The total number of points of non-differentiability of `f(x) = min[|sin x|,|cos x|, (1)/(4)]"in"(0, 2pi)` is

Options

A. 8
B. 9
C. 10
D. 11

Answer

To find the total number of points of non-differentiability of the function \( f(x) = \min[|\sin x|, |\cos x|, \frac{1}{4}] \) in the interval \( (0, 2\pi) \), we can follow these steps:

### Step 1: Identify the Functions
We need to analyze the three functions involved:
1. \( |\sin x| \)
2. \( |\cos x| \)
3. \( \frac{1}{4} \)

### Step 2: Find the Points of Intersection
Next, we find the points where these functions intersect within the interval \( (0, 2\pi) \).

1. **Intersection of \( |\sin x| \) and \( |\cos x| \)**:
- The points of intersection occur when \( |\sin x| = |\cos x| \).
- This happens at \( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \).

2. **Intersection of \( |\sin x| \) and \( \frac{1}{4} \)**:
- Solve \( |\sin x| = \frac{1}{4} \).
- This occurs at \( x = \arcsin\left(\frac{1}{4}\right) \) and \( x = \pi - \arcsin\left(\frac{1}{4}\right) \).
- Also, consider the negative values in the interval \( (0, 2\pi) \):
- \( x = \pi + \arcsin\left(\frac{1}{4}\right) \) and \( x = 2\pi - \arcsin\left(\frac{1}{4}\right) \).

3. **Intersection of \( |\cos x| \) and \( \frac{1}{4} \)**:
- Solve \( |\cos x| = \frac{1}{4} \).
- This occurs at \( x = \arccos\left(\frac{1}{4}\right) \) and \( x = 2\pi - \arccos\left(\frac{1}{4}\right) \).
- Also consider the negative values:
- \( x = \pi - \arccos\left(\frac{1}{4}\right) \).

### Step 3: Count the Points of Non-Differentiability
Now, we need to count the total points of non-differentiability, which occur at the points where the minimum function changes from one function to another (corners or kinks).

From the intersections found:
- \( |\sin x| = |\cos x| \) gives us 4 points.
- \( |\sin x| = \frac{1}{4} \) gives us 4 points.
- \( |\cos x| = \frac{1}{4} \) gives us 4 points.

### Step 4: Total Points
Adding these points together:
- Total points of non-differentiability = \( 4 + 4 + 4 = 12 \).

However, we must check for any overlaps among these points to avoid double counting. After checking, we find that all points are distinct.

Thus, the total number of points of non-differentiability of \( f(x) \) in the interval \( (0, 2\pi) \) is **12**.

### Final Answer
The total number of points of non-differentiability of \( f(x) \) in \( (0, 2\pi) \) is **12**.