Step 1: Understanding Domain Constraints:
For the function \( f(x) \) to be defined, the term inside the square root must be strictly positive (since it's in the denominator), and the term inside the logarithm must be positive.
Conditions:
1. Argument of log must be positive: \( \frac{x-1}{2-x}>0 \).
2. The term under the square root must be positive: \( \log_{\frac{1}{3}}\left(\frac{x-1}{2-x}\right)>0 \).
Step 2: Solving Inequality 1:
\[ \frac{x-1}{2-x}>0 \]
Since \( 2-x = -(x-2) \), this is equivalent to:
\[ \frac{x-1}{-(x-2)}>0 \implies \frac{x-1}{x-2}<0 \]
This holds for \( 1<x<2 \).
Step 3: Solving Inequality 2:
\[ \log_{\frac{1}{3}}\left(\frac{x-1}{2-x}\right)>0 \]
Since the base \( \frac{1}{3} \) is between 0 and 1, the inequality sign flips when removing the logarithm:
\[ 0<\frac{x-1}{2-x}<\left(\frac{1}{3}\right)^0 \]
\[ 0<\frac{x-1}{2-x}<1 \]
We already know \( \frac{x-1}{2-x}>0 \) from Step 2. Now solve:
\[ \frac{x-1}{2-x}<1 \]
\[ \frac{x-1}{2-x} - 1<0 \]
\[ \frac{x-1 - (2-x)}{2-x}<0 \]
\[ \frac{2x - 3}{2-x}<0 \]
Multiplying by -1 (and flipping inequality):
\[ \frac{2x - 3}{x-2}>0 \]
Roots are \( x = \frac{3}{2} \) and \( x = 2 \).
The solution is \( (-\infty, \frac{3}{2}) \cup (2, \infty) \).
Step 4: Finding the Intersection:
We intersect the result from Step 2 (\( 1<x<2 \)) with Step 3 (\( x<1.5 \) or \( x>2 \)).
Intersection: \( (1, 2) \cap ((-\infty, 1.5) \cup (2, \infty)) = (1, 1.5) \).
So, Domain \( (a, b) = (1, \frac{3}{2}) \).
Here, \( a = 1 \) and \( b = \frac{3}{2} \).
Step 5: Calculating \( 2b \):
\[ 2b = 2 \times \frac{3}{2} = 3 \]
Check options:
(A) \( a-1 = 0 \)
(B) \( a = 1 \)
(C) \( a+1 = 2 \)
(D) \( a+2 = 1+2 = 3 \)
Matches Option (D).