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 $[-2,-1]\cup [1,2]$

$f\left(x\right)={\sin }^{-1}\left\{{\log }_2\left(\frac{1}{2}{x}^2\right)\right\}$

For ${\sin }^{-1}y$ function

$y$ should define as, $-1<y<1$

Using this fact,

$-1\le {\log }_2\left(\frac{1}{2}{x}^2\right)\le 1$

Taking antilog with base $2$,

$2^{-1}\le \frac{1}{2}{x}^2\le 2^1$

$\Rightarrow \frac{1}{2}\le \frac{1}{2}{x}^2\le 2$

$\Rightarrow 1\le x^2\le 4$

$\Rightarrow 1\le x^2$ and $x^2\le 4$

$\Rightarrow x^2-1\ge 0$ and $x^2-4\le 0$

$\Rightarrow (x+1)(x-1)\ge 0$ and $(x-2)(x+2)\le 0$

$\Rightarrow x\ge 1andx\le -1$ and $-2\le x\le 2$

taking the intersection of the above inequalities

we get

$\therefore$ Domain is $[-2,-1]\cup [1,2]$