The domain of the function f (x) = 2 ^ - 1 \ _ 2 ((1)/(2) x ^ 2 )\...

For $f\left(x\right)$ to be defined, we must have
- 1 \le lo{g}_{2}\left(\dfrac{1}{2}{x}^{2}\right) \le 1 ⇒ {2}^{- 1} \le \dfrac{1}{2}{x}^{2} \le {2}^{1} \left[asthebase = 2 > 1\right]
$⇒ 1 \le {x}^{2} \le 4 \ldots . \left(i\right)$
Now, $1 \le {x}^{2} ⇒ {x}^{2} - 1 \ge 0$ i.e. \left(x - 1\right)\left(x + 1\right) \ge 0$<br />$⇒ x \le - 1 or $x \ge 1 \ldots . \left(ii\right)$
Also, ${x}^{2} \le 4 ⇒ {x}^{2} - 4 \le 0$ i.e. \left(x - 2\right)\left(x + 2\right) \le 0$<br />$⇒ - 2 \le x \le 2 \ldots . . \left(iii\right)$<br />From Equation \left(ii\right) and \left(iii\right)$, we get,$</p><p>the domain of$f= \left(\left(- \infty, - 1\right] \cup \left[1, \infty\right)\right) \cap \left[- 2,2\right]<br />= \left[- 2, - 1\right]$\cup \left[1,2\right]$.