The function `f(x)=cos(log(x+sqrt(x^2+1)))` is `:`

Options

A. even
B. odd
C. constant
D. None of these

Answer

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

### Step-by-Step Solution:

1. **Find \( f(-x) \)**:
We start by substituting \(-x\) into the function:
\[ f(-x) = \cos(\log(-x + \sqrt{(-x)^2 + 1})) \]
Simplifying the expression inside the logarithm:
\[ f(-x) = \cos(\log(-x + \sqrt{x^2 + 1})) \]

2. **Simplify the logarithmic term**:
We can rewrite the term \(-x + \sqrt{x^2 + 1}\) as follows:
\[ -x + \sqrt{x^2 + 1} = \frac{1}{x + \sqrt{x^2 + 1}} \]
This is because:
\[ (-x + \sqrt{x^2 + 1})(x + \sqrt{x^2 + 1}) = (-x^2 + x^2 + 1) = 1 \]
Thus, we have:
\[ f(-x) = \cos(\log\left(\frac{1}{x + \sqrt{x^2 + 1}}\right)) \]

3. **Use properties of logarithms**:
We can use the property of logarithms that states \(\log\left(\frac{1}{a}\right) = -\log(a)\):
\[ f(-x) = \cos(-\log(x + \sqrt{x^2 + 1})) \]

4. **Apply the cosine function property**:
Since \(\cos(-\theta) = \cos(\theta)\), we can simplify further:
\[ f(-x) = \cos(\log(x + \sqrt{x^2 + 1})) \]

5. **Compare \( f(-x) \) with \( f(x) \)**:
Now, we see that:
\[ f(-x) = \cos(\log(x + \sqrt{x^2 + 1})) = f(x) \]
This shows that \( f(-x) = f(x) \), which is the definition of an even function.

### Conclusion:
Thus, the function \( f(x) = \cos(\log(x + \sqrt{x^2 + 1})) \) is an **even function**.

---