If `f(x)=x(x-1)` is a function from `[1/2,oo)to[-1/4,oo),` then `{x in r :f^(-1)(x)=f(x)}` is a. null set b. `{0,2}` c. `{2}` d. a set containing 3 elements

Options

A. null set
B. `{0,2}`
C. `{2}`
D. a set containing 3 elements

Explanation

To solve the problem, we need to find the set of values \( x \) such that \( f^{-1}(x) = f(x) \) for the function \( f(x) = x(x - 1) \) defined on the interval \([1/2, \infty)\) and mapping to \([-1/4, \infty)\).

### Step-by-Step Solution:

1. **Set Up the Equation**:
We start with the equation \( f^{-1}(x) = f(x) \). We can express this as:
\[ f(f(x)) = x \]
where \( f(x) = x(x - 1) \).

2. **Calculate \( f(f(x)) \)**:
First, we need to find \( f(f(x)) \):
\[ f(x) = x(x - 1) \]
Now, substitute \( f(x) \) into itself:
\[ f(f(x)) = f(x(x - 1)) = x(x - 1)((x(x - 1)) - 1) \]
Simplifying the inner part:
\[ (x(x - 1)) - 1 = x^2 - x - 1 \]
Thus,
\[ f(f(x)) = x(x - 1)(x^2 - x - 1) \]

3. **Set the Equation**:
Now we need to solve:
\[ f(f(x)) = x \]
This leads to:
\[ x(x - 1)(x^2 - x - 1) = x \]

4. **Rearranging the Equation**:
We can rearrange this to:
\[ x(x - 1)(x^2 - x - 1) - x = 0 \]
Factoring out \( x \):
\[ x \left( (x - 1)(x^2 - x - 1) - 1 \right) = 0 \]

5. **Finding Roots**:
The solutions to this equation will be \( x = 0 \) or solving:
\[ (x - 1)(x^2 - x - 1) - 1 = 0 \]
Expanding this:
\[ (x - 1)(x^2 - x - 1) = x^3 - 2x^2 + x - 1 \]
Setting this equal to 1 gives:
\[ x^3 - 2x^2 + x - 2 = 0 \]

6. **Finding the Roots of the Cubic Equation**:
We can use the Rational Root Theorem or synthetic division to find the roots. Testing \( x = 2 \):
\[ 2^3 - 2(2^2) + 2 - 2 = 8 - 8 + 2 - 2 = 0 \]
So \( x = 2 \) is a root. We can factor the cubic polynomial as:
\[ (x - 2)(x^2 + ax + b) = 0 \]
Performing polynomial long division, we can find the other roots.

7. **Conclusion**:
After finding the roots, we find that the only valid solution in the interval \([1/2, \infty)\) is \( x = 2 \). Therefore, the set \( \{ x \in \mathbb{R} : f^{-1}(x) = f(x) \} \) contains only one element, which is \( \{ 2 \} \).

### Final Answer:
The answer is \( \{ 2 \} \).