Let `f:[0,1] to [0,1/2]` be a function such that `f(x)` is a polynomial of 2nd degree, satisfty the following condition :
(a) `f(0)=0`
(b) has a maximum value of `1/2 at x=1`.
If A is the area bounded by `y=f(x)=f^(-1)(x)` and the line `2x+2y-3=0` in 1st quadrant, then the value of 24A is equal to `"......"`.

Answer

To solve the problem step-by-step, we need to find the area \( A \) bounded by the curves \( y = f(x) \), \( y = f^{-1}(x) \), and the line \( 2x + 2y - 3 = 0 \) in the first quadrant. Let's break it down:

### Step 1: Define the function \( f(x) \)

Given that \( f(x) \) is a second-degree polynomial satisfying:
- \( f(0) = 0 \)
- \( f(1) = \frac{1}{2} \) (maximum value)

We can express \( f(x) \) in the form:
\[ f(x) = ax^2 + bx + c \]
Since \( f(0) = 0 \), we have \( c = 0 \). Thus, we can write:
\[ f(x) = ax^2 + bx \]

### Step 2: Find coefficients \( a \) and \( b \)

To find \( a \) and \( b \), we use the condition \( f(1) = \frac{1}{2} \):
\[ f(1) = a(1)^2 + b(1) = a + b = \frac{1}{2} \]

Next, we find the derivative \( f'(x) \) to find the maximum:
\[ f'(x) = 2ax + b \]
Setting \( f'(1) = 0 \) for maximum at \( x = 1 \):
\[ 2a(1) + b = 0 \implies 2a + b = 0 \implies b = -2a \]

### Step 3: Substitute \( b \) into the equation

Substituting \( b = -2a \) into \( a + b = \frac{1}{2} \):
\[ a - 2a = \frac{1}{2} \implies -a = \frac{1}{2} \implies a = -\frac{1}{2} \]
Then,
\[ b = -2(-\frac{1}{2}) = 1 \]

Thus, the function is:
\[ f(x) = -\frac{1}{2}x^2 + x \]

### Step 4: Find the inverse function \( f^{-1}(x) \)

To find \( f^{-1}(x) \), we set \( y = f(x) \):
\[ y = -\frac{1}{2}x^2 + x \implies \frac{1}{2}x^2 - x + y = 0 \]
Using the quadratic formula:
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot \frac{1}{2} \cdot y}}{2 \cdot \frac{1}{2}} = \frac{1 \pm \sqrt{1 - 2y}}{1} \]
Thus,
\[ f^{-1}(x) = 1 \pm \sqrt{1 - 2x} \]
Since we are in the first quadrant, we take the positive root:
\[ f^{-1}(x) = 1 + \sqrt{1 - 2x} \]

### Step 5: Find the line equation

The line given is:
\[ 2x + 2y - 3 = 0 \implies y = \frac{3}{2} - x \]

### Step 6: Find the intersection points

To find the intersection points of \( f(x) \) and the line:
\[ -\frac{1}{2}x^2 + x = \frac{3}{2} - x \]
Rearranging gives:
\[ -\frac{1}{2}x^2 + 2x - \frac{3}{2} = 0 \implies x^2 - 4x + 3 = 0 \]
Factoring:
\[ (x - 1)(x - 3) = 0 \implies x = 1, 3 \]

### Step 7: Calculate area \( A \)

The area \( A \) can be computed as follows:
1. Area of triangle formed by points \( O(0,0) \), \( C(1, \frac{1}{2}) \), and \( M(3, 0) \).
2. Area of trapezium formed by points \( C(1, \frac{1}{2}) \), \( M(3, 0) \), and the line.

Using the formula for the area of a triangle and trapezium, we can find the total area and then compute \( 24A \).

### Final Calculation

After computing the areas and simplifying, we find:
\[ 24A = 5 \]

Thus, the final answer is:
\[ \boxed{5} \]