The function `f (x)={((x ^(2n)))/((x ^(2n) sgn x)^(2n+1))((e ^(1/x)-e ^(-1/x))/(e ^(1/x)+e ^(-(1)/(x))))x ne0 n in N` is:

Options

A. Odd function
B. Even function
C. Neither odd nor even function
D. Constant function

Answer

To determine whether the function

\[ f(x) = \frac{x^{2n}}{(x^{2n} \cdot \text{sgn}(x))^{2n+1}} \cdot \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \]

is an even or odd function, we will follow these steps:

### Step 1: Understand the Definitions
- A function \( f(x) \) is **even** if \( f(-x) = f(x) \).
- A function \( f(x) \) is **odd** if \( f(-x) = -f(x) \).

### Step 2: Find \( f(-x) \)
Substituting \(-x\) into the function:

\[ f(-x) = \frac{(-x)^{2n}}{((-x)^{2n} \cdot \text{sgn}(-x))^{2n+1}} \cdot \frac{e^{1/(-x)} - e^{-1/(-x)}}{e^{1/(-x)} + e^{-1/(-x)}} \]

### Step 3: Simplify \( f(-x) \)
1. Since \( (-x)^{2n} = x^{2n} \) (even power), we have:
\[ f(-x) = \frac{x^{2n}}{((-x)^{2n} \cdot (-1))^{2n+1}} \cdot \frac{e^{-1/x} - e^{1/x}}{e^{-1/x} + e^{1/x}} \]

2. The signum function \( \text{sgn}(-x) = -1 \) when \( x < 0 \), thus:
\[ f(-x) = \frac{x^{2n}}{(x^{2n} \cdot (-1))^{2n+1}} \cdot \frac{e^{-1/x} - e^{1/x}}{e^{-1/x} + e^{1/x}} \]

3. The denominator becomes:
\[ (-x^{2n})^{2n+1} = -x^{2n(2n+1)} \quad \text{(odd power)} \]

4. Therefore:
\[ f(-x) = \frac{x^{2n}}{-x^{2n(2n+1)}} \cdot \frac{e^{-1/x} - e^{1/x}}{e^{-1/x} + e^{1/x}} = -\frac{x^{2n}}{x^{2n(2n+1)}} \cdot \frac{e^{-1/x} - e^{1/x}}{e^{-1/x} + e^{1/x}} \]

### Step 4: Further Simplification
Now, simplify \( \frac{e^{-1/x} - e^{1/x}}{e^{-1/x} + e^{1/x}} \):
- This can be rewritten as:
\[ -\frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \]

### Step 5: Combine Results
Thus, we have:
\[ f(-x) = -\frac{x^{2n}}{x^{2n(2n+1)}} \cdot \left(-\frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}}\right) \]

### Step 6: Conclusion
Since \( f(-x) = -f(x) \), we conclude that the function \( f(x) \) is an **odd function**.