$f(x)=\cos ^{-1}\left(\frac{2 x-5}{11-3 x}\right)+\sin ^{-1}\left(2 x^{2}-3 x+1\right)\\$ $-1 \leq \frac{2 \mathrm{x}-5}{11-3 \mathrm{x}} \leq 1\\$ $-1 \leq 2 \mathrm{x}^{2}-3 \mathrm{x}+1 \leq 1\\$ $2 \mathrm{x}^{2}-3 \mathrm{x}+2 \geq 0,2 \mathrm{x}^{2}-3 \mathrm{x} \leq 0\\$ \mathrm{x} \in\left[0, \frac{3}{2}\right] \tag .......{(1)}\\ $\frac{2 \mathrm{x}-5}{11-3 \mathrm{x}}+1 \geq 0\\$ $\frac{2 \mathrm{x}-5}{11-3 \mathrm{x}}-1 \leq 0\\$ $\frac{2 \mathrm{x}-5+11-3 \mathrm{x}}{11-3 \mathrm{x}} \geq 0\\$ $\frac{5 \mathrm{x}-16}{11-3 \mathrm{x}} \leq 0\\$ $\\ \frac{6-x}{11-3 x} \geq 0\\$ $x \in\left(-\infty, \frac{16}{5}\right] \cup\left(\frac{11}{3}, \infty\right) \\$ $\mathrm{x} \in\left(-\infty, \frac{11}{3}\right) \cup[6, \infty)$ intersection \\ \mathrm{x} \in\left(-\infty, \frac{16}{5}\right] \cup[6, \infty) \tag.......{(2)}\\ Intersection of (i) & (ii) $\mathrm{x} \in\left[0, \frac{3}{2}\right]\\$ $\alpha=0, \beta=\frac{3}{2} \Rightarrow \alpha+2 \beta=3$