A function f defined by fix)=In `(sqrt(x+1-x))` is

Options

A. an even function
B. an odd function
C. Both even and odd function
D. Neither even nor odd function

Answer

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

### Step 1: Understand the Definitions
- An **even function** satisfies the condition \( f(-x) = f(x) \) for all \( x \).
- An **odd function** satisfies the condition \( f(-x) = -f(x) \) for all \( x \).

### Step 2: Calculate \( f(-x) \)
We need to find \( f(-x) \):
\[ f(-x) = \ln(\sqrt{-x + 1 - (-x)}) = \ln(\sqrt{-x + 1 + x}) = \ln(\sqrt{1}) \]
Since \( \sqrt{1} = 1 \), we have:
\[ f(-x) = \ln(1) = 0 \]

### Step 3: Calculate \( f(x) \)
Now, we calculate \( f(x) \):
\[ f(x) = \ln(\sqrt{x + 1 - x}) = \ln(\sqrt{1}) = \ln(1) = 0 \]

### Step 4: Compare \( f(-x) \) and \( f(x) \)
We found that:
\[ f(-x) = 0 \quad \text{and} \quad f(x) = 0 \]
Thus, we have:
\[ f(-x) = f(x) \]

### Step 5: Conclusion
Since \( f(-x) = f(x) \), we conclude that the function \( f(x) \) is an **even function**.

### Final Answer
The function \( f(x) = \ln(\sqrt{x + 1 - x}) \) is an even function.

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