Let `f(x)` is a polynomial satisfying `f(x).f(y) = f(x) +f(y) + f(xy) - 2 ` for all `x, y` and `f(2) = 1025,` then the value of `lim_(x->2) f'(x)` is

Answer

To solve the problem, we need to find the limit of the derivative of the polynomial function \( f(x) \) as \( x \) approaches 2, given the functional equation and the value of \( f(2) \).

### Step-by-step Solution:

1. **Understanding the Functional Equation:**
We start with the equation given:
\[ f(x) f(y) = f(x) + f(y) + f(xy) - 2 \]
for all \( x, y \).

2. **Substituting \( y = \frac{1}{x} \):**
Let's set \( y = \frac{1}{x} \) in the functional equation:
\[ f(x) f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right) + f(1) - 2 \]
Rearranging gives:
\[ f(x) f\left(\frac{1}{x}\right) - f(x) - f\left(\frac{1}{x}\right) + 2 = f(1) \]

3. **Substituting \( x = 1 \):**
Now, set \( x = 1 \) in the original equation:
\[ f(1) f(1) = f(1) + f(1) + f(1) - 2 \]
This simplifies to:
\[ f(1)^2 = 3f(1) - 2 \]
Rearranging gives:
\[ f(1)^2 - 3f(1) + 2 = 0 \]
Factoring, we find:
\[ (f(1) - 1)(f(1) - 2) = 0 \]
Thus, \( f(1) = 1 \) or \( f(1) = 2 \).

4. **Substituting \( y = 1 \) in the Functional Equation:**
Now, set \( y = 1 \) in the original equation:
\[ f(x) f(1) = f(x) + f(1) + f(x) - 2 \]
This simplifies to:
\[ f(x) f(1) = 2f(x) + f(1) - 2 \]
Rearranging gives:
\[ f(1)(f(x) - 1) = 2(f(x) - 1) \]
If \( f(x) \neq 1 \), we can divide both sides by \( f(x) - 1 \):
\[ f(1) = 2 \]
Therefore, we conclude \( f(1) = 2 \).

5. **Finding the Form of \( f(x) \):**
From the previous steps, we have:
\[ f(x) f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right) \]
This implies that \( f(x) \) can be expressed as:
\[ f(x) = x^n + 1 \]
for some integer \( n \).

6. **Using the Given Value \( f(2) = 1025 \):**
We know:
\[ f(2) = 2^n + 1 = 1025 \]
Thus:
\[ 2^n = 1024 \implies n = 10 \]
Therefore, we have:
\[ f(x) = x^{10} + 1 \]

7. **Finding the Derivative \( f'(x) \):**
Now we find the derivative:
\[ f'(x) = 10x^9 \]

8. **Calculating the Limit:**
We need to find:
\[ \lim_{x \to 2} f'(x) = \lim_{x \to 2} 10x^9 = 10 \cdot 2^9 = 10 \cdot 512 = 5120 \]

### Final Answer:
\[ \lim_{x \to 2} f'(x) = 5120 \]