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

Options

A. even function
B. odd function
C. neither even nor odd
D. periodic function

Answer

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

### Step 1: Evaluate \( f(-x) \)

We start by substituting \(-x\) into the function:

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

### Step 2: Simplify \( f(-x) \)

Next, we simplify the expression inside the logarithm:

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

### Step 3: Rationalize the expression

We can multiply the numerator and denominator by the conjugate to simplify further:

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

Using the difference of squares:

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

Thus, we have:

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

### Step 4: Use properties of logarithms

Using the property of logarithms, we can rewrite this as:

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

### Step 5: Compare \( f(-x) \) and \( f(x) \)

Now we need to compare \( f(-x) \) with \( -f(x) \):

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

### Step 6: Check if \( f(-x) = -f(x) \)

We see that:

\[ f(-x) = -f(x) \]

This means that the function is odd.

### Conclusion

Thus, the function \( f(x) = \sin\left| \log\left(x + \sqrt{x^2 + 1}\right) \right| \) is an **odd function**.