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

Question

Accepted Answer

A. 1. For $\cos^{-1}$ to be defined, its argument must lie in $[-1,1]$, so set up two inequalities for the rational expression.
Given $\frac{2x-5}{11-3x}\leq1$ and $\frac{2x-5}{11-3x}\geq-1$, solve each separately.

2. Solve the first inequality by combining into a single fraction.
Given
$$
\frac{5x-16}{11-3x}\leq0 \quad\Rightarrow\quad x\in\left(-\infty,\frac{16}{5}ight]\cup\left(\frac{11}{3},\inftyight)
$$

3. Solve the second inequality similarly.
Given
$$
\frac{x-6}{3x-11}\geq0 \quad\Rightarrow\quad x\in\left(-\infty,\frac{11}{3}ight)\cup[6,\infty)
$$

4. For $\sin^{-1}$ to be defined, its argument must also lie in $[-1,1]$, so set up two more inequalities for the quadratic expression.
Given $2x^2-3x+1\leq1$,
$$
x(2x-3)\leq0 \quad\Rightarrow\quad x\in\left[0,\frac32ight]
$$
Given $2x^2-3x+1\geq-1$,
$
2x^2-3x+2\geq0 \quad\Rightarrow\quad x\in\mathbb{R} 	ext{ (always true, since discriminant is negative)}
$

5. Take the intersection of all four conditions to find the overall domain.
Since the quadratic conditions restrict $x\in[0,3/2]$, and this range is well within the ranges satisfying the rational expression conditions, the final domain is
$$
x\in\left[0,\frac32ight]
$$

6. Identify $\alpha=0$ and $\beta=\frac32$, then compute $\alpha+2\beta$.
Hence,
$$
\alpha+2\beta = 0+2	imes\frac32 = 3
$$

Hence, the answer is option A.