Determine whether the function ` sin { log x + sqrt((x^2 +1))}` is

Options

A. even
B. odd
C. constant
D. none

Answer

To determine whether the function \( f(x) = \sin(\log x + \sqrt{x^2 + 1}) \) is even, odd, or neither, we will follow these steps:

### Step 1: Find \( f(-x) \)
We start by substituting \(-x\) into the function:

\[ f(-x) = \sin(\log(-x) + \sqrt{(-x)^2 + 1}) \]

### Step 2: Simplify \( f(-x) \)
Now, simplify the expression:

1. The term \(\sqrt{(-x)^2 + 1}\) simplifies to \(\sqrt{x^2 + 1}\) because \((-x)^2 = x^2\).
2. The term \(\log(-x)\) is not defined for \(x > 0\). However, for \(x < 0\), we can write \(\log(-x) = \log x + \log(-1) = \log x + i\pi\) (in the complex domain). But we will focus on the real case.

Thus, we have:

\[ f(-x) = \sin(\log(-x) + \sqrt{x^2 + 1}) = \sin(\log(-x) + \sqrt{x^2 + 1}) \]

### Step 3: Analyze the function
Now we need to check if \( f(-x) = f(x) \) or \( f(-x) = -f(x) \).

1. For \( f(x) \):
\[ f(x) = \sin(\log x + \sqrt{x^2 + 1}) \]

2. For \( f(-x) \):
\[ f(-x) = \sin(\log(-x) + \sqrt{x^2 + 1}) = \sin(\log x + i\pi + \sqrt{x^2 + 1}) \]

### Step 4: Determine if the function is even or odd
From the above, we can see that:

- \( f(-x) \) does not equal \( f(x) \), so it is not even.
- To check if it is odd, we need to see if \( f(-x) = -f(x) \).

Using the property of sine:

\[ \sin(a + b) = \sin a \cos b + \cos a \sin b \]

Thus, we can see that \( f(-x) \) does not simplify to \(-f(x)\) either.

### Conclusion
Since \( f(-x) \neq f(x) \) and \( f(-x) \neq -f(x) \), we conclude that the function \( f(x) = \sin(\log x + \sqrt{x^2 + 1}) \) is neither even nor odd.