1\le {{\log }_{2}}\frac{{{x}^{2}}}{2}\le 1\] \[\Rightarrow \] \[{{2}^{-1}}\le \frac{{{x}^{2}}}{2}\le 2\] \[\Rightarrow \] \[\frac{1}{2}\le \frac{{{x}^{2}}}{2}\le 2\] \[\Rightarrow \] \[1\le {{x}^{2}}\le 4\] \[\Rightarrow \] \[|x|\le 2\]and\[|x|\,\,\ge 1\] \[\Rightarrow \] \[x\in \{[-2,\,\,2]-(-1,\,\,1)\}\].
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