If `f''(x) = sec^(2) x` and f (0) = f '(0) = 0 then:

Options

A. f (x) = log sec x
B. `f (x) = sec^(x) x`
C. `f(x) = log sec x + x`
D. None of these

Answer

To solve the problem, we need to find the function \( f(x) \) given that \( f''(x) = \sec^2 x \) and the initial conditions \( f(0) = 0 \) and \( f'(0) = 0 \).

### Step-by-Step Solution:

1. **Start with the second derivative**:
\[ f''(x) = \sec^2 x \]

2. **Integrate to find the first derivative**:
\[ f'(x) = \int f''(x) \, dx = \int \sec^2 x \, dx \]
The integral of \( \sec^2 x \) is \( \tan x \), so we have:
\[ f'(x) = \tan x + C_1 \]
where \( C_1 \) is the constant of integration.

3. **Use the initial condition \( f'(0) = 0 \)**:
\[ f'(0) = \tan(0) + C_1 = 0 \]
Since \( \tan(0) = 0 \), we get:
\[ 0 + C_1 = 0 \implies C_1 = 0 \]
Therefore, we have:
\[ f'(x) = \tan x \]

4. **Integrate again to find the original function \( f(x) \)**:
\[ f(x) = \int f'(x) \, dx = \int \tan x \, dx \]
The integral of \( \tan x \) is \( -\log(\cos x) \) or \( \log(\sec x) \), so we can write:
\[ f(x) = \log(\sec x) + C_2 \]
where \( C_2 \) is another constant of integration.

5. **Use the initial condition \( f(0) = 0 \)**:
\[ f(0) = \log(\sec(0)) + C_2 = 0 \]
Since \( \sec(0) = 1 \), we have:
\[ \log(1) + C_2 = 0 \implies 0 + C_2 = 0 \implies C_2 = 0 \]
Thus, we find:
\[ f(x) = \log(\sec x) \]

### Final Result:
The function \( f(x) \) is:
\[ f(x) = \log(\sec x) \]