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(B) For the domain of $f(x) = \cos^{-1}(\frac{2x-5}{11-3x}) + \sin^{-1}(2x^2-3x+1)$,we must satisfy the conditions for both inverse trigonometric func

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$3$

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(B) For the domain of $f(x) = \cos^{-1}(\frac{2x-5}{11-3x}) + \sin^{-1}(2x^2-3x+1)$,we must satisfy the conditions for both inverse trigonometric functions.$1$) For $\sin^{-1}(2x^2-3x+1)$,we require $-1 \le 2x^2-3x+1 \le 1$.$2x^2-3x+1 \le 1 \implies 2x^2-3x \le 0 \implies x(2x-3) \le 0 \implies x \in [0, \frac{3}{2}]$.$2x^2-3x+1 \ge -1 \implies 2x^2-3x+2 \ge 0$. Since the discriminant $D = (-3)^2 - 4(2)(2) = 9 - 16 = -7 Thus,the first condition gives $x \in [0, \frac{3}{2}]$.$2$) For $\cos^{-1}(\frac{2x-5}{11-3x})$,we require $-1 \le \frac{2x-5}{11-3x} \le 1$.$\frac{2x-5}{11-3x} + 1 \ge 0 \implies \frac{2x-5+11-3x}{11-3x} \ge 0 \implies \frac{6-x}{11-3x} \ge 0 \implies \frac{x-6}{x-11/3} \ge 0$. This gives $x \in (-\infty, 11/3) \cup [6, \infty)$.$\frac{2x-5}{11-3x} - 1 \le 0 \implies \frac{2x-5-11+3x}{11-3x} \le 0 \implies \frac{5x-16}{11-3x} \le 0 \implies \frac{x-16/5}{x-11/3} \ge 0$. This gives $x \in (-\infty, 16/5] \cup (11/3, \infty)$.Taking the intersection of these two conditions,we get $x \in (-\infty, 16/5] \cup [6, \infty)$.$3$) Finally,taking the intersection of the results from $(1)$ and $(2)$:$[0, 3/2] \cap ((-\infty, 16/5] \cup [6, \infty)) = [0, 3/2]$.Thus,$\alpha = 0$ and $\beta = 3/2$.Therefore,$\alpha + 2\beta = 0 + 2(3/2) = 3$.

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

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