JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Inverse Trigonometric Functions

For $f$, require $-1\le \frac{4x+5}{3x-7}\le 1$. The inequality $\frac{4x+5}{3x-7}\le 1$ reduces to $\frac{x+12}{3x-7}\le 0$, giving $-12\le x<\frac{7}{3}$, while $\frac{4x+5}{3x-7}\ge -1$ reduces to $\frac{7x-2}{3x-7}\ge 0$, giving $x\le \frac{2}{7}$ or $x>\frac{7}{3}$. Their intersection is $[-12,\frac{2}{7}]$, so $\alpha =-12,\beta =\frac{2}{7}$. For $g$, need $2x+5>0$ and $2-6{\log }_{27}(2x+5)>0$; the latter gives ${\log }_{27}(2x+5)<\frac{1}{3}$, i.e. $2x+5<27^{1/3}=3$, so $x<-1$. With $x>-\frac{5}{2}$, the domain is $(-\frac{5}{2},-1)$, so $\gamma =-\frac{5}{2},\delta =-1$. Then $|7(\alpha +\beta )+4(\gamma +\delta )|=|7(-\frac{82}{7})+4(-\frac{7}{2})|=|-82-14|=96$.