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
$-1\leq log_{2}\left(\frac{1}{2} x^{2}\right)\leq 1\Rightarrow 2^{- 1}\leq \frac{1}{2}x^{2}\leq 2^{1}$ $\left[as t h e b a s e = 2 > 1\right]$
$\Rightarrow 1\leq x^{2}\leq 4\ldots .\left(i\right)$
Now, $1\leq x^{2}\Rightarrow x^{2}-1\geq 0$ i.e. $\left(x - 1\right)\left(x + 1\right)\geq 0$
$\Rightarrow x\leq -1$ or $x\geq 1\ldots .\left(i i\right)$
Also, $x^{2}\leq 4\Rightarrow x^{2}-4\leq 0$ i.e. $\left(x - 2\right)\left(x + 2\right)\leq 0$
$\Rightarrow -2\leq x\leq 2\ldots ..\left(i i i\right)$
From Equation $\left(i i\right)$ and $\left(i i i\right)$ , we get,
the domain of $f$ $=\left(- \in fty , - 1\right] \cup \left[1 , \in fty\right)\cap \left[- 2,2\right]$
$=\left[- 2 , - 1\right]\cup\left[1,2\right]$ .