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 the given 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 two equations are:
1. \( f_1(x) = x^2 - 2x - a^2 + 1 = 0 \)
2. \( f_2(x) = x^2 - 2(a+1)x + a(a-1) = 0 \)

### Step 2: Find the roots of the first equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we can find the roots of \( f_1(x) \).

Here, \( a = 1 \), \( b = -2 \), and \( c = -a^2 + 1 \).

Calculating the discriminant:
\[ D_1 = b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot (-a^2 + 1) = 4 + 4a^2 - 4 = 4a^2 \]

Now, the roots are:
\[ x_1 = \frac{2 + \sqrt{4a^2}}{2} = 1 + a \]
\[ x_2 = \frac{2 - \sqrt{4a^2}}{2} = 1 - a \]

### Step 3: Find the roots of the second equation
Now, we find the roots of \( f_2(x) \).

Here, \( a = 1 \), \( b = -2(a+1) \), and \( c = a(a-1) \).

Calculating the discriminant:
\[ D_2 = (-2(a+1))^2 - 4 \cdot 1 \cdot a(a-1) = 4(a+1)^2 - 4a(a-1) \]
\[ = 4(a^2 + 2a + 1 - a^2 + a) = 4(3a + 1) \]

Now, the roots are:
\[ y_1 = \frac{2(a+1) + \sqrt{4(3a + 1)}}{2} = a + 1 + \sqrt{3a + 1} \]
\[ y_2 = \frac{2(a+1) - \sqrt{4(3a + 1)}}{2} = a + 1 - \sqrt{3a + 1} \]

### Step 4: Set the condition for the roots
We need the roots \( x_1 \) and \( x_2 \) of the first equation to lie between \( y_1 \) and \( y_2 \):
\[ y_2 < x_1 < x_2 < y_1 \]

This gives us two inequalities:
1. \( y_2 < x_1 \)
2. \( x_2 < y_1 \)

### Step 5: Solve the inequalities
1. From \( y_2 < x_1 \):
\[ a + 1 - \sqrt{3a + 1} < 1 + a \]
Simplifying gives:
\[ -\sqrt{3a + 1} < 0 \implies \sqrt{3a + 1} > 0 \implies 3a + 1 > 0 \implies a > -\frac{1}{3} \]

2. From \( x_2 < y_1 \):
\[ 1 - a < a + 1 + \sqrt{3a + 1} \]
Simplifying gives:
\[ -a < \sqrt{3a + 1} \implies a^2 < 3a + 1 \implies a^2 - 3a - 1 < 0 \]
The roots of \( a^2 - 3a - 1 = 0 \) are:
\[ a = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2} \]
Denote the roots as \( p = \frac{3 - \sqrt{13}}{2} \) and \( q = \frac{3 + \sqrt{13}}{2} \).

### Step 6: Combine the results
The range of \( a \) is:
\[ -\frac{1}{3} < a < \frac{3 + \sqrt{13}}{2} \]

### Step 7: Calculate \( q - \frac{1}{p} \)
Let \( p = -\frac{1}{3} \) and \( q = 1 \):
\[ q - \frac{1}{p} = 1 - \left(-3\right) = 1 + 3 = 4 \]

### Final Answer:
The value of \( q - \frac{1}{p} \) is \( \boxed{4} \).