The value of positive real parameter 'a' such that area of region blunded by parabolas `y=x -ax ^(2), ay = x ^(2)` attains its maximum value is equal to :

Options

A. `1/2`
B. `2`
C. `1/3`
D. `1`

Answer

To find the value of the positive real parameter 'a' such that the area of the region bounded by the parabolas \( y = x - ax^2 \) and \( ay = x^2 \) attains its maximum value, we can follow these steps:

### Step 1: Find the equations of the parabolas
We have two parabolas:
1. \( y = x - ax^2 \)
2. \( ay = x^2 \) which can be rewritten as \( y = \frac{x^2}{a} \)

### Step 2: Find the points of intersection
To find the points of intersection, we set the two equations equal to each other:
\[ x - ax^2 = \frac{x^2}{a} \]
Multiplying through by \( a \) to eliminate the fraction gives:
\[ a(x - ax^2) = x^2 \]
This simplifies to:
\[ ax - a^2x^2 = x^2 \]
Rearranging this, we get:
\[ -a^2x^2 + ax - x^2 = 0 \implies (a^2 + 1)x^2 - ax = 0 \]
Factoring out \( x \):
\[ x((a^2 + 1)x - a) = 0 \]
Thus, the solutions are:
\[ x = 0 \quad \text{or} \quad x = \frac{a}{1 + a^2} \]

### Step 3: Set up the area integral
The area \( A \) between the curves from \( x = 0 \) to \( x = \frac{a}{1 + a^2} \) is given by:
\[ A = \int_0^{\frac{a}{1 + a^2}} \left( (x - ax^2) - \frac{x^2}{a} \right) dx \]
This simplifies to:
\[ A = \int_0^{\frac{a}{1 + a^2}} \left( x - ax^2 - \frac{x^2}{a} \right) dx = \int_0^{\frac{a}{1 + a^2}} \left( x - \left(a + \frac{1}{a}\right)x^2 \right) dx \]

### Step 4: Calculate the integral
Calculating the integral:
\[ A = \int_0^{\frac{a}{1 + a^2}} \left( x - \left(a + \frac{1}{a}\right)x^2 \right) dx \]
This gives:
\[ A = \left[ \frac{x^2}{2} - \frac{\left(a + \frac{1}{a}\right)x^3}{3} \right]_0^{\frac{a}{1 + a^2}} \]
Substituting the limits:
\[ A = \frac{1}{2} \left(\frac{a}{1 + a^2}\right)^2 - \frac{(a + \frac{1}{a})}{3} \left(\frac{a}{1 + a^2}\right)^3 \]

### Step 5: Simplify the area expression
Let \( t = \frac{a}{1 + a^2} \):
\[ A = \frac{1}{2} t^2 - \frac{(a + \frac{1}{a})}{3} t^3 \]

### Step 6: Differentiate the area with respect to 'a'
To find the maximum area, we differentiate \( A \) with respect to \( a \) and set it to zero. After differentiating and simplifying, we find:
\[ 1 - a^2 = 0 \implies a^2 = 1 \implies a = 1 \quad (\text{since } a \text{ is positive}) \]

### Conclusion
Thus, the value of the positive real parameter 'a' such that the area attains its maximum value is:
\[ \boxed{1} \]