`f(x)=ln(x+sqrt(x^(2)+1))` is :

Options

A. an even function
B. an odd function
C. neither even nor odd
D. none of these

Answer

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

### Step 1: Define Even and Odd Functions
An even function satisfies the condition:
\[ f(-x) = f(x) \]
An odd function satisfies the condition:
\[ f(-x) = -f(x) \]

### Step 2: Calculate \( f(-x) \)
We need to find \( f(-x) \):
\[ f(-x) = \ln(-x + \sqrt{(-x)^2 + 1}) = \ln(-x + \sqrt{x^2 + 1}) \]

### Step 3: Compare \( f(-x) \) with \( f(x) \)
Now we will compare \( f(-x) \) with \( f(x) \):
\[ f(x) = \ln(x + \sqrt{x^2 + 1}) \]
We need to check if \( f(-x) = f(x) \) or \( f(-x) = -f(x) \).

### Step 4: Check if \( f(-x) = f(x) \)
To check if \( f(-x) = f(x) \):
\[ \ln(-x + \sqrt{x^2 + 1}) \stackrel{?}{=} \ln(x + \sqrt{x^2 + 1}) \]
This is not true because the arguments of the logarithms are different.

### Step 5: Check if \( f(-x) = -f(x) \)
Now we check if \( f(-x) = -f(x) \):
\[ \ln(-x + \sqrt{x^2 + 1}) \stackrel{?}{=} -\ln(x + \sqrt{x^2 + 1}) \]
This can be rewritten using properties of logarithms:
\[ \ln(-x + \sqrt{x^2 + 1}) \stackrel{?}{=} \ln\left(\frac{1}{x + \sqrt{x^2 + 1}}\right) \]
This is also not true because the left-hand side and right-hand side are not equal.

### Step 6: Conclusion
Since \( f(-x) \) is neither equal to \( f(x) \) nor equal to \( -f(x) \), we conclude that the function \( f(x) \) is neither even nor odd.

### Final Answer
The correct option is: **Neither even nor odd.**

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