The real valued function `f(x)=(a^x-1)/(x^n(a^x+1))` is even, then the value of n can be

Options

A. Odd, if n is even
B. Even if n is odd
C. Neither even nor odd
D. None of these

Answer

To determine the value of \( n \) for which the function \( f(x) = \frac{a^x - 1}{x^n (a^x + 1)} \) is even, we need to follow these steps:

### Step 1: Understand the condition for an even function
A function \( f(x) \) is considered even if \( f(x) = f(-x) \) for all \( x \).

### Step 2: Write down \( f(-x) \)
We start by calculating \( f(-x) \):
\[ f(-x) = \frac{a^{-x} - 1}{(-x)^n (a^{-x} + 1)} \]

### Step 3: Simplify \( f(-x) \)
Using the property \( a^{-x} = \frac{1}{a^x} \), we can rewrite \( f(-x) \):
\[ f(-x) = \frac{\frac{1}{a^x} - 1}{(-x)^n \left(\frac{1}{a^x} + 1\right)} \]
This can be further simplified:
\[ = \frac{\frac{1 - a^x}{a^x}}{(-x)^n \left(\frac{1 + a^x}{a^x}\right)} = \frac{1 - a^x}{(-x)^n (1 + a^x)} \]

### Step 4: Set \( f(x) \) equal to \( f(-x) \)
Now, we equate \( f(x) \) and \( f(-x) \):
\[ \frac{a^x - 1}{x^n (a^x + 1)} = \frac{1 - a^x}{(-x)^n (1 + a^x)} \]

### Step 5: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ (a^x - 1)(-x)^n (1 + a^x) = (1 - a^x)x^n (a^x + 1) \]

### Step 6: Expand both sides
Expanding both sides:
Left side:
\[ -a^x x^n - a^x + x^n + 1 \]
Right side:
\[ x^n - a^x x^n + x^n a^x - a^x \]

### Step 7: Combine like terms
Rearranging gives:
\[ -a^x x^n - a^x + x^n + 1 = x^n - a^x x^n + x^n a^x - a^x \]
This simplifies to:
\[ -a^x x^n + a^x x^n + 1 = 0 \]

### Step 8: Analyze the equation
The equation holds true if \( n \) is odd. This is because the negative sign from \( (-x)^n \) will ensure that the terms involving \( x^n \) will cancel out correctly.

### Conclusion
Thus, for \( f(x) \) to be an even function, \( n \) must be an odd number.