Question

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

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

Answer 1

Answer: (C)

For $f\left(x\right)$ to be defined, we must have
$-1\le lo{g}_2\left(\frac{1}{2}{x}^2\right)\le 1\Rightarrow 2^{-1}\le \frac{1}{2}{x}^2\le 2^1$ $\left[asthebase=2>1\right]$
$\Rightarrow 1\le x^2\le 4\dots .\left(i\right)$
Now, $1\le x^2\Rightarrow x^2-1\ge 0$ i.e. $\left(x-1\right)\left(x+1\right)\ge 0$
$\Rightarrow x\le -1$ or $x\ge 1\dots .\left(ii\right)$
Also, $x^2\le 4\Rightarrow x^2-4\le 0$ i.e. $\left(x-2\right)\left(x+2\right)\le 0$
$\Rightarrow -2\le x\le 2\dots ..\left(iii\right)$
From Equation $\left(ii\right)$ and $\left(iii\right)$ , we get,
the domain of $f$ $=\left(-\in fty,-1\right]\cup \left[1,\in fty\right)\cap \left[-2,2\right]$
$=\left[-2,-1\right]\cup \left[1,2\right]$ .