Let `f : R to R` be a function such that `f(0)=1` and for any `x,y in R, f(xy+1)=f(x)f(y)-f(y)-x+2.` Then `f` is

Options

A. one-one and onto
B. one-one but not onto
C. many one but onto
D. many one and into

Explanation

To solve the problem, we need to analyze the functional equation given and derive the function \( f \).

### Step-by-Step Solution:

1. **Understanding the Functional Equation**:
We start with the functional equation:
\[ f(xy + 1) = f(x)f(y) - f(y) - x + 2 \]
for all \( x, y \in \mathbb{R} \), and we know that \( f(0) = 1 \).

2. **Substituting \( y = 0 \)**:
Let's substitute \( y = 0 \) into the functional equation:
\[ f(x \cdot 0 + 1) = f(x)f(0) - f(0) - x + 2 \]
This simplifies to:
\[ f(1) = f(x) \cdot 1 - 1 - x + 2 \]
Therefore, we have:
\[ f(1) = f(x) - x + 1 \]
Rearranging gives us:
\[ f(x) = f(1) + x - 1 \]

3. **Letting \( f(1) = c \)**:
Let \( f(1) = c \). Then we can express \( f(x) \) as:
\[ f(x) = c + x - 1 \]
Thus, we can write:
\[ f(x) = x + (c - 1) \]

4. **Finding \( c \)**:
We know from the problem statement that \( f(0) = 1 \). Substituting \( x = 0 \) into our expression for \( f(x) \):
\[ f(0) = 0 + (c - 1) = c - 1 \]
Setting this equal to 1 gives:
\[ c - 1 = 1 \implies c = 2 \]

5. **Final Expression for \( f(x) \)**:
Substituting \( c = 2 \) back into the expression for \( f(x) \):
\[ f(x) = x + (2 - 1) = x + 1 \]

6. **Conclusion**:
Therefore, the function \( f \) is:
\[ f(x) = x + 1 \]