JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Sets Relations and Functions

For the inner logarithm's argument $\frac{6+2{\log }_{3}x}{-5x}$ to be positive with $x>0$, the numerator must be negative: $6+2{\log }_{3}x<0$, i.e. ${\log }_{3}x<-3$, so $0<x<\frac{1}{27}$; then the base $3x\in \left(0,\frac{1}{9}\right)$ lies in $(0,1)$. With a base in $(0,1)$, the requirement $-1\le {\log }_{3x}\left(\frac{6+2{\log }_{3}x}{-5x}\right)\le 1$ is equivalent to $3x\le \frac{6+2{\log }_{3}x}{-5x}\le \frac{1}{3x}$. The right inequality gives $-(6+2{\log }_{3}x)\le \frac{5}{3}$, that is ${\log }_{3}x\ge -\frac{23}{6}$, so $x\ge 3^{-23/6}$. Thus the domain lies in $\left[3^{-23/6},\frac{1}{27}\right)\subset (0,1)$, where $[x]=0$ and $g(x)=x-[x]=x$, so the range of $g$ is $(\alpha ,\beta )$ with $\alpha =3^{-23/6}$ and $\beta =\frac{1}{27}$. Here ${\alpha }^2=3^{-23/3}$ is vanishingly small against the integer scale, while $\frac{5}{\beta }=5\times 27=135$, so ${\alpha }^2+\frac{5}{\beta }=135$.