If `f(0) = f'(0) = 0` and `f'' (x) = "tan"^2 x,` then `f(x)` =

Options

A. `log sec x + 1/2 x^2`
B. `log sec x - 1/2 x^2`
C. `log cos x + 1/2 x^2`
D. `log cos x - 1/2 x^2`

Answer

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

### Step-by-step Solution:

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

2. **Integrate to find the first derivative**:
To find \( f'(x) \), we integrate \( f''(x) \):
\[ f'(x) = \int \tan^2 x \, dx \]
We can use the identity \( \tan^2 x = \sec^2 x - 1 \):
\[ f'(x) = \int (\sec^2 x - 1) \, dx = \int \sec^2 x \, dx - \int 1 \, dx \]
The integrals yield:
\[ f'(x) = \tan x - x + C \]
where \( C \) is the constant of integration.

3. **Use the condition \( f'(0) = 0 \)**:
Substitute \( x = 0 \) into \( f'(x) \):
\[ f'(0) = \tan(0) - 0 + C = 0 \implies C = 0 \]
Therefore, we have:
\[ f'(x) = \tan x - x \]

4. **Integrate to find the original function**:
Now, we integrate \( f'(x) \) to find \( f(x) \):
\[ f(x) = \int (\tan x - x) \, dx \]
This can be split into two parts:
\[ f(x) = \int \tan x \, dx - \int x \, dx \]
The integral of \( \tan x \) is \( -\log(\cos x) \) (or \( \log(\sec x) \)), and the integral of \( x \) is \( \frac{x^2}{2} \):
\[ f(x) = -\log(\cos x) - \frac{x^2}{2} + K \]
where \( K \) is another constant of integration.

5. **Use the condition \( f(0) = 0 \)**:
Substitute \( x = 0 \) into \( f(x) \):
\[ f(0) = -\log(\cos(0)) - \frac{0^2}{2} + K = -\log(1) + K = K \]
Since \( f(0) = 0 \), we have \( K = 0 \).

6. **Final expression for \( f(x) \)**:
Thus, we conclude that:
\[ f(x) = -\log(\cos x) - \frac{x^2}{2} \]
This can also be expressed as:
\[ f(x) = \log(\sec x) - \frac{x^2}{2} \]

### Final Answer:
\[ f(x) = \log(\sec x) - \frac{x^2}{2} \]