If the range of the values of a for which the roots of the equation `x ^(2) -2x - a ^(2) +1=0` lie between the roots of the equation `x ^(2) -2 (a+1)x +a(a -1) =0` is (p,q), then find the value of `(q- (1)/(p)).`

Answer

To solve the problem, we need to analyze two quadratic equations and find the range of values for \( a \) such that the roots of the first equation lie between the roots of the second equation.

### Step 1: Identify the equations
The first equation is:
\[ x^2 - 2x - a^2 + 1 = 0 \]
The second equation is:
\[ x^2 - 2(a+1)x + a(a-1) = 0 \]

### Step 2: Find the roots of the first equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = -2 \), and \( c = -a^2 + 1 \).
- The roots \( x_1 \) and \( x_2 \) are given by:
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-a^2 + 1)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ = \frac{2 \pm \sqrt{4 + 4a^2 - 4}}{2} = \frac{2 \pm \sqrt{4a^2}}{2} = 1 \pm a \]
Thus, the roots are:
\[ x_1 = 1 - a, \quad x_2 = 1 + a \]

### Step 3: Find the roots of the second equation
Using the quadratic formula again:
- Here, \( a = 1 \), \( b = -2(a+1) \), and \( c = a(a-1) \).
- The roots \( y_1 \) and \( y_2 \) are given by:
\[ y = \frac{2(a+1) \pm \sqrt{(-2(a+1))^2 - 4 \cdot 1 \cdot a(a-1)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ = \frac{2(a+1) \pm \sqrt{4(a+1)^2 - 4a(a-1)}}{2} \]
Simplifying the discriminant:
\[ = \frac{2(a+1) \pm \sqrt{4(a^2 + 2a + 1 - a^2 + a)}}{2} = \frac{2(a+1) \pm \sqrt{4(3a + 1)}}{2} \]
Thus, the roots are:
\[ y_1 = a + 1 - \sqrt{3a + 1}, \quad y_2 = a + 1 + \sqrt{3a + 1} \]

### Step 4: Set the conditions for the roots
We need the roots \( x_1 \) and \( x_2 \) to lie between \( y_1 \) and \( y_2 \):
\[ y_1 < x_1 < x_2 < y_2 \]
This gives us two inequalities:
1. \( y_1 < 1 - a \)
2. \( 1 + a < y_2 \)

### Step 5: Solve the inequalities
**For the first inequality:**
\[ a + 1 - \sqrt{3a + 1} < 1 - a \]
Rearranging gives:
\[ 2a < \sqrt{3a + 1} \]
Squaring both sides:
\[ 4a^2 < 3a + 1 \]
Rearranging gives:
\[ 4a^2 - 3a - 1 < 0 \]
Factoring or using the quadratic formula:
\[ a = \frac{3 \pm \sqrt{9 + 16}}{8} = \frac{3 \pm 5}{8} \]
Thus, the roots are:
\[ a = 1 \quad \text{and} \quad a = -\frac{1}{4} \]
The inequality holds for:
\[ -\frac{1}{4} < a < 1 \]

**For the second inequality:**
\[ 1 + a < a + 1 + \sqrt{3a + 1} \]
This simplifies to:
\[ 0 < \sqrt{3a + 1} \]
This is always true for \( a > -\frac{1}{3} \).

### Step 6: Find the intersection of the ranges
From the first inequality, we have \( -\frac{1}{4} < a < 1 \).
From the second inequality, we have \( a > -\frac{1}{3} \).
The intersection of these two ranges is:
\[ -\frac{1}{4} < a < 1 \]

### Step 7: Identify \( p \) and \( q \)
Here, \( p = -\frac{1}{4} \) and \( q = 1 \).

### Step 8: Calculate \( q - \frac{1}{p} \)
\[ q - \frac{1}{p} = 1 - \frac{1}{-\frac{1}{4}} = 1 + 4 = 5 \]

### Final Answer
The value of \( q - \frac{1}{p} \) is:
\[ \boxed{5} \]