If a function satisfies the relation `f(x) f''(x)-f(x)f'(x)=(f'(x))^(2) AA x in R and f(0)=f'(0)=1,` then
Number of roots of the equation `f(x)=e^(x)` is

Options

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

Explanation

To solve the problem, we start with the given relation:

\[ f(x) f''(x) - f(x) f'(x) = (f'(x))^2 \]

and the initial conditions:

\[ f(0) = 1 \quad \text{and} \quad f'(0) = 1. \]

### Step 1: Rearranging the Equation

We can rearrange the given equation as follows:

\[ f(x) f''(x) - f(x) f'(x) - (f'(x))^2 = 0. \]

### Step 2: Factoring

We can factor out \( f'(x) \):

\[ f'(x) (f''(x) - f'(x)) = f(x) f'(x). \]

Dividing both sides by \( f'(x) \) (assuming \( f'(x) \neq 0 \)) gives us:

\[ f''(x) - f'(x) = \frac{f(x)}{f'(x)}. \]

### Step 3: Integrating Both Sides

Now we integrate both sides with respect to \( x \):

\[ \int (f''(x) - f'(x)) \, dx = \int \frac{f(x)}{f'(x)} \, dx. \]

The left side simplifies to:

\[ f'(x) - f(x) + C_1, \]

where \( C_1 \) is a constant of integration.

### Step 4: Substituting for the Right Side

For the right side, we can use the substitution \( f'(x) = t \), which gives us \( f''(x) dx = dt \):

\[ \int \frac{f(x)}{t} \, dt. \]

This leads us to:

\[ \ln |f'(x)| - x + C_2, \]

where \( C_2 \) is another constant of integration.

### Step 5: Applying Initial Conditions

Using the initial conditions \( f(0) = 1 \) and \( f'(0) = 1 \):

At \( x = 0 \):

\[ f'(0) - f(0) + C_1 = 0 \implies 1 - 1 + C_1 = 0 \implies C_1 = 0. \]

Thus, we have:

\[ f'(x) - f(x) = \ln |f'(x)| - x. \]

### Step 6: Finding the Function

From the previous results, we can express:

\[ \frac{f'(x)}{f(x)} = e^x. \]

Integrating gives us:

\[ \ln |f(x)| = e^x + C. \]

Exponentiating both sides yields:

\[ f(x) = e^{e^x + C} = e^{e^x} e^C. \]

Using \( f(0) = 1 \):

\[ e^{e^0 + C} = 1 \implies e^{1 + C} = 1 \implies 1 + C = 0 \implies C = -1. \]

Thus, we have:

\[ f(x) = e^{e^x - 1}. \]

### Step 7: Finding the Number of Roots

Now we need to find the number of roots of the equation:

\[ f(x) = e^x. \]

This translates to:

\[ e^{e^x - 1} = e^x. \]

Taking the natural logarithm on both sides gives:

\[ e^x - 1 = x. \]

### Step 8: Analyzing the Function

Let \( g(x) = e^x - 1 - x \). We need to find the roots of \( g(x) = 0 \).

Calculating \( g(0) \):

\[ g(0) = e^0 - 1 - 0 = 0. \]

Calculating the derivative:

\[ g'(x) = e^x - 1. \]

Since \( g'(0) = 0 \) and \( g'(x) > 0 \) for \( x > 0 \), \( g(x) \) is increasing for \( x > 0 \).

### Conclusion

Since \( g(0) = 0 \) and \( g(x) \) is increasing for \( x > 0 \), there is only one root at \( x = 0 \).

Thus, the number of roots of the equation \( f(x) = e^x \) is:

**1 root.**