If `f(x+y)=f(x)+f(y) -xy -1` for all `x, y in R` and `f(1)=1` then `f(n)=n, n in N` is true if

Options

A. `n=1`
B. `n=1, and n=2`
C. n is odd
D. any value of n

Answer

To solve the functional equation \( f(x+y) = f(x) + f(y) - xy - 1 \) for all \( x, y \in \mathbb{R} \) with the condition \( f(1) = 1 \), we will proceed step by step.

### Step 1: Substitute \( x = 1 \) and \( y = 1 \)

We start by substituting \( x = 1 \) and \( y = 1 \) into the functional equation:

\[ f(1 + 1) = f(1) + f(1) - 1 \cdot 1 - 1 \]

### Step 2: Simplify the equation

Substituting \( f(1) = 1 \):

\[ f(2) = 1 + 1 - 1 - 1 \]

This simplifies to:

\[ f(2) = 0 \]

### Step 3: Substitute \( x = 2 \) and \( y = 1 \)

Next, we substitute \( x = 2 \) and \( y = 1 \) into the functional equation:

\[ f(2 + 1) = f(2) + f(1) - 2 \cdot 1 - 1 \]

### Step 4: Simplify the equation

Substituting \( f(2) = 0 \) and \( f(1) = 1 \):

\[ f(3) = 0 + 1 - 2 - 1 \]

This simplifies to:

\[ f(3) = -2 \]

### Step 5: Generalize for \( n \in \mathbb{N} \)

Now, we can observe a pattern. Let's assume \( f(n) \) for \( n \in \mathbb{N} \):

- \( f(1) = 1 \)
- \( f(2) = 0 \)
- \( f(3) = -2 \)

### Step 6: Check for \( n = 4 \)

Let's find \( f(4) \) by substituting \( x = 3 \) and \( y = 1 \):

\[ f(3 + 1) = f(3) + f(1) - 3 \cdot 1 - 1 \]

Substituting the known values:

\[ f(4) = -2 + 1 - 3 - 1 \]

This simplifies to:

\[ f(4) = -5 \]

### Step 7: Conclusion

From the calculations, we see that \( f(n) \) does not yield natural numbers for \( n = 2, 3, 4 \) and so on. The only value of \( n \) for which \( f(n) = n \) holds true is \( n = 1 \).

Thus, we conclude that \( f(n) = n \) is true only for \( n = 1 \).