If `f(x)=(x)/(x-1)`, then the points of discontinuity of the function `f^(15)(x)`, where `f^(n)=fof………..of` (n times), are

Options

A. `x=2, 1`
B. `x=0, 1`
C. `x=1,2,0`
D. continuous everywhere except x = 1

Answer

To find the points of discontinuity of the function \( f^{(15)}(x) \), where \( f(x) = \frac{x}{x-1} \) and \( f^{(n)} \) denotes the \( n \)-th iterate of \( f \), we can follow these steps:

### Step 1: Understand the Function
The function is given as:
\[ f(x) = \frac{x}{x-1} \]
This function is defined for all \( x \) except \( x = 1 \), where it becomes undefined.

### Step 2: Find the First Iteration
To find \( f^{(2)}(x) = f(f(x)) \), we substitute \( f(x) \) into itself:
\[ f(f(x)) = f\left(\frac{x}{x-1}\right) = \frac{\frac{x}{x-1}}{\frac{x}{x-1} - 1} \]
Calculating the denominator:
\[ \frac{x}{x-1} - 1 = \frac{x - (x-1)}{x-1} = \frac{1}{x-1} \]
Thus,
\[ f(f(x)) = \frac{\frac{x}{x-1}}{\frac{1}{x-1}} = x \]

### Step 3: Identify the Pattern
From the calculations:
- \( f^{(1)}(x) = f(x) = \frac{x}{x-1} \) (odd)
- \( f^{(2)}(x) = x \) (even)

Continuing this pattern, we can see that:
- For odd \( n \), \( f^{(n)}(x) = f(x) = \frac{x}{x-1} \)
- For even \( n \), \( f^{(n)}(x) = x \)

### Step 4: Determine \( f^{(15)}(x) \)
Since \( 15 \) is odd, we have:
\[ f^{(15)}(x) = f(x) = \frac{x}{x-1} \]

### Step 5: Find Points of Discontinuity
The function \( f^{(15)}(x) = \frac{x}{x-1} \) is discontinuous at \( x = 1 \) because it is undefined there. Therefore, the only point of discontinuity is:
\[ x = 1 \]

### Final Answer
The points of discontinuity of the function \( f^{(15)}(x) \) are:
\[ \{1\} \]