Question

Find the domain of the following function:

$f\left(x\right)={\sin }^{-1}\left[{\log }_2\left(\frac{x^2}{2}\right)\right]$, where $[\cdot ]$ denotes the greatest integer function

• A. $[-2,-1]\cup [1,2]$
• B. $[-\sqrt{\frac{7}{2},}-1]\cup [1,\sqrt{\frac{7}{2}}]$
• C. $[-\sqrt{\frac{1}{2},}-1]\cup [1,\sqrt{\frac{1}{2}}]$
• D. $(-2\sqrt{2},-1]\cup [1,2\sqrt{2})$

Answer 1

The correct option is D $(-2\sqrt{2},-1]\cup [1,2\sqrt{2})$

Given function is, $f\left(x\right)={\sin }^{-1}\left[{\log }_2\left(\frac{x^2}{2}\right)\right]$

For this function to be defined, $-1\le \left[{\log }_2\left(\frac{x^2}{2}\right)\right]\le 1$

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

$\Rightarrow -{\log }_{2}2\le {\log }_2\left(\frac{x^2}{2}\right)<2{\log }_{2}2$ ...... $[\because {\log }_{a}a=1]$

$\Rightarrow {\log }_2\left(\frac{1}{2}\right)\le {\log }_2\left(\frac{x^2}{2}\right)<{\log }_2{2}^2$ ......... $[-\log (a)=\log \left(\frac{1}{a}\right)\text{and}a\log b=\log b^a]$

$\Rightarrow \frac{1}{2}\le \frac{x^2}{2}<4\Rightarrow 1\le x^2<8$

$\Rightarrow x\in (-2\sqrt{2},-1]\cup [1,2\sqrt{2})$