`f(x)+f(1-1/x)=1+x ` for `x in R-{0,1}.` Find the value of `4f(2).`

Answer

To solve the equation \( f(x) + f(1 - \frac{1}{x}) = 1 + x \) for \( x \in \mathbb{R} - \{0, 1\} \) and find the value of \( 4f(2) \), we can follow these steps:

### Step 1: Substitute \( x = 2 \)

We start by substituting \( x = 2 \) into the given equation:

\[ f(2) + f(1 - \frac{1}{2}) = 1 + 2 \]

This simplifies to:

\[ f(2) + f(\frac{1}{2}) = 3 \quad \text{(Equation 1)} \]

### Step 2: Substitute \( x = \frac{1}{2} \)

Next, we substitute \( x = \frac{1}{2} \):

\[ f(\frac{1}{2}) + f(1 - \frac{1}{\frac{1}{2}}) = 1 + \frac{1}{2} \]

This simplifies to:

\[ f(\frac{1}{2}) + f(-1) = \frac{3}{2} \quad \text{(Equation 2)} \]

### Step 3: Substitute \( x = -1 \)

Now, we substitute \( x = -1 \):

\[ f(-1) + f(1 - \frac{1}{-1}) = 1 - 1 \]

This simplifies to:

\[ f(-1) + f(2) = 0 \quad \text{(Equation 3)} \]

### Step 4: Solve the equations

Now we have three equations:

1. \( f(2) + f(\frac{1}{2}) = 3 \) (Equation 1)
2. \( f(\frac{1}{2}) + f(-1) = \frac{3}{2} \) (Equation 2)
3. \( f(-1) + f(2) = 0 \) (Equation 3)

From Equation 3, we can express \( f(-1) \) in terms of \( f(2) \):

\[ f(-1) = -f(2) \quad \text{(Equation 4)} \]

### Step 5: Substitute Equation 4 into Equation 2

Substituting Equation 4 into Equation 2 gives:

\[ f(\frac{1}{2}) - f(2) = \frac{3}{2} \]

Rearranging this, we find:

\[ f(\frac{1}{2}) = f(2) + \frac{3}{2} \quad \text{(Equation 5)} \]

### Step 6: Substitute Equation 5 into Equation 1

Now we substitute Equation 5 into Equation 1:

\[ f(2) + (f(2) + \frac{3}{2}) = 3 \]

This simplifies to:

\[ 2f(2) + \frac{3}{2} = 3 \]

### Step 7: Solve for \( f(2) \)

Subtract \( \frac{3}{2} \) from both sides:

\[ 2f(2) = 3 - \frac{3}{2} \]

This simplifies to:

\[ 2f(2) = \frac{3}{2} \]

Dividing both sides by 2 gives:

\[ f(2) = \frac{3}{4} \]

### Step 8: Find \( 4f(2) \)

Now, we find \( 4f(2) \):

\[ 4f(2) = 4 \times \frac{3}{4} = 3 \]

Thus, the value of \( 4f(2) \) is:

\[ \boxed{3} \]