A function f : ` R to R ` Satisfies the following conditions
(i) `f (x) ne 0 AA x in R `
(ii) `f(x +y)= f(x) f(y) AA x, y, in R `
(iii) f(x) is differentiable
(iv ) f'(0) =2
`lim_(xto 0) (f(x)-f(-x))/x`

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem step by step, we will analyze the given conditions and find the required limit.

### Step 1: Analyze the conditions
We have a function \( f: \mathbb{R} \to \mathbb{R} \) that satisfies:
1. \( f(x) \neq 0 \) for all \( x \in \mathbb{R} \)
2. \( f(x+y) = f(x)f(y) \) for all \( x, y \in \mathbb{R} \)
3. \( f(x) \) is differentiable
4. \( f'(0) = 2 \)

### Step 2: Identify the form of the function
From the second condition, \( f(x+y) = f(x)f(y) \), we can infer that \( f(x) \) is likely of the exponential form. Let's assume:
\[ f(x) = a^x \]
for some constant \( a \).

### Step 3: Verify the assumption
We need to check if this form satisfies the conditions:
- For \( f(x+y) = a^{x+y} = a^x a^y = f(x)f(y) \), this holds true.
- Since \( a^x \) is differentiable for any real \( a \), the third condition is satisfied.
- Now, we need to find \( f'(0) \).

### Step 4: Differentiate the function
The derivative of \( f(x) = a^x \) is given by:
\[ f'(x) = a^x \ln(a) \]
Now, substituting \( x = 0 \):
\[ f'(0) = a^0 \ln(a) = \ln(a) \]

### Step 5: Use the condition \( f'(0) = 2 \)
From the fourth condition, we have:
\[ \ln(a) = 2 \]
This implies:
\[ a = e^2 \]

### Step 6: Write the function
Thus, the function is:
\[ f(x) = (e^2)^x = e^{2x} \]

### Step 7: Find the limit
We need to evaluate:
\[ \lim_{x \to 0} \frac{f(x) - f(-x)}{x} \]
Substituting the function:
\[ \lim_{x \to 0} \frac{e^{2x} - e^{-2x}}{x} \]

### Step 8: Simplify the limit
As \( x \to 0 \), both \( e^{2x} \) and \( e^{-2x} \) approach 1, leading to an indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule:
Differentiate the numerator and denominator:
- Numerator: \( \frac{d}{dx}(e^{2x} - e^{-2x}) = 2e^{2x} + 2e^{-2x} \)
- Denominator: \( \frac{d}{dx}(x) = 1 \)

Now apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{2e^{2x} + 2e^{-2x}}{1} \]

### Step 9: Evaluate the limit
Substituting \( x = 0 \):
\[ 2e^{0} + 2e^{0} = 2 \cdot 1 + 2 \cdot 1 = 4 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{f(x) - f(-x)}{x} = 4 \]