JEE Main 2023 — Functions Question with Solution

We have been given

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

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

$\Rightarrow {\log }_{3x}\left(\frac{6+2{\log }_{3}x}{-5x}\right)\in \underset{-\left(1\right)}{\left[-1 1\right]}, \underset{-\left(2\right)}{3x>0}, \underset{-\left(3\right)}{3x\ne 1}, \underset{-\left(4\right)}{\frac{6+2{\log }_{3}x}{-5x}}>0$

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

From $\left(4\right), 6+2{\log }_{3}x<0 \left(\because x>0\right)$

$\Rightarrow {\log }_{3}x<-3$

$\Rightarrow x<3^{-3}$

$\Rightarrow x<\frac{1}{27}$ 

From $\left(2\right), \left(3\right), \left(4\right), 0<x<\frac{1}{27} ...\left(5\right)$

From $\left(1\right), -1\le {\log }_{3x}\left(\frac{6+2{\log }_{3}x}{-5x}\right)\le 1$

$\because 3x\in \left(0, \frac{1}{9}\right)$

$\Rightarrow \frac{1}{3x}\ge \frac{6+2{\log }_{3}x}{-5x}\ge 3x$

$\Rightarrow -\frac{5}{3}\le 6+2{\log }_{3}x\le -15x^2$

$\Rightarrow -\frac{23}{6}\le {\log }_{3}x\le \frac{-15x^2-6}{2}$

$\Rightarrow 3^{-\frac{23}{6}}\le x\le 3^{\frac{-15x^2-6}{2}}$

$\Rightarrow \frac{1}{3^{{\displaystyle \frac{23}{6}}}}\le x\le \frac{1}{3^{{\displaystyle \frac{15x^2+6}{2}}}}$

$\Rightarrow 3^{\frac{-23}{6}}\le x<\frac{1}{3^{{\displaystyle \frac{15x^2+6}{2}}}} ......\left(6\right)$, here $3^{\frac{-23}{6}}$ is very small positive quantity,

From $\left(5\right)$ and $\left(6\right), 3^{\frac{-23}{6}}<x<\frac{1}{27}, \left[x\right]=0$

$\Rightarrow g\left(x\right)=x$

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

$\Rightarrow g\left(x\right)\in \left(3^{\frac{-23}{6}},\frac{1}{27}\right)\equiv \left(\alpha ,\beta \right)$.

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

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