Consider the cubic equation in `x , x ^(3) - x^(2) + (x- x ^(2)) sin theta + (x -x ^(2)) cos theta + (x-1) sin theta cos theta =0` whose roots are `alpha, beta , gamma. `
The value of `((alpha)/(2))^(2) + ((beta)/(2 )) ^(2) + ((gamma)/(2 ))^(2) =`

Options

A. 1
B. `1/2`
C. `2 cos theta`
D. `1/2(sin theta + cos theta + sin theta cos theta)`

Answer

To solve the cubic equation given by:

\[ x^3 - x^2 + (x - x^2) \sin \theta + (x - x^2) \cos \theta + (x - 1) \sin \theta \cos \theta = 0 \]

with roots \(\alpha\), \(\beta\), and \(\gamma\), we need to find the value of:

\[ \left(\frac{\alpha}{2}\right)^2 + \left(\frac{\beta}{2}\right)^2 + \left(\frac{\gamma}{2}\right)^2 \]

### Step 1: Rearranging the Equation

First, we rearrange the cubic equation by grouping the terms:

\[ x^3 - (1 + \sin \theta + \cos \theta)x^2 + (1 + \sin \theta + \cos \theta)x - \sin \theta \cos \theta = 0 \]

### Step 2: Identifying Coefficients

From the standard form of a cubic equation \(ax^3 + bx^2 + cx + d = 0\), we identify:

- \(a = 1\)
- \(b = -(1 + \sin \theta + \cos \theta)\)
- \(c = 1 + \sin \theta + \cos \theta\)
- \(d = -\sin \theta \cos \theta\)

### Step 3: Using Vieta's Formulas

According to Vieta's formulas for a cubic equation:

1. The sum of the roots \(\alpha + \beta + \gamma = -\frac{b}{a} = 1 + \sin \theta + \cos \theta\)
2. The sum of the product of the roots taken two at a time \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = 1 + \sin \theta + \cos \theta\)
3. The product of the roots \(\alpha\beta\gamma = -\frac{d}{a} = \sin \theta \cos \theta\)

### Step 4: Finding \(\left(\frac{\alpha}{2}\right)^2 + \left(\frac{\beta}{2}\right)^2 + \left(\frac{\gamma}{2}\right)^2\)

We can express this as:

\[ \left(\frac{\alpha}{2}\right)^2 + \left(\frac{\beta}{2}\right)^2 + \left(\frac{\gamma}{2}\right)^2 = \frac{1}{4}(\alpha^2 + \beta^2 + \gamma^2) \]

Using the identity:

\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \]

Substituting the values from Vieta's formulas:

\[ \alpha^2 + \beta^2 + \gamma^2 = (1 + \sin \theta + \cos \theta)^2 - 2(1 + \sin \theta + \cos \theta) \]

### Step 5: Simplifying the Expression

Calculating \((1 + \sin \theta + \cos \theta)^2\):

\[ = 1 + 2(\sin \theta + \cos \theta) + (\sin^2 \theta + \cos^2 \theta) \]
Since \(\sin^2 \theta + \cos^2 \theta = 1\):

\[ = 1 + 2(\sin \theta + \cos \theta) + 1 = 2 + 2(\sin \theta + \cos \theta) \]

Now substituting back:

\[ \alpha^2 + \beta^2 + \gamma^2 = (2 + 2(\sin \theta + \cos \theta)) - 2(1 + \sin \theta + \cos \theta) \]

This simplifies to:

\[ = 2 + 2(\sin \theta + \cos \theta) - 2 - 2(\sin \theta + \cos \theta) = 0 \]

### Step 6: Final Calculation

Thus,

\[ \left(\frac{\alpha}{2}\right)^2 + \left(\frac{\beta}{2}\right)^2 + \left(\frac{\gamma}{2}\right)^2 = \frac{1}{4} \cdot 0 = 0 \]

### Conclusion

The value of \(\left(\frac{\alpha}{2}\right)^2 + \left(\frac{\beta}{2}\right)^2 + \left(\frac{\gamma}{2}\right)^2\) is:

\[ \boxed{0} \]