If `f(x)=x^m n ,n in N ,` is an even function, then `m` is even integer (b) odd integer any integer (d) `f(x)-e v e ni snotpos s i b l e`

Options

A. even integer
B. odd integer
C. any integer
D. `f(x)`-even is not possible

Explanation

To determine the value of \( m \) such that the function \( f(x) = x^{mn} \) is an even function, we can follow these steps:

### Step 1: Understanding the definition of an even function
An even function satisfies the condition:
\[ f(-x) = f(x) \]
for all \( x \).

### Step 2: Calculate \( f(-x) \)
Given \( f(x) = x^{mn} \), we can find \( f(-x) \):
\[ f(-x) = (-x)^{mn} \]

### Step 3: Simplifying \( f(-x) \)
Using the property of exponents, we can rewrite \( f(-x) \):
\[ f(-x) = (-1)^{mn} \cdot x^{mn} \]

### Step 4: Setting the condition for evenness
For \( f(x) \) to be an even function, we must have:
\[ f(-x) = f(x) \]
This gives us:
\[ (-1)^{mn} \cdot x^{mn} = x^{mn} \]

### Step 5: Analyzing the equation
For the above equation to hold true for all \( x \), the term \( (-1)^{mn} \) must equal \( 1 \). This occurs when \( mn \) is an even integer.

### Step 6: Conclusion about \( m \)
Since \( n \) is a natural number (and thus positive), \( m \) must be an even integer for \( mn \) to be even. If \( m \) were odd, \( mn \) would be odd, contradicting our requirement.

### Final Answer
Thus, we conclude that \( m \) must be an even integer.