If point `P(alpha, alpha^2-2)` lies inside the triangle formed by the lines `x + y=1 , y=x + 1 and y=-1` then `alpha in`

Options

A. `(-sqrt3 , sqrt3)`
B. ` ((1 - sqrt(13))/(2) , -1) cup (1 , (-1 + sqrt(13))/(2))`
C. [-1,1]
D. `((1- sqrt(13))/(2) , (-1 + sqrt(13))/(2))`

Explanation

To determine the values of \( \alpha \) such that the point \( P(\alpha, \alpha^2 - 2) \) lies inside the triangle formed by the lines \( x + y = 1 \), \( y = x + 1 \), and \( y = -1 \), we will follow these steps:

### Step 1: Identify the vertices of the triangle
First, we need to find the points of intersection of the given lines to determine the vertices of the triangle.

1. **Intersection of \( x + y = 1 \) and \( y = -1 \)**:
\[ x - 1 = 1 \implies x = 2 \implies (2, -1) \]

2. **Intersection of \( y = x + 1 \) and \( y = -1 \)**:
\[ -1 = x + 1 \implies x = -2 \implies (-2, -1) \]

3. **Intersection of \( x + y = 1 \) and \( y = x + 1 \)**:
\[ x + (x + 1) = 1 \implies 2x + 1 = 1 \implies 2x = 0 \implies x = 0 \implies (0, 1) \]

Thus, the vertices of the triangle are \( A(0, 1) \), \( B(2, -1) \), and \( C(-2, -1) \).

### Step 2: Determine the conditions for point \( P \) to be inside the triangle
For point \( P(\alpha, \alpha^2 - 2) \) to lie inside the triangle, it must satisfy the following inequalities based on the sides of the triangle.

1. **Condition with line \( y = -1 \)**:
The point \( P \) must be above the line \( y = -1 \):
\[ \alpha^2 - 2 > -1 \implies \alpha^2 - 1 > 0 \implies (\alpha - 1)(\alpha + 1) > 0 \]
This inequality holds when:
\[ \alpha < -1 \quad \text{or} \quad \alpha > 1 \]

2. **Condition with line \( x + y = 1 \)**:
The point \( P \) must be below the line \( x + y = 1 \):
\[ \alpha + (\alpha^2 - 2) < 1 \implies \alpha^2 + \alpha - 3 < 0 \]
Factoring gives:
\[ (\alpha - 1)(\alpha + 3) < 0 \]
This inequality holds when:
\[ -3 < \alpha < 1 \]

3. **Condition with line \( y = x + 1 \)**:
The point \( P \) must be below the line \( y = x + 1 \):
\[ \alpha^2 - 2 < \alpha + 1 \implies \alpha^2 - \alpha - 3 < 0 \]
Factoring gives:
\[ (\alpha - 3)(\alpha + 1) < 0 \]
This inequality holds when:
\[ -1 < \alpha < 3 \]

### Step 3: Combine the conditions
Now we combine the intervals obtained from the three conditions:

1. From \( y = -1 \): \( \alpha < -1 \) or \( \alpha > 1 \)
2. From \( x + y = 1 \): \( -3 < \alpha < 1 \)
3. From \( y = x + 1 \): \( -1 < \alpha < 3 \)

The overlapping intervals are:
- For \( \alpha < -1 \): No overlap with the second condition.
- For \( \alpha > 1 \): The overlap is \( 1 < \alpha < 3 \).

### Final Result
Thus, the values of \( \alpha \) for which the point \( P(\alpha, \alpha^2 - 2) \) lies inside the triangle are:
\[ \alpha \in (1, 3) \]