Question

The domain of $f\left(x\right)={\sin }^{-1}\left\{{\log }_3\left(\frac{x^2}{3}\right)\right\}$ is

• A. $(-\infty ,3]$
• B. $[3,\infty )$
• C. $[-3,-1]\cup [1,3]$
• D. $(-9,-1)\cup (1,9)$

Answer 1

The correct option is C $[-3,-1]\cup [1,3]$

${\sin }^{-1}x$ is defined if $x\in [-1,1]$

$\Rightarrow -1\le {\log }_3\frac{x^2}{3}\le 1$

$\Rightarrow \frac{1}{3}\le \frac{x^2}{3}\le 3$

$1\le x^2\le 9$

$\Rightarrow x\in [-1,-3]\cup [1,3]$

$\therefore x\in [-3,-1]\cup [1,3]$