The domain of the function f(x) = ^ - 1 [2 x ^ 2 - 3] + _ 2...

Given $f$\left(x\right) = \sin^{- 1}$\left[2{x}^{2} - 3\right] + \log_{2}\left(log\dfrac{1}{2}\left({x}^{2} - 5x + 5\right)\right)$

For $f\left(x\right)$ to be defined

$- 1 \le \left[2{x}^{2} - 3\right] < 1$

$⇒ - 1 \le 2{x}^{2} - 3 < 2$

$⇒ 2 < 2{x}^{2} < 5$

$⇒ 1 < {x}^{2} < \dfrac{5}{2}$

i.e. $x \in \left(- \sqrt{\dfrac{5}{2}}, - 1\right) \cup \left(1, \sqrt{\dfrac{5}{2}}\right) . . . \left(i\right)$

and ${x}^{2} - 5x + 5 > 0$

$⇒ \left(x - \left(\dfrac{5 - \sqrt{5}}{2}\right)\right)\left(x - \left(\dfrac{5 + \sqrt{5}}{2}\right)\right) > 0$

i.e. \left(- \infty , \dfrac{5 - \sqrt{5}}{2}\right) \cup \left(\dfrac{5 + \sqrt{5}}{2}, \infty\right) . . . \left(ii\right)$</p><p>Also$log\dfrac{1}{2}\left({x}^{2} - 5x + 5\right) > 0$</p><p>$⇒ {x}^{2} - 5x + 5 < 1$</p><p>$⇒ {x}^{2} - 5x + 4 < 0$</p><p>i.e.$x \in \left(1, 4\right) . . . \left(iii\right)$</p><p>From \left(i\right), \left(ii\right) \text{ and } \left(iii\right)$

So, $x \in \left(1, \dfrac{5 - \sqrt{5}}{2}\right)$