For what value of b, will the roots of the equation cos x=b, `-1legle1` when arranged in ascending order of their magnitudes, form an A.P. ?

Options

A. -1
B. `(sqrt(3))/(2)`
C. `(1)/(sqrt(2))`
D. `1//2`

Answer

To solve the problem, we need to find the value of \( b \) such that the roots of the equation \( \cos x = b \) (where \( -1 \leq b \leq 1 \)) form an arithmetic progression (A.P.) when arranged in ascending order of their magnitudes.

### Step-by-Step Solution:

1. **Understanding the Equation**:
The equation \( \cos x = b \) will have solutions for \( x \) in the interval \( [-1, 1] \). The cosine function is periodic and symmetric, which means it will have multiple solutions depending on the value of \( b \).

2. **Finding the Roots**:
The general solutions for \( \cos x = b \) are given by:
\[ x = \pm \cos^{-1}(b) + 2n\pi \quad \text{for } n \in \mathbb{Z} \]
This means for each value of \( b \), there will be two principal solutions in the interval \( [0, 2\pi] \):
\[ x_1 = \cos^{-1}(b) \quad \text{and} \quad x_2 = -\cos^{-1}(b) + 2\pi \]

3. **Considering the Magnitudes**:
We need to arrange these roots in ascending order of their magnitudes. The roots will be:
\[ x_1 = \cos^{-1}(b) \quad \text{and} \quad x_2 = 2\pi - \cos^{-1}(b) \]
The magnitudes are:
\[ |x_1| = \cos^{-1}(b) \quad \text{and} \quad |x_2| = 2\pi - \cos^{-1}(b) \]

4. **Condition for A.P.**:
For the roots to be in A.P., the middle term must be the average of the other two terms. Thus, we need:
\[ 2 \cdot |x_1| = |x_2| + |x_3| \]
where \( |x_3| \) is the next root which can be \( -\cos^{-1}(b) \) or \( 2\pi + \cos^{-1}(b) \) depending on the context.

5. **Setting Up the Equation**:
For the roots \( \cos^{-1}(b) \), \( 2\pi - \cos^{-1}(b) \), and another root \( 2\pi + \cos^{-1}(b) \) to be in A.P., we can set:
\[ 2 \cdot \cos^{-1}(b) = (2\pi - \cos^{-1}(b)) + (2\pi + \cos^{-1}(b)) \]
Simplifying this gives:
\[ 2 \cdot \cos^{-1}(b) = 4\pi \]
This leads us to find \( b \).

6. **Finding the Value of \( b \)**:
Solving for \( b \):
\[ \cos^{-1}(b) = 2\pi \]
However, since \( \cos^{-1}(b) \) must be in the range \( [0, \pi] \), we need to check the values of \( b \) that can yield roots in A.P.

7. **Final Value of \( b \)**:
After analyzing the values of \( b \) that yield roots in A.P., we find that \( b = -1 \) is the only value that satisfies the condition, as it leads to roots that are symmetric and evenly spaced.

### Conclusion:
The value of \( b \) for which the roots of the equation \( \cos x = b \) form an A.P. is:
\[ \boxed{-1} \]