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(D) For the function $f(x)$ to be defined, the expression inside the square root must be positive and the denominator must not be zero. Thus, $\log_{\

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$a+2$

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(D) For the function $f(x)$ to be defined, the expression inside the square root must be positive and the denominator must not be zero. Thus, $\log_{\frac{1}{3}}\left(\frac{x-1}{2-x}\right) > 0$. Since the base $\frac{1}{3} Subtracting $1$ from both sides: $\frac{x-1}{2-x} - 1 Multiplying by $-1$ gives $\frac{2x-3}{x-2} > 0$. The critical points are $x = 1.5$ and $x = 2$. Testing intervals, the inequality holds for $x \in (-\infty, 1.5) \cup (2, \infty)$. Additionally, the argument of the logarithm must be positive: $\frac{x-1}{2-x} > 0$. This holds for $x \in (1, 2)$. Intersecting these conditions, the domain is $(1, 1.5)$. Thus, $a = 1$ and $b = 1.5$. Then $2b = 2(1.5) = 3$. Since $a = 1$, $a+2 = 1+2 = 3$. Therefore, $2b = a+2$.

If the domain of the real valued function $f(x) = \frac{1}{\sqrt{\log_{\frac{1}{3}}\left(\frac{x-1}{2-x}\right)}}$ is $(a, b)$, then $2b =$

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