Let `alpha + beta = 1, 2 alpha^(2) + 2beta^(2) = 1 and f(x)` be a continuous function such that `f(2 + x) + f(x) = 2` for all `x in [0, 2] and p = int_(0)^(4) f(x) dx - 4, q = (alpha)/(beta)`. Then, find the least positive integral value of 'a' for which the equation `ax^(2) - bx + c = 0` has both roots lying between p and q, where `a, b, c in N`.

Answer

To solve the problem step by step, we will start by analyzing the given equations and conditions.

### Step 1: Analyze the given equations
We have:
1. \( \alpha + \beta = 1 \)
2. \( 2\alpha^2 + 2\beta^2 = 1 \)

From the first equation, we can express \( \beta \) in terms of \( \alpha \):
\[ \beta = 1 - \alpha \]

### Step 2: Substitute \(\beta\) in the second equation
Substituting \(\beta\) into the second equation:
\[ 2\alpha^2 + 2(1 - \alpha)^2 = 1 \]
Expanding the equation:
\[ 2\alpha^2 + 2(1 - 2\alpha + \alpha^2) = 1 \]
\[ 2\alpha^2 + 2 - 4\alpha + 2\alpha^2 = 1 \]
Combining like terms:
\[ 4\alpha^2 - 4\alpha + 2 = 1 \]
\[ 4\alpha^2 - 4\alpha + 1 = 0 \]

### Step 3: Solve the quadratic equation for \(\alpha\)
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 4, b = -4, c = 1 \):
\[ \alpha = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} \]
\[ \alpha = \frac{4 \pm \sqrt{16 - 16}}{8} \]
\[ \alpha = \frac{4}{8} = \frac{1}{2} \]

### Step 4: Find \(\beta\)
Since \( \alpha + \beta = 1 \):
\[ \beta = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 5: Calculate \( q \)
Now, we can find \( q = \frac{\alpha}{\beta} \):
\[ q = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \]

### Step 6: Analyze the function \( f(x) \)
We have the condition:
\[ f(2 + x) + f(x) = 2 \quad \text{for all } x \in [0, 2] \]
Let’s check \( x = 0 \):
\[ f(2) + f(0) = 2 \]
Let’s check \( x = 2 \):
\[ f(4) + f(2) = 2 \]

### Step 7: Find \( p \)
We need to compute \( p = \int_0^4 f(x) \, dx - 4 \).
Using the functional equation, we can find \( f(x) \) is constant. Assume \( f(x) = 1 \) (since \( f(x) \) must be continuous and satisfy the equation):
\[ \int_0^4 f(x) \, dx = \int_0^4 1 \, dx = 4 \]
Thus,
\[ p = 4 - 4 = 0 \]

### Step 8: Set up the quadratic equation
We need to find the least positive integral value of \( a \) such that the equation \( ax^2 - bx + c = 0 \) has both roots between \( p \) and \( q \):
\[ p = 0, \quad q = 1 \]
Thus, we need both roots to lie in the interval \( (0, 1) \).

### Step 9: Conditions for the roots
For the roots to lie in \( (0, 1) \):
1. The sum of the roots \( \frac{b}{a} < 2 \) (since both roots are less than 1).
2. The product of the roots \( \frac{c}{a} > 0 \) (since both roots are positive).

### Step 10: Choose values for \( a, b, c \)
Let’s choose \( a = 1 \):
1. Then \( b < 2 \) implies \( b = 1 \).
2. \( c > 0 \) implies \( c = 1 \).

Thus, the quadratic equation becomes:
\[ x^2 - x + 1 = 0 \]
The discriminant \( b^2 - 4ac = 1 - 4 < 0 \) does not yield real roots.

### Step 11: Increase \( a \)
Let’s try \( a = 2 \):
1. Then \( b < 4 \) implies \( b = 3 \).
2. \( c > 0 \) implies \( c = 1 \).

The quadratic equation becomes:
\[ 2x^2 - 3x + 1 = 0 \]
The discriminant:
\[ (-3)^2 - 4 \cdot 2 \cdot 1 = 9 - 8 = 1 \geq 0 \]
The roots are:
\[ x = \frac{3 \pm 1}{4} = 1 \quad \text{and} \quad \frac{2}{4} = \frac{1}{2} \]
Both roots \( \frac{1}{2} \) and \( 1 \) lie in \( (0, 1) \).

### Conclusion
The least positive integral value of \( a \) for which the equation has both roots lying between \( p \) and \( q \) is:
\[ \boxed{2} \]