Question

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 $2b =$

• A. $a-1$
• B. $a$
• C. $a+1$
• D. $a+2$

Hint:

When solving inequalities involving logarithms with a base between 0 and 1, always remember to reverse the inequality sign. Also, don't forget the implicit condition that the argument of the logarithm must always be positive. It's often safer to solve these two conditions separately and then find the intersection of the solution sets.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For the function $f(x)$ to be defined, we need to satisfy three conditions based on its structure: 1. The denominator cannot be zero.
2. The expression inside the square root must be non-negative.
3. The argument of the logarithm must be positive.
Combining the first two conditions, the expression inside the square root, $\log_{\frac{1}{3}}\left(\frac{x-1}{2-x}\right)$, must be strictly positive. 
Step 2: Key Formula or Approach 
We need to solve the inequality $\log_{\frac{1}{3}}\left(\frac{x-1}{2-x}\right)>0$.
Recall the property of logarithms: if $\log_b(A)>C$ and the base $b$ is between 0 and 1 ($0<b<1$), then the inequality sign reverses, i.e., $A<b^C$. Also, the argument $A$ must be positive, $A>0$. 
Step 3: Detailed Explanation 
The main condition for the function to be defined is:
\[ \log_{\frac{1}{3}}\left(\frac{x-1}{2-x}\right)>0 \] Since the base of the logarithm is $\frac{1}{3}$ (which is between 0 and 1), the inequality sign flips when we remove the logarithm.
\[ \frac{x-1}{2-x}<\left(\frac{1}{3}\right)^0 \] \[ \frac{x-1}{2-x}<1 \] Additionally, the argument of any logarithm must be positive:
\[ \frac{x-1}{2-x}>0 \] So, we need to solve the combined inequality: $0<\frac{x-1}{2-x}<1$. 
We solve this in two parts.
Part 1: $\frac{x-1}{2-x}>0$ 
The critical points are $x=1$ and $x=2$. We can use a sign chart or test intervals.
The numerator $x-1$ is positive for $x>1$ and negative for $x<1$.
The denominator $2-x$ is positive for $x<2$ and negative for $x>2$.
The fraction is positive when both have the same sign. This does not happen. It is positive when (num $>0$ and den $>0$) or (num $<0$ and den $<0$).
Case (i): $x-1>0$ and $2-x>0 \implies x>1$ and $x<2$. So, $1<x<2$.
Case (ii): $x-1<0$ and $2-x<0 \implies x<1$ and $x>2$. This is impossible.
So, the solution to Part 1 is $1<x<2$.
Part 2: $\frac{x-1}{2-x}<1$ 
\[ \frac{x-1}{2-x} - 1<0 \] \[ \frac{(x-1) - (2-x)}{2-x}<0 \] \[ \frac{x-1-2+x}{2-x}<0 \] \[ \frac{2x-3}{2-x}<0 \] The critical points are $x=3/2$ and $x=2$. The fraction is negative when the numerator and denominator have opposite signs.
Case (i): $2x-3>0$ and $2-x<0 \implies x>3/2$ and $x>2$. The intersection is $x>2$.
Case (ii): $2x-3<0$ and $2-x>0 \implies x<3/2$ and $x<2$. The intersection is $x<3/2$.
So, the solution to Part 2 is $x<3/2$ or $x>2$.
Now, we must find the intersection of the solutions from Part 1 and Part 2.
We need $x$ such that $(1<x<2)$ AND $(x<3/2 \text{ or } x>2)$.
The common interval is $1<x<3/2$. 
Step 4: Final Answer 
The domain of the function is $(1, 3/2)$. 
The question states the domain is $(a, b)$. So, $a=1$ and $b=3/2$. 
We need to find the value of $2b$. 
\[ 2b = 2 \times \frac{3}{2} = 3 \] Now we express this result in terms of $a=1$:
(A) $a-1 = 1-1 = 0$
(B) $a = 1$
(C) $a+1 = 1+1 = 2$
(D) $a+2 = 1+2 = 3$ 
The correct option is (D).