The jump value of the function at the point of the discontinuity of the function `f(x)=(1-k^(1//x))/(1+k^(1//x))(kgt0)` is (a) 4 (b) 2 (c) 3 (d) none of these

Options

A. 4
B. 2
C. 3
D. None of these

Answer

To find the jump value of the function \( f(x) = \frac{1 - k^{1/x}}{1 + k^{1/x}} \) at the point of discontinuity (which occurs as \( x \) approaches 0), we will calculate the right-hand limit and the left-hand limit at this point.

### Step 1: Calculate the Right-Hand Limit
We need to evaluate:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1 - k^{1/x}}{1 + k^{1/x}} \]

As \( x \to 0^+ \), \( \frac{1}{x} \to +\infty \). Therefore, we consider two cases based on the value of \( k \):

1. **If \( k > 1 \)**:
- \( k^{1/x} \to +\infty \)
- Thus, \( f(x) \to \frac{1 - \infty}{1 + \infty} = \frac{-\infty}{+\infty} = -1 \)

2. **If \( 0 < k < 1 \)**:
- \( k^{1/x} \to 0 \)
- Thus, \( f(x) \to \frac{1 - 0}{1 + 0} = 1 \)

### Step 2: Calculate the Left-Hand Limit
Now we evaluate:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1 - k^{1/x}}{1 + k^{1/x}} \]

As \( x \to 0^- \), \( \frac{1}{x} \to -\infty \). Again, we consider two cases based on the value of \( k \):

1. **If \( k > 1 \)**:
- \( k^{1/x} \to 0 \)
- Thus, \( f(x) \to \frac{1 - 0}{1 + 0} = 1 \)

2. **If \( 0 < k < 1 \)**:
- \( k^{1/x} \to +\infty \)
- Thus, \( f(x) \to \frac{1 - \infty}{1 + \infty} = \frac{-\infty}{+\infty} = -1 \)

### Step 3: Calculate the Jump Value
The jump value is given by:
\[ \text{Jump} = | \text{Right-hand limit} - \text{Left-hand limit} | \]

- **If \( k > 1 \)**:
- Right-hand limit = -1
- Left-hand limit = 1
- Jump = \( | -1 - 1 | = | -2 | = 2 \)

- **If \( 0 < k < 1 \)**:
- Right-hand limit = 1
- Left-hand limit = -1
- Jump = \( | 1 - (-1) | = | 1 + 1 | = 2 \)

### Conclusion
In both cases, the jump value is \( 2 \).

Thus, the answer is:
\[ \boxed{2} \]