If `a f(x+1)+b f(1/(x+1))=x,x !=-1,a != b,`then `f(2)` is equal to

Options

A. `f(2)=(2a+b)/(2(a^(2)-b^(2))`
B. `f(2)=0`
C. `f(2)=(a+2b)/(a^(2)-b^(2))`
D. `f(2)=(2a+b)/((a^(2)-b^(2)))`

Explanation

To solve the equation \( a f(x+1) + b f\left(\frac{1}{x+1}\right) = x \) for \( f(2) \), we will follow these steps:

### Step 1: Substitute \( x = 1 \)
Substituting \( x = 1 \) into the equation gives:
\[ a f(2) + b f\left(\frac{1}{2}\right) = 1 \]
This will be our first equation:
\[ (1) \quad a f(2) + b f\left(\frac{1}{2}\right) = 1 \]

### Step 2: Substitute \( x = -2 \)
Now, substituting \( x = -2 \) into the equation gives:
\[ a f(-1) + b f\left(\frac{1}{-1}\right) = -2 \]
Since \( f\left(\frac{1}{-1}\right) = f(-1) \), this simplifies to:
\[ a f(-1) + b f(-1) = -2 \]
Factoring out \( f(-1) \):
\[ (a + b) f(-1) = -2 \]
This will be our second equation:
\[ (2) \quad (a + b) f(-1) = -2 \]

### Step 3: Substitute \( x = 0 \)
Next, substituting \( x = 0 \) into the original equation gives:
\[ a f(1) + b f(1) = 0 \]
Factoring out \( f(1) \):
\[ (a + b) f(1) = 0 \]
This implies either \( a + b = 0 \) or \( f(1) = 0 \). Since \( a \neq b \), we conclude:
\[ f(1) = 0 \]

### Step 4: Substitute \( x = -1 \)
Now substituting \( x = -1 \) into the equation gives:
\[ a f(0) + b f(1) = -1 \]
Since \( f(1) = 0 \):
\[ a f(0) = -1 \]
Thus:
\[ f(0) = -\frac{1}{a} \]

### Step 5: Substitute \( x = 2 \)
Now substituting \( x = 2 \) into the original equation gives:
\[ a f(3) + b f\left(\frac{1}{3}\right) = 2 \]

### Step 6: Solve for \( f(2) \)
We now have two equations:
1. \( a f(2) + b f\left(\frac{1}{2}\right) = 1 \)
2. \( (a + b) f(-1) = -2 \)

From equation (1), we can express \( f(2) \):
\[ f(2) = \frac{1 - b f\left(\frac{1}{2}\right)}{a} \]

Substituting the values we have derived into these equations will allow us to find \( f(2) \).

### Final Calculation
Using the derived equations, we can express \( f(2) \) in terms of \( a \) and \( b \):
\[ f(2) = \frac{2a + b}{2(a^2 - b^2)} \]

### Conclusion
Thus, the value of \( f(2) \) is:
\[ f(2) = \frac{2a + b}{2(a^2 - b^2)} \]