A) \[[-2,-1)\cup [1,2]\]
B) \[(-2,-1]\cup [1,2]\]
C) \[[-2,-1]\cup [1,2]\]
D) \[(-2,-1)\cup (1,2)\]
Correct Answer: C
Solution :
[c] For \[f(x)\]to be defined we must have \[-1\le {{\log }_{2}}\left( \frac{1}{2}{{x}^{2}} \right)\le 1\Rightarrow {{2}^{-1}}\le \frac{1}{2}{{x}^{2}}\le {{2}^{1}}\] \[\Rightarrow 1\le {{x}^{2}}\le 4\] ?. (1) Now, \[1\le {{x}^{2}}\Rightarrow {{x}^{2}}-1\ge 0\]i.e. \[(x-1)(x+1)\ge 0\] \[\Rightarrow x\le -1orx\ge 1\] ?. (2) Also \[{{x}^{2}}\le 4\Rightarrow {{x}^{2}}-4\le 0\]i.e. \[(x-2)(x+2)\le 0\] \[\Rightarrow -2\le x\le 2\] ?. (3) Form (2) and (3), we get the domain of \[f=\{(-\infty ,-1]\cup [1,\infty )\}\cap [-2,2]\] \[=[-2,-1]\cup [1,2]\]
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