JEE Main 2025 — Mathematics Algebra

$$\begin{aligned} & f(x)=\cos ^{-1}\left(\frac{4 x+5}{3 x-7}\right) \\ & \Rightarrow-1 \leq\left(\frac{4 x+5}{3 x-7}\right) \leq 1 \\ & \left(\frac{4 x+5}{3 x-7}\right) \geq-1 \\ & \frac{4 x+5+3 x-7}{3 x-7} \geq 0 \\ & \Rightarrow \frac{7 x-2}{3 x-7} \geq 0\end{aligned}$$​f(x)=cos−1(3x−74x+5​)⇒−1≤(3x−74x+5​)≤1(3x−74x+5​)≥−13x−74x+5+3x−7​≥0⇒3x−77x−2​≥0​

$\begin{aligned} & x \in\left(-\infty, \frac{2}{7}\right] \cup\left(\frac{7}{3}, \infty\right) \\ & \& \frac{4 x+5}{3 x-7} \leq 1 \Rightarrow \frac{x+12}{3 x-7} \leq 0\end{aligned}$​x∈(−∞,72​]∪(37​,∞)&3x−74x+5​≤1⇒3x−7x+12​≤0​

$\therefore$∴ Domain of $f(x)$f(x) is

$\begin{aligned}
& {\left[-12, \frac{2}{7}\right] \alpha=-12, \beta=\frac{2}{7}} \
& \mathrm{~g}(\mathrm{x})=\log _2\left(2-6 \log _{27}(2 \mathrm{x}+5)\right)
\end{aligned}$

Domain

$$2-6 \log _{27}(2 x+5) \gt 0$$2−6log27​(2x+5)>0

$$\begin{array}{ll}\Rightarrow & 6 \log _{27}(2 \mathrm{x}+5) \lt 2 \\ \Rightarrow & \log _{27}(2 \mathrm{x}+5) \lt \frac{1}{3} \\ \Rightarrow & 2 \mathrm{x}+5 \lt 3 \\ \Rightarrow & \mathrm{x} \lt -1\end{array}$$⇒⇒⇒⇒​6log27​(2x+5)<2log27​(2x+5)<31​2x+5<3x<−1​

$$\& 2 x+5 \gt 0 \Rightarrow x \gt -\frac{5}{2}$$&2x+5>0⇒x>−25​

Domain is $\mathrm{x} \in\left(-\frac{5}{2},-1\right)$x∈(−25​,−1)

$\begin{aligned}
& \gamma=-\frac{5}{2}, \delta=-1 \
& |7(\alpha+\beta)+4(\gamma+\delta)|=\left\lvert, 7\left(\left.-12+\frac{2}{7}+4\left(-\frac{5}{2}-1\right) \right\rvert,\right.\right. \
& |-82-14|=96
\end{aligned}$