Question

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

$${- 1 \leq \frac{2x - 5}{11 - 3x} \leq 1 }{- 1 \leq 2x^{2} - 3x + 1 \leq 1 }{2x^{2} - 3x + 2 \geq 0,2x^{2} - 3x \leq 0}$$

$$\begin{array}{r} x \in \left\lbrack 0,\frac{3}{2} \right\rbrack\#(i) \end{array}$$

$${\frac{2x - 5}{11 - 3x} + 1 \geq 0 }{\frac{2x - 5}{11 - 3x} - 1 \leq 0 }{\frac{2x - 5 + 11 - 3x}{11 - 3x} \geq 0 }{\frac{5x - 16}{11 - 3x} \leq 0 }$$

$$\frac{6 - x}{11 - 3x} \geq 0 $$

$$x \in \left( - \infty,\frac{16}{5} \right\rbrack \cup \left( \frac{11}{3},\infty \right)$$

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

$$\begin{array}{r} x \in \left( - \infty,\frac{16}{5} \right\rbrack \cup \lbrack 6,\infty)\#(ii) \end{array}$$

Intersection of (i) & (ii) $x \in \left\lbrack 0,\frac{3}{2} \right\rbrack$

$$\alpha = 0,\beta = \frac{3}{2} \Rightarrow \alpha + 2\beta = 3$$