If `f(x)={((alpha-1)x,xepsilonQ^(c)),(-alpha^(2)x+alpha+3x-1,xepsilonQ):}` and `g(x)={(x,xepsilonQ^(c)),(1-x,xepsilonQ):}` are inverse to each other then find all possible values of `alpha`.

Explanation

To solve the problem, we need to find the values of \( \alpha \) such that the functions \( f(x) \) and \( g(x) \) are inverses of each other.

Given:
\[ f(x) = \begin{cases} (\alpha - 1)x & \text{if } x \in Q^c \\ -\alpha^2 x + \alpha + 3x - 1 & \text{if } x \in Q \end{cases} \]
\[ g(x) = \begin{cases} x & \text{if } x \in Q^c \\ 1 - x & \text{if } x \in Q \end{cases} \]

### Step 1: Find the inverse of \( g(x) \)

Since \( g(x) \) is defined piecewise, we will find its inverse piecewise as well.

1. For \( x \in Q^c \):
\[ g(x) = x \implies g^{-1}(x) = x \]

2. For \( x \in Q \):
\[ g(x) = 1 - x \implies y = 1 - x \implies x = 1 - y \implies g^{-1}(x) = 1 - x \]

Thus, we have:
\[ g^{-1}(x) = \begin{cases} x & \text{if } x \in Q^c \\ 1 - x & \text{if } x \in Q \end{cases} \]

### Step 2: Set \( f(x) \) equal to \( g^{-1}(x) \)

Since \( f(x) \) and \( g(x) \) are inverses, we have:
1. For \( x \in Q^c \):
\[ f(x) = g^{-1}(x) \implies (\alpha - 1)x = x \]
Simplifying this gives:
\[ \alpha - 1 = 1 \implies \alpha = 2 \]

2. For \( x \in Q \):
\[ f(x) = g^{-1}(x) \implies -\alpha^2 x + \alpha + 3x - 1 = 1 - x \]
Rearranging this gives:
\[ -\alpha^2 x + \alpha + 3x - 1 = 1 - x \]
\[ -\alpha^2 x + 3x + x = 1 - \alpha + 1 \]
\[ (-\alpha^2 + 4)x = 2 - \alpha \]

### Step 3: Analyze the equation

For the equation to hold for all \( x \in Q \), the coefficients must be equal:
1. Coefficient of \( x \):
\[ -\alpha^2 + 4 = 0 \implies \alpha^2 = 4 \implies \alpha = 2 \text{ or } \alpha = -2 \]

2. Constant term:
\[ 2 - \alpha = 0 \implies \alpha = 2 \]

### Conclusion

The values of \( \alpha \) that satisfy both conditions are:
- From the first part, we found \( \alpha = 2 \).
- From the second part, we found \( \alpha = 2 \) or \( \alpha = -2 \).

However, since \( \alpha = -2 \) does not satisfy the first part, the only valid solution is:
\[ \alpha = 2 \]

### Final Answer
The only possible value of \( \alpha \) is \( 2 \).