Let `f : R – {0, 1} rarr R` be a function such that `f(x) + f(1/(1-x))= 1 + x`. then `f(2)` is equal to

Options

A. `(7)/(3)`
B. `(9)/(2)`
C. `(9)/(4)`
D. `(7)/(4)`

Answer

To solve the problem, we need to find the value of \( f(2) \) given the functional equation:

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

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

Let's start by substituting \( x = 2 \) into the functional equation:

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

This simplifies to:

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

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

Next, we substitute \( x = -1 \) into the functional equation:

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

This simplifies to:

\[ f(-1) + f\left(\frac{1}{2}\right) = 0 \tag{2} \]

### Step 3: Substitute \( x = \frac{1}{2} \)

Now, we substitute \( x = \frac{1}{2} \) into the functional equation:

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

This simplifies to:

\[ f\left(\frac{1}{2}\right) + f(2) = \frac{3}{2} \tag{3} \]

### Step 4: Solve the equations

Now we have three equations:

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

From Equation (2), we can express \( f(-1) \):

\[ f(-1) = -f\left(\frac{1}{2}\right) \tag{4} \]

Substituting Equation (4) into Equation (1):

\[ f(2) - f\left(\frac{1}{2}\right) = 3 \tag{5} \]

Now we have two equations (3) and (5):

1. \( f\left(\frac{1}{2}\right) + f(2) = \frac{3}{2} \) (Equation 3)
2. \( f(2) - f\left(\frac{1}{2}\right) = 3 \) (Equation 5)

### Step 5: Add Equations (3) and (5)

Adding Equations (3) and (5):

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

This simplifies to:

\[ 2f(2) = \frac{3}{2} + 3 = \frac{3}{2} + \frac{6}{2} = \frac{9}{2} \]

Thus,

\[ f(2) = \frac{9}{4} \]

### Final Answer

The value of \( f(2) \) is:

\[ \boxed{\frac{9}{4}} \]