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

Options

A. an odd function
B. an even function
C. neither an odd nor an even function
D. a constant function

Answer

To determine the nature of the function \( f(x) = \sec[\log(x + \sqrt{1+x^2})] \), we will check if it is an even function, odd function, or neither.

### Step-by-step Solution:

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

2. **Calculate \( f(-x) \)**:
We need to find \( f(-x) \):
\[ f(-x) = \sec[\log(-x + \sqrt{1+(-x)^2})] \]
Simplifying \( \sqrt{1+(-x)^2} \) gives:
\[ f(-x) = \sec[\log(-x + \sqrt{1+x^2})] \]

3. **Rationalize the Argument**:
We can rewrite the argument of the secant function:
\[ -x + \sqrt{1+x^2} = \frac{(-x + \sqrt{1+x^2})(\sqrt{1+x^2} + x)}{\sqrt{1+x^2} + x} \]
The numerator simplifies as follows:
\[ (-x + \sqrt{1+x^2})(\sqrt{1+x^2} + x) = (1+x^2) - x^2 = 1 \]
Thus,
\[ f(-x) = \sec[\log\left(\frac{1}{\sqrt{1+x^2} + x}\right)] \]

4. **Using Logarithmic Properties**:
We can use the property of logarithms:
\[ \log\left(\frac{1}{a}\right) = -\log(a) \]
Therefore, we have:
\[ f(-x) = \sec[-\log(\sqrt{1+x^2} + x)] \]

5. **Using the Property of Secant**:
The secant function has the property:
\[ \sec(-\theta) = \sec(\theta) \]
Thus:
\[ f(-x) = \sec[\log(\sqrt{1+x^2} + x)] \]

6. **Conclusion**:
Since \( f(-x) = f(x) \), we conclude that:
\[ f(x) \text{ is an even function.} \]

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