QJEEM23Ma8772: JEE Main 2023 Mathematics - Class 11 JEEM23Ma8772 Solution: Let D be the domain of the function f ( x ) = sin − 1 ⁡ ( log 3 …

First, the function $f(x) = \sin^{-1}(\log_{3x}(\frac{6 + 2 \log_3{x}}{-5x}))$ has several restrictions :

1. 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.

2. For the logarithm to be defined, 3x > 0 \Rightarrow x > 0 .......(1)
   
   because the base of a logarithm must be greater than 0 and not equal to 1. Also, $x \neq \frac{1}{3}$ .......(2)
   
   as the base cannot be 1.

3. Moreover, the inner function of the logarithm $\frac{6 + 2 \log_3{x}}{-5x}$ must be greater than 0.
   
   $$\Rightarrow$$ 6+2 \log _3 x<0 \quad(\because x>0)
   
   $$\begin{array}{rl}\Rightarrow & {\log }_{3}x<-3\\ \\ \Rightarrow & x<3^{-3}\\ \\ \Rightarrow & x<\frac{1}{27}.........(3)\end{array}$$
   
   $$\Rightarrow \text{ From (1), (2), (3), }0<x<\frac{1}{27}$$

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 $3x \leq \frac{6+2 \log_3 x}{-5x} \leq \frac{1}{3x}$ .

Solving the inequality $15x^2 + 6 + 2 \log_3 x \geq 0$ , we find that $x \in(0, \frac{1}{27})$ ........(4)

Likewise, solving the inequality $6+2 \log_3 x + \frac{5}{3} \geq 0$ , we find that $x \geq 3^{-\frac{23}{6}}$ ........(5)

Combining Equations (3), (4), and (5), the intersection of all these intervals is $x \in[3^{-\frac{23}{6}}, \frac{1}{27})$ .

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 $\beta = \frac{1}{27}$ .

Finally, substitute these values into the equation $\alpha^{2}+\frac{5}{\beta}$ :

$\alpha^{2}+\frac{5}{\beta} = (3^{-\frac{23}{6}})^2 + \frac{5}{\frac{1}{27}} = 3^{-\frac{23}{3}} + 135$ = 135.0002198.

Since $3^{-\frac{23}{3}}$ = 0.0002198 is an extremely small number, it's approximately 0. So, $\alpha^{2}+\frac{5}{\beta} \approx 135$ .