Let f(x)=`[(g(x)-g(-x))/(f(x)+f(-x))]^(m)` such that m =2n `in` N and f(-x) `ne` -f(x) then f(x) is

Options

A. an odd function
B. an even fiinction
C. periodic function
D. None of these

Answer

To solve the problem, we need to analyze the function \( f(x) = \left( \frac{g(x) - g(-x)}{f(x) + f(-x)} \right)^{m} \) under the given conditions. Let's break it down step by step.

### Step 1: Understand the given function
We have:
\[ f(x) = \left( \frac{g(x) - g(-x)}{f(x) + f(-x)} \right)^{m} \]
where \( m = 2n \) and \( n \in \mathbb{N} \).

### Step 2: Substitute \( -x \) into the function
To find \( f(-x) \), we replace \( x \) with \( -x \) in the function:
\[ f(-x) = \left( \frac{g(-x) - g(x)}{f(-x) + f(x)} \right)^{m} \]
Notice that \( g(-x) - g(x) = - (g(x) - g(-x)) \).

### Step 3: Rewrite \( f(-x) \)
Using the substitution from Step 2, we can rewrite \( f(-x) \):
\[ f(-x) = \left( \frac{- (g(x) - g(-x))}{f(-x) + f(x)} \right)^{m} \]
This can be simplified to:
\[ f(-x) = \left( - \frac{g(x) - g(-x)}{f(-x) + f(x)} \right)^{m} \]

### Step 4: Analyze the implications of \( m = 2n \)
Since \( m = 2n \) is an even number, we can note that raising a negative number to an even power results in a positive number. Therefore:
\[ f(-x) = \left( \frac{g(x) - g(-x)}{f(-x) + f(x)} \right)^{m} \]
This means:
\[ f(-x) = f(x) \]

### Step 5: Conclusion about the function
Since we have found that \( f(-x) = f(x) \), this means that the function \( f(x) \) is an **even function**.

### Final Answer
Thus, the answer to the question is that \( f(x) \) is an **even function**.

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