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 the nature of the function \( f(x) \) given by

\[ f(x) = \frac{x^{2n}}{(x^{2n} \cdot \text{sgn}(x))^{2n+1}} \cdot \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \quad (x \neq 0, n \in \mathbb{N}), \]

we will analyze the function step by step.

### Step 1: Understanding the Sign Function
The sign function, \( \text{sgn}(x) \), is defined as follows:
- \( \text{sgn}(x) = 1 \) if \( x > 0 \)
- \( \text{sgn}(x) = 0 \) if \( x = 0 \) (not applicable here since \( x \neq 0 \))
- \( \text{sgn}(x) = -1 \) if \( x < 0 \)

### Step 2: Splitting the Function into Cases
We will consider two cases based on the sign of \( x \).

#### Case 1: \( x > 0 \)
For \( x > 0 \):
- \( \text{sgn}(x) = 1 \)

Thus, the function becomes:

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

This simplifies to:

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

#### Case 2: \( x < 0 \)
For \( x < 0 \):
- \( \text{sgn}(x) = -1 \)

Thus, the function becomes:

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

This simplifies to:

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

### Step 3: Compare \( f(-x) \) and \( -f(x) \)
Now, we will compare \( f(-x) \) and \( -f(x) \):

- For \( x > 0 \), \( f(-x) = -\frac{1}{(-x)^{2n^2 + 2n}} \cdot \frac{e^{-1/x} - e^{1/x}}{e^{-1/x} + e^{1/x}} = -f(x) \)

This shows that \( f(-x) = -f(x) \) for \( x > 0 \).

### Conclusion
Since \( f(-x) = -f(x) \), the function \( f(x) \) is an **odd function**.