JEE Main 2025 — Functions Question with Solution

$\begin{array}{rl}& f(x)={\cos }^{-1}\left(\frac{4x+5}{3x-7}\right)\\ & \Rightarrow -1\le \left(\frac{4x+5}{3x-7}\right)\le 1\\ & \left(\frac{4x+5}{3x-7}\right)\ge -1\\ & \frac{4x+5+3x-7}{3x-7}\ge 0\\ & \Rightarrow \frac{7x-2}{3x-7}\ge 0\end{array}$

$\begin{array}{rl}& x\in \left(-\infty ,\frac{2}{7}\right]\cup \left(\frac{7}{3},\infty \right)\\ & \&\frac{4x+5}{3x-7}\le 1\Rightarrow \frac{x+12}{3x-7}\le 0\end{array}$
$\therefore$ Domain of $f(x)$ is
$\begin{aligned}
& {\left[-12, \frac{2}{7}\right] \alpha=-12, \beta=\frac{2}{7}} \\ & \mathrm{~g}(\mathrm{x})=\log _2\left(2-6 \log _{27}(2 \mathrm{x}+5)\right)
\end{aligned}$<br/>Domain<br/>$2-6 \log _{27}(2 x+5) \gt 0$<br/>$\begin{array}{ll}\Rightarrow & 6 \log _{27}(2 \mathrm{x}+5) \lt 2 \\ \Rightarrow & \log _{27}(2 \mathrm{x}+5) \lt \frac{1}{3} \\ \Rightarrow & 2 \mathrm{x}+5 \lt 3 \\ \Rightarrow & \mathrm{x} \lt -1\end{array}$<br/>$\& 2 x+5 \gt 0 \Rightarrow x \gt -\frac{5}{2}$<br/>Domainis$\mathrm{x} \in\left(-\frac{5}{2},-1\right)$<br/>$\begin{aligned}
& \gamma=-\frac{5}{2}, \delta=-1 \\ & |7(\alpha+\beta)+4(\gamma+\delta)|=\left\lvert\, 7\left(\left.-12+\frac{2}{7}+4\left(-\frac{5}{2}-1\right) \right\rvert\,\right.\right. \\ & |-82-14|=96
\end{aligned}$