Question

The domain of ${\sin }^{-1}\left[{\log }_2\right(x^2/2\left)\right]$ is

• A. $[2,1]$
• B. $[1,2]$
• C. $[-2,-1]\cup [1,2]$
• D. $[-2,0]$

Answer 1

Answer: (C)

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

${\sin }^{-1}\left(x\right)domain:x\in [-1,1]$

$\therefore$domain of ${\sin }^{-1}\left[{\log }_2\frac{x^2}{2}\right]$

$\therefore {\log }_2\frac{x^2}{2}\in [-1,1]$

$-1\le {\log }_2\frac{x^2}{2}\le 1$ and $\frac{1}{2}\le \frac{x^2}{2}\le 2$

$1\le x^2\le 4$

For $x^2\ge 1$

$x\in [-\infty ,-1]\cup [1,\infty ]$...........(i)

For $x^2\le 4$

$x\in [-2,2]$.................(ii)

By taking intersection of (i) & (ii)

$\therefore xϵ[-2,-1]\cup [1,2]$