`f(x) =(x-1)^(1/(2-x))` is not defined at x = 2. If f(x) is continuous, then f(2) is equal to:

Options

A. e
B. `e^(-1)`
C. `e^(-2)`
D. 1

Answer

To solve the problem, we need to find the value of \( f(2) \) such that the function \( f(x) = (x - 1)^{\frac{1}{2 - x}} \) is continuous at \( x = 2 \).

### Step-by-Step Solution

1. **Understanding Continuity**:
For \( f(x) \) to be continuous at \( x = 2 \), we need:
\[ f(2) = \lim_{x \to 2} f(x) \]

2. **Finding the Limit**:
We need to evaluate:
\[ \lim_{x \to 2} (x - 1)^{\frac{1}{2 - x}} \]
As \( x \) approaches 2, \( (x - 1) \) approaches \( 1 \) and \( (2 - x) \) approaches \( 0 \). This results in the expression taking the form \( 1^{\infty} \), which is indeterminate.

3. **Using the Exponential Form**:
We can rewrite the expression using the exponential function:
\[ (x - 1)^{\frac{1}{2 - x}} = e^{\frac{\ln(x - 1)}{2 - x}} \]
Therefore, we need to evaluate:
\[ \lim_{x \to 2} e^{\frac{\ln(x - 1)}{2 - x}} \]

4. **Finding the Limit of the Exponent**:
We focus on:
\[ \lim_{x \to 2} \frac{\ln(x - 1)}{2 - x} \]
As \( x \to 2 \), \( \ln(x - 1) \) approaches \( \ln(1) = 0 \) and \( (2 - x) \) approaches \( 0 \), leading to the indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule.

5. **Applying L'Hôpital's Rule**:
Differentiate the numerator and the denominator:
- Derivative of \( \ln(x - 1) \) is \( \frac{1}{x - 1} \)
- Derivative of \( 2 - x \) is \( -1 \)

Thus, we have:
\[ \lim_{x \to 2} \frac{\ln(x - 1)}{2 - x} = \lim_{x \to 2} \frac{\frac{1}{x - 1}}{-1} = -\lim_{x \to 2} \frac{1}{x - 1} \]
As \( x \to 2 \), \( x - 1 \to 1 \), so:
\[ -\frac{1}{1} = -1 \]

6. **Final Limit Calculation**:
Now substituting back, we get:
\[ \lim_{x \to 2} e^{\frac{\ln(x - 1)}{2 - x}} = e^{-1} \]

7. **Conclusion**:
Therefore, since \( f(2) = \lim_{x \to 2} f(x) \):
\[ f(2) = e^{-1} \]

### Final Answer:
\[ f(2) = \frac{1}{e} \]