Let `f(x) = sin(log_(3)(x+sqrt(x^(2)+1))), x in R`, then

Options

A. f is an even function
B. f is an odd function
C. f is a periodic function
D. f is neither even nor odd functions

Answer

To determine whether the function \( f(x) = \sin(\log_3(x + \sqrt{x^2 + 1})) \) is even, odd, or periodic, we will follow these steps:

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

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

\[ f(-x) = \sin\left(\log_3(-x + \sqrt{(-x)^2 + 1}}\right) \]

Since \((-x)^2 = x^2\), we have:

\[ f(-x) = \sin\left(\log_3(-x + \sqrt{x^2 + 1}}\right) \]

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

Next, we simplify the expression inside the logarithm:

\[ f(-x) = \sin\left(\log_3(-x + \sqrt{x^2 + 1}}\right) \]

We can rewrite \(-x + \sqrt{x^2 + 1}\) as:

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

### Step 3: Use the property of logarithms

Using the property of logarithms, we can express:

\[ \log_3(-x + \sqrt{x^2 + 1}) = \log_3\left(\frac{1}{x + \sqrt{x^2 + 1}}\right) \]

This can be rewritten as:

\[ \log_3(-x + \sqrt{x^2 + 1}) = -\log_3(x + \sqrt{x^2 + 1}) \]

### Step 4: Substitute back into \( f(-x) \)

Now we substitute this back into our expression for \( f(-x) \):

\[ f(-x) = \sin\left(-\log_3(x + \sqrt{x^2 + 1})\right) \]

Using the property of sine, we know that:

\[ \sin(-\theta) = -\sin(\theta) \]

Thus, we have:

\[ f(-x) = -\sin\left(\log_3(x + \sqrt{x^2 + 1})\right) = -f(x) \]

### Step 5: Conclusion

Since we have shown that \( f(-x) = -f(x) \), we conclude that the function \( f(x) \) is an **odd function**.

### Final Answer:
The function \( f(x) = \sin(\log_3(x + \sqrt{x^2 + 1})) \) is an odd function.

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