If the domain of the function $$f(x)=\cos^{-1}\left(\frac{2x-5}{11-3x}\right)+\sin^{-1}(2x^{2}-3x+1)$$ is the interval $$[\alpha, \beta]$$, then $$\alpha+2\beta$$ is equal to:

We need both inverse trig arguments to lie in ([-1,1]).

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

For the cosine inverse term:
$$-1\le\frac{2x-5}{11-3x}\le1$$

Solving,
$$\frac{2x-5}{11-3x}\ge-1$$
$$\Rightarrow\frac{x+3}{11-3x}\ge0$$
$$and \frac{2x-5}{11-3x}\le1$$
$$\Rightarrow\frac{x-16}{11-3x}\le0$$

Combining gives
$$-3\le x<\frac{11}{3}$$

Now for the sine inverse term:
$$-1\le2x^2-3x+1\le1$$

First,
$$2x^2-3x+1\ge-1$$
$$\Rightarrow2x^2-3x+2\ge0$$
always true.

Next,
$$2x^2-3x+1\le1$$
$$\Rightarrow2x^2-3x\le0$$
$$\Rightarrow x(2x-3)\le0$$

$$0\le x\le\frac{3}{2}$$

Intersecting both conditions:
$$[\alpha,\beta,]=\left[0,\frac{3}{2}\right]$$

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