Let `f(x) = secx*f'(x), f(0) = 1,` then `f(pi/6)` is equal to

Options

A. `(1)/(sqrt(e))`
B. `sqrt(e)`
C. `e^((3)/(2))`
D. `(1)/(2sqrt(e))`

Answer

To solve the differential equation given by \( f(x) = \sec x \cdot f'(x) \) with the initial condition \( f(0) = 1 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ f(x) = \sec x \cdot f'(x) \]
We can rearrange this to isolate \( f'(x) \):
\[ f'(x) = \cos x \cdot f(x) \]

### Step 2: Separate the variables
Next, we can separate the variables:
\[ \frac{f'(x)}{f(x)} = \cos x \]

### Step 3: Integrate both sides
Now, we integrate both sides. The left side integrates to:
\[ \int \frac{f'(x)}{f(x)} \, dx = \ln |f(x)| \]
The right side integrates to:
\[ \int \cos x \, dx = \sin x + C \]
Thus, we have:
\[ \ln |f(x)| = \sin x + C \]

### Step 4: Exponentiate to solve for \( f(x) \)
Exponentiating both sides gives us:
\[ |f(x)| = e^{\sin x + C} = e^C \cdot e^{\sin x} \]
Let \( K = e^C \), so we can write:
\[ f(x) = K e^{\sin x} \]

### Step 5: Use the initial condition to find \( K \)
We know that \( f(0) = 1 \):
\[ f(0) = K e^{\sin(0)} = K e^0 = K \]
Thus, \( K = 1 \). Therefore, we have:
\[ f(x) = e^{\sin x} \]

### Step 6: Find \( f(\pi/6) \)
Now we need to find \( f(\pi/6) \):
\[ f\left(\frac{\pi}{6}\right) = e^{\sin\left(\frac{\pi}{6}\right)} = e^{\frac{1}{2}} \]

### Final Answer
Thus, the value of \( f\left(\frac{\pi}{6}\right) \) is:
\[ \boxed{e^{1/2}} \]