Find the values of a in the domain of the definition of the function, `f(a)=sqrt(2a^(2)-a)` for which the roots of the equation `x^(2)+(a+1)x+(a-1)=0` lie between `-2` & 1.

Explanation

To solve the problem, we need to find the values of \( a \) in the domain of the function \( f(a) = \sqrt{2a^2 - a} \) such that the roots of the equation \( x^2 + (a+1)x + (a-1) = 0 \) lie between -2 and 1.

### Step 1: Determine the domain of \( f(a) \)

The function \( f(a) \) is defined when the expression under the square root is non-negative:

\[ 2a^2 - a \geq 0 \]

Factoring gives:

\[ a(2a - 1) \geq 0 \]

### Step 2: Find the critical points

The critical points occur when \( a = 0 \) or \( 2a - 1 = 0 \) (i.e., \( a = \frac{1}{2} \)). We will analyze the sign of the expression in the intervals determined by these points:

- \( (-\infty, 0) \)
- \( [0, \frac{1}{2}] \)
- \( (\frac{1}{2}, \infty) \)

### Step 3: Test the intervals

1. For \( a < 0 \) (e.g., \( a = -1 \)):
\[ 2(-1)^2 - (-1) = 2 + 1 = 3 \quad (\text{positive}) \]

2. For \( 0 < a < \frac{1}{2} \) (e.g., \( a = \frac{1}{4} \)):
\[ 2\left(\frac{1}{4}\right)^2 - \frac{1}{4} = \frac{2}{16} - \frac{4}{16} = -\frac{2}{16} \quad (\text{negative}) \]

3. For \( a > \frac{1}{2} \) (e.g., \( a = 1 \)):
\[ 2(1)^2 - 1 = 2 - 1 = 1 \quad (\text{positive}) \]

### Step 4: Conclusion for the domain

The function \( f(a) \) is defined for:

\[ (-\infty, 0] \cup \left[\frac{1}{2}, \infty\right) \]

### Step 5: Analyze the roots of the quadratic equation

The roots of the quadratic equation \( x^2 + (a + 1)x + (a - 1) = 0 \) can be found using the quadratic formula:

\[ x = \frac{-(a + 1) \pm \sqrt{(a + 1)^2 - 4(a - 1)}}{2} \]

### Step 6: Calculate the discriminant

The discriminant \( D \) must be non-negative for the roots to be real:

\[ D = (a + 1)^2 - 4(a - 1) \geq 0 \]

Expanding this gives:

\[ a^2 + 2a + 1 - 4a + 4 \geq 0 \implies a^2 - 2a + 5 \geq 0 \]

This quadratic has no real roots (discriminant \( < 0 \)), so it is always positive.

### Step 7: Roots must lie between -2 and 1

We need to ensure that both roots lie between -2 and 1. For this, we will check the following conditions:

1. The vertex of the parabola (given by \( x = -\frac{b}{2a} = -\frac{a + 1}{2} \)) must lie between -2 and 1.
2. The function values at -2 and 1 must be positive.

### Step 8: Set up inequalities

1. The vertex condition:
\[ -2 < -\frac{a + 1}{2} < 1 \]

This leads to two inequalities:
- From \( -2 < -\frac{a + 1}{2} \):
\[ 4 < -a - 1 \implies a < -5 \]
- From \( -\frac{a + 1}{2} < 1 \):
\[ -a - 1 < 2 \implies -a < 3 \implies a > -3 \]

2. The function values:
- For \( x = -2 \):
\[ f(-2) = 4 - 2a + a - 1 > 0 \implies 3 - a > 0 \implies a < 3 \]
- For \( x = 1 \):
\[ f(1) = 1 + a + 1 + a - 1 > 0 \implies 1 + 2a > 0 \implies a > -\frac{1}{2} \]

### Step 9: Combine the conditions

Combining all conditions:
- From the domain: \( (-\infty, 0] \cup [\frac{1}{2}, \infty) \)
- From the inequalities: \( -3 < a < 3 \) and \( a > -\frac{1}{2} \)

### Final Step: Intersection of intervals

The final values of \( a \) that satisfy all conditions are:

\[ [-\frac{1}{2}, 0] \cup [\frac{1}{2}, 3) \]