JEE Main 2026 — 21 January, Morning Shift — Question 1

$f(x)=\cos ^{-1}\left(\frac{2 x-5}{11-3 x}\right)+\sin ^{-1}\left(2 x^{2}-3 x+1\right)$

$-1 \leq \frac{2 \mathrm{x}-5}{11-3 \mathrm{x}} \leq 1$

$-1 \leq 2 \mathrm{x}^{2}-3 \mathrm{x}+1 \leq 1$

$2 \mathrm{x}^{2}-3 \mathrm{x}+2 \geq 0,2 \mathrm{x}^{2}-3 \mathrm{x} \leq 0$

$$\begin{gathered} \mathrm{x} \in\left[0, \frac{3}{2}\right] \end{gathered}$$ $$\begin{aligned} & \frac{2 x-5}{11-3 x}+1 \geq 0 & \frac{2 x-5+11-3 x}{11-3 x} \geq 0 \end{aligned}$$ $$\frac{5 x-16}{11-3 x} \leq 0$$ $$\frac{6-x}{11-3 x} \geq 0$$ $$x \in\left(-\infty, \frac{16}{5}\right] \cup\left(\frac{11}{3}, \infty\right)$$

$\mathrm{x} \in\left(-\infty, \frac{11}{3}\right) \cup[6, \infty)$

intersection $\begin{gathered} x \in\left(-\infty, \frac{16}{5}\right] \cup[6, \infty) \end{gathered}$

Intersection of (i) & (ii) $\mathrm{x} \in\left[0, \frac{3}{2}\right]$ $\alpha=0, \beta=\frac{3}{2} \Rightarrow \alpha+2 \beta=3$