The function `f(x) = sec[log(x + sqrt(1+x^2))]` is

Options

A. Even
B. odd
C. Constant
D. None of these

Answer

To determine the nature of the function \( f(x) = \sec[\log(x + \sqrt{1+x^2})] \), we will analyze whether it is an even function, odd function, constant, or none of these.

### Step-by-Step Solution:

1. **Define the Function**:
\[ f(x) = \sec[\log(x + \sqrt{1+x^2})] \]

2. **Find \( f(-x) \)**:
We need to compute \( f(-x) \):
\[ f(-x) = \sec[\log(-x + \sqrt{1+(-x)^2})] \]
Simplifying the expression inside the logarithm:
\[ f(-x) = \sec[\log(-x + \sqrt{1+x^2})] \]

3. **Simplify \( -x + \sqrt{1+x^2} \)**:
Notice that:
\[ -x + \sqrt{1+x^2} = \frac{1+x^2 - x^2}{-x + \sqrt{1+x^2} + x} = \frac{1}{x + \sqrt{1+x^2}} \]
Thus:
\[ \log(-x + \sqrt{1+x^2}) = \log\left(\frac{1}{x + \sqrt{1+x^2}}\right) = -\log(x + \sqrt{1+x^2}) \]

4. **Substituting Back**:
Now substituting back into \( f(-x) \):
\[ f(-x) = \sec[-\log(x + \sqrt{1+x^2})] \]

5. **Using the Property of Secant**:
We know that \( \sec(-\theta) = \sec(\theta) \):
\[ f(-x) = \sec[\log(x + \sqrt{1+x^2})] = f(x) \]

6. **Conclusion**:
Since \( f(-x) = f(x) \), we conclude that the function \( f(x) \) is an **even function**.

### Final Answer:
The function \( f(x) = \sec[\log(x + \sqrt{1+x^2})] \) is an **even function**.