Let f be a function that is differentiable everywhere and have the following properties :
(i) `f(x) gt 0`
(ii) `f'(0) = -1`
(iii) `f(-x) = (1)/(f(x))and f(x+h)=f(x).f(h)`
A standard result is `(f'(x))/(f(x))dx = log|f(x)| + C`
The function y = f(x) is

Options

A. odd
B. even
C. increasing
D. decreasing

Answer

To solve the problem step-by-step, we will analyze the properties given for the function \( f \) and derive the necessary conclusions.

### Step 1: Analyze the properties of the function
Given properties:
1. \( f(x) > 0 \) for all \( x \)
2. \( f'(0) = -1 \)
3. \( f(-x) = \frac{1}{f(x)} \)
4. \( f(x+h) = f(x) \cdot f(h) \)

### Step 2: Use the property \( f(x+h) = f(x) \cdot f(h) \)
This property indicates that \( f \) is a multiplicative function. We can analyze the derivative of \( f \) using this property.

### Step 3: Differentiate \( f(x+h) = f(x) \cdot f(h) \)
Using the definition of the derivative:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Substituting the property:
\[ f'(x) = \lim_{h \to 0} \frac{f(x) \cdot f(h) - f(x)}{h} \]
Factoring out \( f(x) \):
\[ f'(x) = f(x) \cdot \lim_{h \to 0} \frac{f(h) - 1}{h} \]

### Step 4: Evaluate the limit
Let \( L = \lim_{h \to 0} \frac{f(h) - 1}{h} \). Then we have:
\[ f'(x) = f(x) \cdot L \]

### Step 5: Find \( L \) using \( f'(0) = -1 \)
From the property \( f'(0) = -1 \):
\[ f'(0) = f(0) \cdot L \]
Since \( f(0) \) is positive (from property 1), we can solve for \( L \):
\[ -1 = f(0) \cdot L \implies L = -\frac{1}{f(0)} \]

### Step 6: Substitute \( L \) back into the derivative
Thus, we have:
\[ f'(x) = -\frac{f(x)}{f(0)} \]

### Step 7: Solve the differential equation
This is a separable differential equation:
\[ \frac{f'(x)}{f(x)} = -\frac{1}{f(0)} \]
Integrating both sides:
\[ \int \frac{f'(x)}{f(x)} \, dx = -\frac{1}{f(0)} \int dx \]
This gives:
\[ \log |f(x)| = -\frac{x}{f(0)} + C \]
Exponentiating both sides:
\[ f(x) = e^{C} e^{-\frac{x}{f(0)}} \]
Let \( e^{C} = k \), where \( k > 0 \) since \( f(x) > 0 \):
\[ f(x) = k e^{-\frac{x}{f(0)}} \]

### Step 8: Determine the function
Using the property \( f(-x) = \frac{1}{f(x)} \):
\[ f(-x) = k e^{\frac{x}{f(0)}} = \frac{1}{k e^{-\frac{x}{f(0)}}} \]
This leads to:
\[ k e^{\frac{x}{f(0)}} = \frac{1}{k} e^{\frac{x}{f(0)}} \]
Thus, \( k^2 = 1 \) implies \( k = 1 \) (since \( k > 0 \)):
\[ f(x) = e^{-\frac{x}{f(0)}} \]

### Conclusion
The function \( f(x) \) is of the form:
\[ f(x) = e^{-x} \]
This function is positive and decreasing, confirming the properties given.