If `f(x)=1/(1-x),` then the set of points discontinuity of the function `f(f(f(x)))i s` `(a){1} (b) {0,1} (c) {-1,1}` (d) none of these

Explanation

To solve the problem, we need to find the points of discontinuity of the function \( f(f(f(x))) \) where \( f(x) = \frac{1}{1-x} \).

### Step 1: Find \( f(f(x)) \)

1. Start with the function \( f(x) = \frac{1}{1-x} \).
2. Substitute \( f(x) \) into itself:
\[ f(f(x)) = f\left(\frac{1}{1-x}\right) = \frac{1}{1 - \frac{1}{1-x}}. \]
3. Simplify the expression:
\[ 1 - \frac{1}{1-x} = \frac{(1-x) - 1}{1-x} = \frac{-x}{1-x}. \]
4. Thus, we have:
\[ f(f(x)) = \frac{1}{\frac{-x}{1-x}} = \frac{1-x}{-x} = \frac{x-1}{x}. \]

### Step 2: Find \( f(f(f(x))) \)

1. Now substitute \( f(f(x)) \) into \( f(x) \):
\[ f(f(f(x))) = f\left(\frac{x-1}{x}\right) = \frac{1}{1 - \frac{x-1}{x}}. \]
2. Simplify this expression:
\[ 1 - \frac{x-1}{x} = \frac{x - (x-1)}{x} = \frac{1}{x}. \]
3. Therefore, we have:
\[ f(f(f(x))) = \frac{1}{\frac{1}{x}} = x. \]

### Step 3: Determine Points of Discontinuity

1. The function \( f(f(f(x))) = x \) is continuous for all \( x \) except where \( f(x) \) is discontinuous.
2. Identify points of discontinuity of \( f(x) \):
- \( f(x) \) is discontinuous at \( x = 1 \) because \( f(1) \) leads to division by zero.
3. Identify points of discontinuity of \( f(f(x)) \):
- \( f(f(x)) = \frac{x-1}{x} \) is discontinuous at \( x = 0 \) (division by zero).
4. Therefore, the points of discontinuity for \( f(f(f(x))) \) are \( x = 0 \) and \( x = 1 \).

### Final Answer

The set of points of discontinuity of the function \( f(f(f(x))) \) is:
\[ \{0, 1\}. \]