Consider `alpha gt 1` and `f:[1/alpha,alpha] rarr [1/alpha,alpha]` be bijective function. Suppose that `f^(-1)(x)=1/f(x), " for all " in [1/alpha,alpha]`.
Then f(1) is equal to

Options

A. 1
B. 0
C. -1
D. does'nt attain a unique value

Answer

To solve the problem, we need to find the value of \( f(1) \) given the properties of the function \( f \) and its inverse. Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Understand the properties of the function**: We are given that \( f: [\frac{1}{\alpha}, \alpha] \to [\frac{1}{\alpha}, \alpha] \) is a bijective function. This means that \( f \) is both one-to-one (injective) and onto (surjective).

2. **Use the inverse function property**: We know that \( f^{-1}(x) = \frac{1}{f(x)} \) for all \( x \) in the interval \( [\frac{1}{\alpha}, \alpha] \).

3. **Relate \( f \) and \( f^{-1} \)**: Since \( f^{-1}(x) = \frac{1}{f(x)} \), we can express this in terms of \( y \):
\[ y = f(x) \implies f^{-1}(y) = \frac{1}{y} \]
This means that if \( y = f(x) \), then \( x = f^{-1}(y) = \frac{1}{y} \).

4. **Substituting \( x = 1 \)**: To find \( f(1) \), we can set \( x = 1 \) in the relationship \( f^{-1}(x) = \frac{1}{f(x)} \):
\[ f^{-1}(1) = \frac{1}{f(1)} \]

5. **Finding \( f^{-1}(1) \)**: Since \( f \) is bijective, there exists some \( c \) in the interval \( [\frac{1}{\alpha}, \alpha] \) such that \( f(c) = 1 \). Therefore, we have:
\[ f^{-1}(1) = c \]

6. **Equating the expressions**: From our earlier step, we have:
\[ c = \frac{1}{f(1)} \]
Since \( f(c) = 1 \), we can substitute \( c \) back into the equation:
\[ f(c) = 1 \implies c = f^{-1}(1) = \frac{1}{f(1)} \]

7. **Finding \( f(1) \)**: Now, we can express \( f(1) \) in terms of \( c \):
\[ c = f^{-1}(1) = \frac{1}{f(1)} \implies f(1) = \frac{1}{c} \]
Since \( f(c) = 1 \), we can also say \( f(1) = 1 \).

### Conclusion:
Thus, the value of \( f(1) \) is:
\[ \boxed{1} \]