If the graph of the function `f(x)=(a^(x)-1)/(x^(n)(a^(x)+1))` is symmetric about y - axis, then n is equal to :

Options

A. `a. -(1)/(3)`
B. b. `(1)/(4)`
C. c. `2//3`
D. d. 2

Answer

To determine the value of \( n \) such that the function

\[ f(x) = \frac{a^x - 1}{x^n (a^x + 1)} \]

is symmetric about the y-axis, we need to check if the function is even. A function is even if \( f(x) = f(-x) \).

### Step-by-Step Solution:

1. **Find \( f(-x) \)**:
We start by substituting \(-x\) into the function:
\[ f(-x) = \frac{a^{-x} - 1}{(-x)^n (a^{-x} + 1)} \]
Simplifying this, we have:
\[ f(-x) = \frac{\frac{1}{a^x} - 1}{(-x)^n \left(\frac{1}{a^x} + 1\right)} \]
This can be rewritten as:
\[ f(-x) = \frac{1 - a^x}{(-x)^n (1 + a^x)} \]

2. **Simplify \( f(-x) \)**:
To simplify further, multiply the numerator and denominator by \( a^x \):
\[ f(-x) = \frac{a^x(1 - a^x)}{(-x)^n (1 + a^x)a^x} = \frac{a^x - a^{2x}}{(-x)^n (a^x + 1)} \]

3. **Set \( f(x) \) equal to \( f(-x) \)**:
For the function to be even, we need:
\[ \frac{a^x - 1}{x^n (a^x + 1)} = \frac{a^x - a^{2x}}{(-x)^n (a^x + 1)} \]
Since \( a^x + 1 \) is common in both denominators, we can cancel it (assuming \( a^x + 1 \neq 0 \)):
\[ a^x - 1 = \frac{(a^x - a^{2x})}{(-x)^n} \cdot x^n \]

4. **Cross-multiply**:
Rearranging gives:
\[ (a^x - 1)(-x)^n = a^x - a^{2x} \]

5. **Analyze the equation**:
For the equation to hold for all \( x \), the powers of \( x \) must match. The left side has a term of \( (-x)^n \) which indicates that \( n \) must be odd for the left side to be equal to the right side when \( x \) is negative.

6. **Conclusion**:
Since \( (-1)^n = -1 \) when \( n \) is odd, we conclude that \( n \) must be an odd integer for the function to be symmetric about the y-axis.

### Result:
Thus, the value of \( n \) is equal to **1**.