Let `f(x)` be a function such that `f(x).f(y)=f(x+y)`, `f(0)=1`,`f(1)=4`. If `2g(x)=f(x).(1-g(x))`

Options

A. g(x) - g(1-x) = 0
B. g(x) = 1- g(1-x)
C. `underset(k = 1)overset(9)sumg((k)/(10)) = (9)/(2)`
D. `underset(k = 1)overset(18)sum g((k)/(19)) = 9`

Answer

To solve the problem step by step, we need to analyze the given function \( f(x) \) and the relationship with \( g(x) \).

### Step 1: Understand the given function properties
We are given that:
1. \( f(x) \cdot f(y) = f(x+y) \) (This indicates that \( f \) is a multiplicative function).
2. \( f(0) = 1 \)
3. \( f(1) = 4 \)

### Step 2: Find the general form of \( f(x) \)
Using the property of \( f \):
- Let’s find \( f(2) \):
\[ f(2) = f(1+1) = f(1) \cdot f(1) = 4 \cdot 4 = 16 \]
- Let’s find \( f(3) \):
\[ f(3) = f(2+1) = f(2) \cdot f(1) = 16 \cdot 4 = 64 \]
- Let’s find \( f(4) \):
\[ f(4) = f(2+2) = f(2) \cdot f(2) = 16 \cdot 16 = 256 \]

From the calculations, we can observe that \( f(n) = 4^n \) for integer values of \( n \).

### Step 3: Verify the general form of \( f(x) \)
To verify, we can check:
- \( f(0) = 4^0 = 1 \) (correct)
- \( f(1) = 4^1 = 4 \) (correct)
- \( f(2) = 4^2 = 16 \) (correct)

Thus, we can conclude that \( f(x) = 4^x \) for all \( x \).

### Step 4: Substitute \( f(x) \) into the equation for \( g(x) \)
We are given:
\[ 2g(x) = f(x)(1 - g(x)) \]
Substituting \( f(x) = 4^x \):
\[ 2g(x) = 4^x(1 - g(x)) \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 2g(x) + 4^x g(x) = 4^x \]
Factoring out \( g(x) \):
\[ g(x)(2 + 4^x) = 4^x \]
Thus,
\[ g(x) = \frac{4^x}{2 + 4^x} \]

### Step 6: Find \( g(1-x) \)
Now, we need to find \( g(1-x) \):
\[ g(1-x) = \frac{4^{1-x}}{2 + 4^{1-x}} = \frac{\frac{4}{4^x}}{2 + \frac{4}{4^x}} = \frac{4}{4 + 2 \cdot 4^x} \]

### Step 7: Summing \( g(x) \) and \( g(1-x) \)
Now, we calculate \( g(x) + g(1-x) \):
\[ g(x) + g(1-x) = \frac{4^x}{2 + 4^x} + \frac{4}{4 + 2 \cdot 4^x} \]
Finding a common denominator:
\[ = \frac{4^x(4 + 2 \cdot 4^x) + 4(2 + 4^x)}{(2 + 4^x)(4 + 2 \cdot 4^x)} \]
Simplifying the numerator:
\[ = \frac{4^{x+1} + 2 \cdot 4^{2x} + 8 + 4 \cdot 4^x}{(2 + 4^x)(4 + 2 \cdot 4^x)} \]
Combining like terms:
\[ = \frac{2 \cdot 4^{2x} + 8 + 8 \cdot 4^x}{(2 + 4^x)(4 + 2 \cdot 4^x)} \]
This simplifies to:
\[ = 1 \]

### Conclusion
Thus, we conclude that:
\[ g(x) + g(1-x) = 1 \]
This means \( g(x) = 1 - g(1-x) \).

### Final Answer
The correct option is \( \text{B} \).