Let D be the domain of the function f(x) = ^ - 1 ( _ 3x ( 6 + 2 _ 3...

We have been given

$f\left(x\right) = \sin^{- 1}\left(\log_{3x}\left(\dfrac{6 + 2\log_{3}x}{- 5x}\right)\right)$

We know that the domain of $\sin^{- 1}x$ is $- 1 \le x \le 1$.

$⇒ \log_{3x}\left(\dfrac{6 + 2\log_{3}x}{- 5x}\right) \in \mathop{\left[- 11\right]}_{- \left(1\right)}, \mathop{3x > 0}_{- \left(2\right)}, \mathop{3x \ne 1}_{- \left(3\right)}, \mathop{\dfrac{6 + 2\log_{3}x}{- 5x}}_{- \left(4\right)} > 0$

$⇒$ From \left(2\right), x > 0,$from \left(3\right), x \ne \dfrac{1}{3}$

From \left(4\right), 6 + 2\log_{3}x 0)$</p><p>$⇒ \log_{3}x < - 3$</p><p>$⇒ x < {3}^{- 3}$</p><p>$⇒ x < \dfrac{1}{27}$</p><p>From \left(2\right), \left(3\right), \left(4\right), 0 < x < \dfrac{1}{27} . . . \left(5\right)$

From \left(1\right), - 1 \le \log_{3x}\left(\dfrac{6 + 2\log_{3}x}{- 5x}\right) \le 1$</p><p>$∵ 3x \in \left(0, \dfrac{1}{9}\right)$</p><p>$⇒ \dfrac{1}{3x} \ge \dfrac{6 + 2\log_{3}x}{- 5x} \ge 3x$</p><p>$⇒ - \dfrac{5}{3} \le 6 + 2\log_{3}x \le - 15{x}^{2}$</p><p>$⇒ - \dfrac{23}{6} \le \log_{3}x \le \dfrac{- 15{x}^{2} - 6}{2}$</p><p>$⇒ {3}^{- \dfrac{23}{6}} \le x \le 3\dfrac{- 15{x}^{2} - 6}{2}$</p><p>$⇒ \dfrac{1}{{3}^{\dfrac{23}{6}}} \le x \le \dfrac{1}{{3}^{\dfrac{15{x}^{2} + 6}{2}}}$</p><p>$⇒ 3\dfrac{- 23}{6} \le x < \dfrac{1}{{3}^{\dfrac{15{x}^{2} + 6}{2}}} . . . . . . \left(6\right)$, here$3\dfrac{- 23}{6}$is very small positive quantity,</p><p>From \left(5\right)$ and \left(6\right), 3\dfrac{- 23}{6} < x < \dfrac{1}{27}, $\left[x\right] = 0$

$⇒ g\left(x\right) = x$

$⇒$ Range $g\left(x\right)$ is domain of $f\left(x\right)$

$⇒ g\left(x\right) \in \left(3\dfrac{- 23}{6}, \dfrac{1}{27}\right) ≡ \left(\alpha, \beta\right)$.

$⇒ {\alpha}^{2} + \dfrac{5}{\beta} = {\left(3\dfrac{- 23}{6}\right)}^{2} + \dfrac{5}{\dfrac{1}{27}} = {\left(3\dfrac{- 23}{6}\right)}^{2} + 135 \approx 135$

Note: This question was bonus in Jee Main April 2023 session.