JEE Main 2023 — Functions Question with Solution

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 $-1\le x\le 1$, this means that ${\log }_{3x}(\frac{6+2{\log }_{3}x}{-5x})$ 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, $x\ne \frac{1}{3}$ .......(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.
   
   $$\Rightarrow$$ $6+2{\log }_{3}x<0(\because x>0)$
   
   $$\begin{array}{rl}\Rightarrow & {\log }_{3}x<-3\\ & \\ \Rightarrow & x<3^{-3}\\ & \\ \Rightarrow & x<\frac{1}{27}.........(3)\end{array}$$
   
   $$\Rightarrow \text{ From (1), (2), (3), }0<x<\frac{1}{27}$$

The next step is to solve the inequality for $-1\le {\log }_{3x}(\frac{6+2{\log }_{3}x}{-5x})\le 1$. To do this, we make the observation that for the logarithmic part to be within $[-1,1]$, it must be true that $3x\le \frac{6+2{\log }_{3}x}{-5x}\le \frac{1}{3x}$.

Solving the inequality $15x^2+6+2{\log }_{3}x\ge 0$, we find that $x\in (0,\frac{1}{27})$ ........(4)

Likewise, solving the inequality $6+2{\log }_{3}x+\frac{5}{3}\ge 0$, we find that $x\ge 3^{-\frac{23}{6}}$ ........(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, ${\alpha }^2+\frac{5}{\beta }\approx 135$.