If `f:R rarr (0,oo)`be a differentiable function `f(x)` satisfying
`f(x+y)-f(x-y)=f(x)*{f(y)-f(-y)}, AA x, y in R, (f(y)!= f(-y) " for all " y in R)` and `f'(0)=2010`.
Now, answer the following questions.
Which of the following is true for `f(x)`

Options

A. `f(x)` is one-one and into
B. `{f(x)}` is non-periodic, where `{*}` denotes fractional part of `x`.
C. `f(x)=4` has only two solutions.
D. `f(x)=f^(-1)(x)` has only one solution .

Answer

To solve the problem, we need to analyze the given functional equation and the properties of the function \( f(x) \). Let's go through the steps systematically.

### Step 1: Analyze the Functional Equation
The functional equation given is:
\[ f(x+y) - f(x-y) = f(x) \cdot (f(y) - f(-y)) \]
for all \( x, y \in \mathbb{R} \).

### Step 2: Differentiate the Functional Equation
We differentiate both sides of the equation with respect to \( y \) at \( y = 0 \):
\[ \frac{d}{dy}[f(x+y) - f(x-y)] \bigg|_{y=0} = \frac{d}{dy}[f(x) \cdot (f(y) - f(-y))] \bigg|_{y=0} \]

Using the chain rule on the left side:
\[ f'(x+0) + f'(x-0) = f'(x) \cdot (f(0) - f(0)) + f(x) \cdot (f'(0) + f'(0)) \]
This simplifies to:
\[ 2f'(x) = 2f(x)f'(0) \]

### Step 3: Simplify the Equation
Dividing both sides by 2:
\[ f'(x) = f(x)f'(0) \]

### Step 4: Substitute the Given Value
We know that \( f'(0) = 2010 \):
\[ f'(x) = 2010 f(x) \]

### Step 5: Solve the Differential Equation
This is a separable differential equation. We can rewrite it as:
\[ \frac{f'(x)}{f(x)} = 2010 \]
Integrating both sides:
\[ \int \frac{1}{f(x)} df = \int 2010 \, dx \]
This gives us:
\[ \ln |f(x)| = 2010x + C \]
Exponentiating both sides:
\[ f(x) = e^{2010x + C} = Ce^{2010x} \]
where \( C = e^C \) is a constant.

### Step 6: Determine the Constant \( C \)
To find \( C \), we use the condition \( f(0) \):
\[ f(0) = Ce^{2010 \cdot 0} = C \]
Since \( f(x) \) must be positive for all \( x \), we can set \( C = 1 \):
\[ f(x) = e^{2010x} \]

### Step 7: Analyze the Properties of \( f(x) \)
The function \( f(x) = e^{2010x} \) is an increasing function. The fractional part of an increasing function is non-periodic.

### Conclusion
Thus, the correct statement regarding \( f(x) \) is:
- The fractional part of \( f(x) \) is non-periodic.