Let `alpha` and `beta` be the roots of the equation `x^2+(2-costheta)x-(1+costheta)=0`, then

Options

A. least value of `alpha^2 + beta^2` is 5
B. `alpha^2+beta^2` is least when `theta=2npi,n in I`
C. least value of `alpha^2+beta^2` is 6
D. `alpha^2+beta^2` is least when `theta=(2n+1)pi/2, n in I`

Answer

To solve the problem, we need to analyze the quadratic equation given and find the least value of \( \alpha^2 + \beta^2 \) where \( \alpha \) and \( \beta \) are the roots of the equation:

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

### Step 1: Identify the coefficients
From the equation, we can identify:
- Coefficient of \( x^2 \) (a) = 1
- Coefficient of \( x \) (b) = \( 2 - \cos \theta \)
- Constant term (c) = \( -(1 + \cos \theta) \)

### Step 2: Use Vieta's formulas
According to Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -(2 - \cos \theta) = \cos \theta - 2 \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = -(-1 - \cos \theta) = 1 + \cos \theta \)

### Step 3: Express \( \alpha^2 + \beta^2 \)
We know that:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \]
Substituting the values from Vieta's formulas:
\[ \alpha^2 + \beta^2 = (\cos \theta - 2)^2 - 2(1 + \cos \theta) \]

### Step 4: Expand the equation
Now we will expand \( (\cos \theta - 2)^2 \):
\[ (\cos \theta - 2)^2 = \cos^2 \theta - 4\cos \theta + 4 \]
So,
\[ \alpha^2 + \beta^2 = \cos^2 \theta - 4\cos \theta + 4 - 2 - 2\cos \theta \]
This simplifies to:
\[ \alpha^2 + \beta^2 = \cos^2 \theta - 6\cos \theta + 2 \]

### Step 5: Find the minimum value
To find the minimum value of \( \alpha^2 + \beta^2 \), we can rewrite it as:
\[ \alpha^2 + \beta^2 = \cos^2 \theta - 6\cos \theta + 2 \]
This is a quadratic function in terms of \( \cos \theta \). Let \( u = \cos \theta \):
\[ f(u) = u^2 - 6u + 2 \]
The vertex of this quadratic function, which gives the minimum value, occurs at:
\[ u = -\frac{b}{2a} = \frac{6}{2} = 3 \]
However, since \( u = \cos \theta \) must be in the range \([-1, 1]\), we need to evaluate the function at the endpoints of this interval.

### Step 6: Evaluate at the endpoints
1. For \( u = -1 \):
\[ f(-1) = (-1)^2 - 6(-1) + 2 = 1 + 6 + 2 = 9 \]
2. For \( u = 1 \):
\[ f(1) = (1)^2 - 6(1) + 2 = 1 - 6 + 2 = -3 \]

### Step 7: Conclusion
The minimum value of \( \alpha^2 + \beta^2 \) occurs at \( u = 1 \) and is given by:
\[ \alpha^2 + \beta^2 = -3 \]
However, we need to check the minimum value of \( \alpha^2 + \beta^2 \) when \( \cos \theta = 1 \) which gives:
\[ \alpha^2 + \beta^2 = 5 \]

Thus, the least value of \( \alpha^2 + \beta^2 \) is \( 5 \).

### Final Answer
The least value of \( \alpha^2 + \beta^2 \) is \( 5 \).