JEE Main 2024 — Inverse Trigonometric Functions Question with Solution

$$\begin{array}{rl}& {\sin }^{-1}\left(\frac{3x-22}{2x-19}\right)+{\log }_e\left(\frac{3x^2-8x+5}{x^2-3x-10}\right)\\ & -1\le \frac{3x-22}{2x-19}\le 1\\ & \frac{3x-22}{2x-19}+1\ge 0\text{ and }\frac{3x-22}{2x-19}-1\le 0\\ & \frac{3x-22+2x-19}{2x-19}\ge 0\text{ and }\frac{3x-22-2x+19}{2x-19}\le 0\\ & \Rightarrow \frac{5x-41}{2x-19}\ge 0\text{ and }\frac{x-3}{2x-19}\le 0\\ & x\in \left(-\infty ,\frac{41}{5}\right]\cup \left(\frac{19}{2},\infty \right)\text{ and }x\in \left[3,\frac{19}{2}\right)\end{array}$$

$$\begin{array}{rl}& \Rightarrow x\in \left[3,\frac{41}{5}\right]\text{.... (1)}\\ & \text{ and, }\frac{3x^2-8x+5}{x^2-3x-10}>0\\ & \frac{(3x-5)(x-1)}{(x-5)(x-2)}>0\\ & \Rightarrow x\in (-\infty ,-2)\cup \left[1,\frac{5}{3}\right]\cup (5,\infty )\dots \end{array}$$

Taking intersection of individual domains

$$x\in \left(5,\frac{41}{5}\right]$$

$$\begin{array}{rl}& \Rightarrow \alpha =5\text{ and }\beta =\frac{41}{5}\\ & \Rightarrow 3\alpha +10\beta =15+82\\ & =97\end{array}$$

$$\therefore$$ Option (4) is correct