Let `f :R to R` be a continous and differentiable function such that `f (x+y ) =f (x). F(y)AA x, y, f(x) ne 0 and f (0 )=1 and f '(0) =2.`
Let `g (xy) =g (x). g (y) AA x, y and g'(1)=2.g (1) ne =0`
The number of values of x, where `f (x) g(x):`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) based on the given conditions. Let's break it down step by step.

### Step 1: Analyze the function \( f(x) \)

We are given that:
1. \( f(x+y) = f(x)f(y) \) for all \( x, y \)
2. \( f(0) = 1 \)
3. \( f'(0) = 2 \)

From the functional equation \( f(x+y) = f(x)f(y) \), we can deduce that \( f(x) \) is an exponential function. We can assume:
\[ f(x) = e^{g(x)} \]
for some function \( g(x) \).

### Step 2: Find \( g(x) \)

Substituting into the functional equation:
\[ e^{g(x+y)} = e^{g(x)} e^{g(y)} \]
This simplifies to:
\[ g(x+y) = g(x) + g(y) \]
This is Cauchy's functional equation, which has solutions of the form \( g(x) = kx \) for some constant \( k \).

### Step 3: Determine \( k \)

Since \( f'(0) = 2 \), we can find \( f'(x) \):
\[ f'(x) = f(x) g'(x) \]
At \( x = 0 \):
\[ f'(0) = f(0) g'(0) = 1 \cdot g'(0) = g'(0) \]
Thus, \( g'(0) = 2 \).

Since \( g(x) = kx \), we have \( g'(x) = k \). Therefore, \( k = 2 \) and:
\[ g(x) = 2x \]

### Step 4: Write \( f(x) \)

Now substituting back, we find:
\[ f(x) = e^{g(x)} = e^{2x} \]

### Step 5: Analyze the function \( g(x) \)

We are also given:
1. \( g(xy) = g(x)g(y) \) for all \( x, y \)
2. \( g'(1) = 2 \)
3. \( g(1) \neq 0 \)

Using a similar approach as before, we can assume:
\[ g(x) = h(x) \]
where \( h(x) \) satisfies the same functional equation. The solutions to this equation are also of the form \( h(x) = c \ln(x) \) for some constant \( c \).

### Step 6: Determine \( c \)

Given \( g'(1) = 2 \):
\[ g'(x) = \frac{c}{x} \]
Thus, \( g'(1) = c \). Therefore, \( c = 2 \).

So, we have:
\[ g(x) = 2 \ln(x) \]

### Step 7: Find the intersection of \( f(x) \) and \( g(x) \)

We need to find the values of \( x \) such that:
\[ f(x) = g(x) \]
This translates to:
\[ e^{2x} = 2 \ln(x) \]

### Step 8: Solve the equation

Let \( h(x) = e^{2x} - 2 \ln(x) \). We need to find the roots of \( h(x) = 0 \).

### Step 9: Analyze the function \( h(x) \)

1. As \( x \to 0^+ \), \( e^{2x} \to 1 \) and \( 2 \ln(x) \to -\infty \), so \( h(x) \to \infty \).
2. As \( x \to \infty \), \( e^{2x} \to \infty \) and \( 2 \ln(x) \to \infty \), but \( e^{2x} \) grows much faster than \( 2 \ln(x) \), so \( h(x) \to \infty \).

### Step 10: Check for critical points

We can find the derivative \( h'(x) \) to check for critical points:
\[ h'(x) = 2e^{2x} - \frac{2}{x} \]

Setting \( h'(x) = 0 \):
\[ 2e^{2x} = \frac{2}{x} \]
\[ e^{2x} = \frac{1}{x} \]

This equation will have a unique solution since \( e^{2x} \) is increasing and \( \frac{1}{x} \) is decreasing.

### Conclusion

Since \( h(x) \) has one critical point and changes from positive to negative, we conclude that there is exactly **one value of \( x \)** where \( f(x) = g(x) \).

### Final Answer

The number of values of \( x \) where \( f(x) = g(x) \) is **1**.

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