Question

The domain of the function $f\left(x\right)={\sin }^{-1}\left\{{\log }_2\left(\frac{1}{2}{x}^2\right)\right\}$ is

• A. $[-2,-1)\cup [1,2]$
• B. $(-2,-1]\cup [1,2]$
• C. $[-2,-1]\cup [1,2]$
• D. $(-2,-1)\cup (1,2)$

Answer 1

The correct option is C $[-2,-1]\cup [1,2]$

For $f\left(x\right)={\sin }^{-1}\left\{{\log }_2\left(\frac{1}{2}{x}^2\right)\right\}$ 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) $[\because$ the base $=2>1]$
Now, $1\le x^2\Rightarrow x^2-1\ge 0\Rightarrow (x-1)(x+1)\ge 0\Rightarrow x\le -1$ or $x\ge 1$ ...(2)
Also, $x^2\le 4\Rightarrow x^2-4\le 0\Rightarrow (x-2)(x+2)\le 0\Rightarrow -2\le x\le 2$ ...(3)
From (2) and (3), we get the domain of $f$
$=\left(\right(-\infty ,-1]\cup [1,\infty \left)\right)\cap [-2,2]=[-2,-1]\cup [1,2]$