If f(x) is a real-valued function discontinuous at all integral points lying in [0,n] and if `(f(x))^(2)=1, forall x in [0,n],` then number of functions f(x) are

Options

A. `2^(n+1)`
B. `6 times 3^(n)`
C. `2 times 3^(n-1)`
D. `3^(n+1)`

Answer

To solve the problem, we need to determine the number of functions \( f(x) \) that satisfy the given conditions. Let's break down the solution step by step.

### Step 1: Understanding the Function
Given that \( (f(x))^2 = 1 \) for all \( x \) in the interval \([0, n]\), we can deduce that:
\[ f(x) = 1 \quad \text{or} \quad f(x) = -1 \]
for all \( x \) in \([0, n]\).

**Hint:** Recognize that the square of a function being equal to 1 implies the function can only take values of 1 or -1.

### Step 2: Discontinuity at Integer Points
The function \( f(x) \) is discontinuous at all integer points in the interval \([0, n]\). This means that at each integer point \( k \) (where \( k = 0, 1, 2, \ldots, n \)), the value of \( f(k) \) must differ from the values of \( f(x) \) immediately before and after \( k \).

**Hint:** Identify the integer points in the interval and remember that the function must switch values at these points.

### Step 3: Analyzing the Intervals
The interval \([0, n]\) contains \( n \) integer points: \( 0, 1, 2, \ldots, n \). This creates \( n \) segments:
- From \( 0 \) to \( 1 \)
- From \( 1 \) to \( 2 \)
- From \( 2 \) to \( 3 \)
- ...
- From \( n-1 \) to \( n \)

Each of these segments can independently take values of either \( 1 \) or \( -1 \), but the values at the endpoints must differ.

**Hint:** Count the segments created by the integer points and consider how the function can alternate between 1 and -1.

### Step 4: Counting the Configurations
For each segment between two consecutive integers, we have two choices for the value of \( f(x) \) (either \( 1 \) or \( -1 \)). However, at each integer point, the function must switch values.

For \( n \) integer points, there are \( n-1 \) intervals. The value of \( f(x) \) can be \( 1 \) or \( -1 \) in each of these intervals, leading to:
- \( 2 \) choices for the first interval
- \( 2 \) choices for the last interval
- \( 3 \) choices for each of the \( n-1 \) intervals in between (since it can switch from \( 1 \) to \( -1 \) or from \( -1 \) to \( 1 \)).

Thus, the total number of functions is:
\[ \text{Total Functions} = 2 \times 3^{n-1} \]

**Hint:** Remember to consider the choices for the first and last intervals separately from the middle intervals.

### Final Answer
Therefore, the total number of functions \( f(x) \) that satisfy the given conditions is:
\[ \boxed{2 \times 3^{n-1}} \]