Let `alpha` and ` beta` be any two positve values of x for which `2 cos x, | cos x|` and `1-3 cos^(2)x` are in GP. The minium value of `|alpha+beta|` is

Options

A. `pi/3`
B. `pi/4`
C. `pi/2`
D. None of these

Answer

To solve the problem, we need to find the minimum value of \(|\alpha + \beta|\) where \(\alpha\) and \(\beta\) are the angles for which \(2 \cos x\), \(|\cos x|\), and \(1 - 3 \cos^2 x\) are in geometric progression (GP).

### Step-by-Step Solution:

1. **Understanding the Condition for GP**:
Given three terms \(a\), \(b\), and \(c\) are in GP, the condition is:
\[ b^2 = ac \]
Here, we have:
- \(a = 2 \cos x\)
- \(b = |\cos x|\)
- \(c = 1 - 3 \cos^2 x\)

2. **Setting Up the Equation**:
Plugging in the values, we get:
\[ |\cos x|^2 = (2 \cos x)(1 - 3 \cos^2 x) \]
Since \(\cos x\) is positive in the range we are considering, we can drop the absolute value:
\[ \cos^2 x = 2 \cos x (1 - 3 \cos^2 x) \]

3. **Expanding and Rearranging**:
Expanding the right side:
\[ \cos^2 x = 2 \cos x - 6 \cos^3 x \]
Rearranging gives:
\[ 6 \cos^3 x + \cos^2 x - 2 \cos x = 0 \]

4. **Factoring the Equation**:
We can factor out \(\cos x\):
\[ \cos x (6 \cos^2 x + \cos x - 2) = 0 \]
This gives us one solution:
\[ \cos x = 0 \quad \text{(not valid since we need positive values)} \]
Now we solve the quadratic:
\[ 6 \cos^2 x + \cos x - 2 = 0 \]

5. **Using the Quadratic Formula**:
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 6\), \(b = 1\), and \(c = -2\):
\[ \cos x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 6 \cdot (-2)}}{2 \cdot 6} \]
\[ = \frac{-1 \pm \sqrt{1 + 48}}{12} = \frac{-1 \pm 7}{12} \]
This gives:
\[ \cos x = \frac{6}{12} = \frac{1}{2} \quad \text{or} \quad \cos x = \frac{-8}{12} = -\frac{2}{3} \]

6. **Finding Angles**:
- For \(\cos x = \frac{1}{2}\):
\[ x = \alpha = \frac{\pi}{3} \]
- For \(\cos x = -\frac{2}{3}\):
\[ x = \beta = \cos^{-1}\left(-\frac{2}{3}\right) \]

7. **Calculating \(|\alpha + \beta|\)**:
We need to find:
\[ |\alpha + \beta| = \left|\frac{\pi}{3} + \cos^{-1}\left(-\frac{2}{3}\right)\right| \]

8. **Finding Minimum Value**:
Since both \(\alpha\) and \(\beta\) are positive angles, the minimum value of \(|\alpha + \beta|\) is simply:
\[ \text{Minimum value of } |\alpha + \beta| = \frac{\pi}{3} + \cos^{-1}\left(-\frac{2}{3}\right) \]

### Final Answer:
The minimum value of \(|\alpha + \beta|\) is:
\[ \frac{\pi}{3} + \cos^{-1}\left(-\frac{2}{3}\right) \]