Let $\mathrm{D}$ be the domain of the function $f(x)=\sin ^{-1}\left(\log _{3 x}\left(\frac{6+2 \log _{3} x}{-5 x}\right)\right)$. If the range of the function $\mathrm{g}: \mathrm{D} \rightarrow \mathbb{R}$ defined by $\mathrm{g}(x)=x-[x],([x]$ is the greatest integer function), is $(\alpha, \beta)$, then $\alpha^{2}+\frac{5}{\beta}$ is equal to

<p>First, the function $f(x) = \sin^{-1}(\log_{3x}(\frac{6 + 2 \log_3{x}}{-5x}))$ has several restrictions :</p> <ol> <li><p>Since the arcsine function $\sin^{-1}(x)$ is only defined for $-1\leq x\leq 1$, this means that $\log_{3x}(\frac{6 + 2 \log _3 x}{-5 x})$ must be between -1 and 1.</p> </li> <li><p>For the logarithm to be defined, $3x &gt; 0 \Rightarrow x &gt; 0$ .......(1)<br/><br/> because the base of a logarithm must be greater than 0 and not equal to 1. Also, $x \neq \frac{1}{3}$ .......(2)<br/><br/> as the base cannot be 1.</p> </li> <li><p>Moreover, the inner function of the logarithm $\frac{6 + 2 \log_3{x}}{-5x}$ must be greater than 0. <br/><br/>$\Rightarrow$ $6+2 \log _3 x<0 \quad(\because x>0)$ <br/><br/>$$ \begin{aligned} \Rightarrow & \log _3 x<-3 \\\\ \Rightarrow & x<3^{-3} \\\\ \Rightarrow & x<\frac{1}{27} .........(3) \end{aligned} $$ <br/><br/>$\Rightarrow \text { From (1), (2), (3), } 0 < x < \frac{1}{27}$</p> </li> </ol> <p>The next step is to solve the inequality for $-1 \leq \log_{3x}(\frac{6+2 \log _3 x}{-5 x}) \leq 1$. To do this, we make the observation that for the logarithmic part to be within $[-1,1]$, it must be true that $3x \leq \frac{6+2 \log_3 x}{-5x} \leq \frac{1}{3x}$.</p> <p>Solving the inequality $15x^2 + 6 + 2 \log_3 x \geq 0$, we find that $x \in(0, \frac{1}{27})$ ........(4)</p> <p>Likewise, solving the inequality $6+2 \log_3 x + \frac{5}{3} \geq 0$, we find that $x \geq 3^{-\frac{23}{6}}$ ........(5)</p> <p>Combining Equations (3), (4), and (5), the intersection of all these intervals is $x \in[3^{-\frac{23}{6}}, \frac{1}{27})$.</p> <p>Now, consider the function $g(x) = x - [x]$, where $[x]$ is the greatest integer function. For this function, the range is the fractional part of $x$. In this case, the range $(\alpha, \beta)$ is given by the minimum and maximum possible values of $x$ in its domain. Hence, $\alpha = 3^{-\frac{23}{6}}$ and $\beta = \frac{1}{27}$.</p> <p>Finally, substitute these values into the equation $\alpha^{2}+\frac{5}{\beta}$:</p> <p>$\alpha^{2}+\frac{5}{\beta} = (3^{-\frac{23}{6}})^2 + \frac{5}{\frac{1}{27}} = 3^{-\frac{23}{3}} + 135$ = 135.0002198.</p> <p>Since $3^{-\frac{23}{3}}$ = 0.0002198 is an extremely small number, it&#39;s approximately 0. So, $\alpha^{2}+\frac{5}{\beta} \approx 135$.</p>