Let `f(x)` be a continuous and differentiable function satisfying `f(x + y) = f(x)f(y) AA x,y in R` if `f(x)` an be expressed as `f(x) = 1 + x P(x) + x^2Q(x)` where `lim_(x->0) P(x) = a and lim_(x->0) Q(x) = b,` then `f'(x)` is equal to :

Options

A. `af (x)`
B. `bf (x)`
C. `(a+b) f (x)`
D. `(a+2b) f(x)`

Answer

To solve the problem, we need to find the derivative \( f'(x) \) of the function \( f(x) \) given that it satisfies the functional equation \( f(x+y) = f(x)f(y) \) and can be expressed in the form \( f(x) = 1 + xP(x) + x^2Q(x) \).

### Step-by-Step Solution:

1. **Understanding the Functional Equation**:
The equation \( f(x+y) = f(x)f(y) \) suggests that \( f(x) \) is an exponential function. A common solution to this type of equation is \( f(x) = e^{g(x)} \) for some function \( g(x) \).

2. **Substituting \( y = h \)**:
Let \( h \to 0 \). Then, we have:
\[ f(x+h) = f(x)f(h) \]

3. **Finding the Derivative**:
The derivative \( f'(x) \) can be expressed as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Substituting from the functional equation:
\[ f'(x) = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h} \]

4. **Calculating \( \lim_{h \to 0} \frac{f(h) - 1}{h} \)**:
Given \( f(h) = 1 + hP(h) + h^2Q(h) \), we can find:
\[ f(h) - 1 = hP(h) + h^2Q(h) \]
Thus,
\[ \frac{f(h) - 1}{h} = P(h) + hQ(h) \]
Taking the limit as \( h \to 0 \):
\[ \lim_{h \to 0} \left( P(h) + hQ(h) \right) = P(0) = a \]
since \( \lim_{h \to 0} Q(h) = b \) and \( h \to 0 \) makes \( hQ(h) \to 0 \).

5. **Final Expression for \( f'(x) \)**:
Thus, we have:
\[ f'(x) = f(x) \cdot a \]

6. **Conclusion**:
Therefore, the derivative \( f'(x) \) can be expressed as:
\[ f'(x) = a f(x) \]

### Final Answer:
\[ f'(x) = a f(x) \]