If `alpha, beta` (where `alpha lt beta`) are the points of discontinuity of the function g(x) = f(f(f(x))), where `f(x) = (1)/(1-x), and P(a, a^(2))` is any point on XY - plane. Then,
The domain of f(g(x)), is

Options

A. `x in R`
B. `x in R - {1}`
C. `x in R - {0, 1}`
D. `x in R - {0, 1, -1}`

Answer

To solve the problem, we need to analyze the function \( g(x) = f(f(f(x))) \) where \( f(x) = \frac{1}{1-x} \) and determine the domain of \( f(g(x)) \).

### Step-by-Step Solution:

1. **Define the function \( f(x) \)**:
\[ f(x) = \frac{1}{1-x} \]
The function \( f(x) \) is discontinuous when its denominator is zero, i.e., when \( 1 - x = 0 \) or \( x = 1 \).

2. **Find the points of discontinuity of \( f(x) \)**:
The point of discontinuity for \( f(x) \) is \( x = 1 \).

3. **Calculate \( f(f(x)) \)**:
To find \( f(f(x)) \):
\[ f(f(x)) = f\left(\frac{1}{1-x}\right) = \frac{1}{1 - \frac{1}{1-x}} = \frac{1}{\frac{(1-x)-1}{1-x}} = \frac{1-x}{-x} = \frac{x-1}{x} \]
The function \( f(f(x)) \) is discontinuous when its denominator is zero, i.e., when \( x = 0 \).

4. **Calculate \( g(x) = f(f(f(x))) \)**:
Now we need to find \( f(f(f(x))) \):
\[ g(x) = f(f(f(x))) = f\left(\frac{x-1}{x}\right) = \frac{1}{1 - \frac{x-1}{x}} = \frac{1}{\frac{x - (x-1)}{x}} = \frac{x}{1} = x \]
The function \( g(x) \) is continuous everywhere except where \( f(f(x)) \) is discontinuous, which is at \( x = 0 \).

5. **Identify the points of discontinuity of \( g(x) \)**:
The points of discontinuity of \( g(x) \) are \( x = 0 \) and \( x = 1 \) (from \( f(x) \)).

6. **Determine the domain of \( f(g(x)) \)**:
We need to find the domain of \( f(g(x)) = f(x) \). The function \( f(x) \) is undefined at \( x = 1 \). Therefore, the domain of \( f(g(x)) \) excludes the points of discontinuity:
\[ \text{Domain of } f(g(x)) = \mathbb{R} - \{1\} \]

### Final Answer:
The domain of \( f(g(x)) \) is:
\[ \mathbb{R} - \{1\} \]