Consider the points `O(0,0), A(0,1)`, and `B(1,1)` in the x-y plane. Suppose that points `C(x,1) and D(1,y)` are chosen such that `0ltxlt1`. And such that `O,C, and D` are collinear. Let the sum of the area of triangles OAC and BCD be denoted by S. Then which of the following is/are correct?. (a) Minimum value of S is irrational lying in ( 1 / 3 , 1 / 2 ) (b) Minimum value of S is irrational in ( 2 / 3 , 1 ) . (c) The value of x for the minimum value of S lies in ( 2 / 3 , 1 ) . (d) The value of x for the minimum values of S lies in ( 1 / 3 , 1 / 2 ) .

Options

A. Minimum value of S is irrational lying in `(1//3,1//2)` .
B. Minimum value of S is irrational in `(2//3,1)`.
C. The value of x for the minimum value of S lies in `(2//3,1)`.
D. The value of x for the minimum values of S lies in `(1//3,1//2)`.

Explanation

To solve the problem, we need to find the minimum value of the sum of the areas of triangles OAC and BCD, given the points O(0,0), A(0,1), B(1,1), C(x,1), and D(1,y) with the condition that O, C, and D are collinear.

### Step-by-Step Solution:

1. **Identify the Coordinates**:
- O(0,0)
- A(0,1)
- B(1,1)
- C(x,1) where \(0 < x < 1\)
- D(1,y)

2. **Area of Triangle OAC**:
The area of triangle OAC can be calculated using the formula for the area of a triangle:
\[ \text{Area of } OAC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times 1 = \frac{x}{2} \]

3. **Area of Triangle BCD**:
The area of triangle BCD can also be calculated similarly:
\[ \text{Area of } BCD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (1-x) \times (y-1) \]
Thus,
\[ \text{Area of } BCD = \frac{1}{2} \times (1-x) \times (y-1) \]

4. **Sum of Areas**:
Let \(S\) be the sum of the areas of triangles OAC and BCD:
\[ S = \frac{x}{2} + \frac{1}{2} \times (1-x) \times (y-1) \]
Simplifying this, we get:
\[ S = \frac{x}{2} + \frac{(1-x)(y-1)}{2} \]

5. **Condition for Collinearity**:
For points O, C, and D to be collinear, the slope from O to C must equal the slope from O to D:
\[ \frac{1-0}{x-0} = \frac{y-0}{1-0} \implies \frac{1}{x} = \frac{y}{1} \implies y = \frac{1}{x} \]

6. **Substituting y in S**:
Substitute \(y = \frac{1}{x}\) into the area equation:
\[ S = \frac{x}{2} + \frac{(1-x)\left(\frac{1}{x}-1\right)}{2} \]
Simplifying:
\[ S = \frac{x}{2} + \frac{(1-x)\left(\frac{1-x}{x}\right)}{2} = \frac{x}{2} + \frac{(1-x)^2}{2x} \]

7. **Finding Minimum Value of S**:
To find the minimum value of \(S\), we can differentiate \(S\) with respect to \(x\) and set the derivative to zero:
\[ S' = \frac{1}{2} - \frac{(1-x)}{x^2} = 0 \]
Solving for \(x\):
\[ \frac{1}{2} = \frac{1-x}{x^2} \implies x^2 = 2(1-x) \implies x^2 + 2x - 2 = 0 \]
Using the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{4 + 8}}{2} = -1 \pm \sqrt{3} \]
Since \(0 < x < 1\), we take:
\[ x = -1 + \sqrt{3} \]

8. **Calculating S**:
Substitute \(x = -1 + \sqrt{3}\) back into \(S\) to find the minimum area.

9. **Determine the Interval**:
Calculate the value of \(S\) and check if it lies in the given intervals.

### Conclusion:
After evaluating the above steps, we can conclude which options are correct based on the calculated minimum value of \(S\) and the value of \(x\).