Let `alphaa n dbeta` be any two positive values of `x` for which `2cosx ,|cosx|,` and `1-3cos^2x` are in G.P. The minimum value of `|alpha-beta|` is `pi/3` (b) `pi/4` (c) `pi/2` (d) none of these

Options

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

Explanation

To solve the problem, we need to find the positive values of \( x \) for which the terms \( 2 \cos x \), \( |\cos x| \), and \( 1 - 3 \cos^2 x \) are in geometric progression (G.P.).

### Step 1: Set up the condition for G.P.
For three terms \( a, b, c \) to be in G.P., the condition is:
\[ b^2 = ac \]
Here, let \( a = 2 \cos x \), \( b = |\cos x| \), and \( c = 1 - 3 \cos^2 x \). Therefore, we have:
\[ |\cos x|^2 = (2 \cos x)(1 - 3 \cos^2 x) \]

### Step 2: Simplify the equation
Since \( \cos x \) is positive in the first quadrant, we can drop the absolute value:
\[ \cos^2 x = 2 \cos x (1 - 3 \cos^2 x) \]
Expanding the right side:
\[ \cos^2 x = 2 \cos x - 6 \cos^3 x \]

### Step 3: Rearranging the equation
Rearranging gives us:
\[ 6 \cos^3 x + \cos^2 x - 2 \cos x = 0 \]
Factoring out \( \cos x \):
\[ \cos x (6 \cos^2 x + \cos x - 2) = 0 \]

### Step 4: Solve for \( \cos x \)
This gives us two cases:
1. \( \cos x = 0 \)
2. \( 6 \cos^2 x + \cos x - 2 = 0 \)

For the first case, \( \cos x = 0 \) gives \( x = \frac{\pi}{2} \).

For the second case, we can use the quadratic formula:
\[ \cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 6, b = 1, c = -2 \):
\[ \cos x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 6 \cdot (-2)}}{2 \cdot 6} = \frac{-1 \pm \sqrt{1 + 48}}{12} = \frac{-1 \pm 7}{12} \]
Calculating the roots:
1. \( \cos x = \frac{6}{12} = \frac{1}{2} \) (valid)
2. \( \cos x = \frac{-8}{12} = -\frac{2}{3} \) (not valid since we need positive values)

### Step 5: Find the angles
The valid solutions for \( \cos x \) are:
1. \( \cos x = \frac{1}{2} \) gives \( x = \frac{\pi}{3} \)
2. \( \cos x = 0 \) gives \( x = \frac{\pi}{2} \)

### Step 6: Calculate \( |\alpha - \beta| \)
Let \( \alpha = \frac{\pi}{2} \) and \( \beta = \frac{\pi}{3} \):
\[ |\alpha - \beta| = \left| \frac{\pi}{2} - \frac{\pi}{3} \right| = \left| \frac{3\pi}{6} - \frac{2\pi}{6} \right| = \left| \frac{\pi}{6} \right| \]

### Conclusion
The minimum value of \( |\alpha - \beta| \) is \( \frac{\pi}{6} \), which is not listed in the options provided. Therefore, the answer is:
**(d) none of these.**