The number of points of non- differentiability of function f(x) = max `{ sin ^(-1)|sin x|, cos ^(-1) |sin x|} ,0 lt x lt 2 pi ,` is ________.

Answer

To find the number of points of non-differentiability of the function \( f(x) = \max \{ \sin^{-1} |\sin x|, \cos^{-1} |\sin x| \} \) for \( 0 < x < 2\pi \), we will follow these steps:

### Step 1: Understand the Functions Involved
The function \( f(x) \) is defined as the maximum of two functions:
- \( \sin^{-1} |\sin x| \)
- \( \cos^{-1} |\sin x| \)

### Step 2: Identify the Domains of the Functions
1. The function \( \sin^{-1} y \) is defined for \( -1 \leq y \leq 1 \) and takes values in \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
2. The function \( \cos^{-1} y \) is defined for \( 0 \leq y \leq 1 \) and takes values in \( [0, \pi] \).

Since \( |\sin x| \) ranges from 0 to 1 for \( 0 < x < 2\pi \), both functions are defined in this interval.

### Step 3: Find Points of Non-Differentiability
The points of non-differentiability occur where the two functions intersect, as well as at points where either function has a corner or cusp.

1. **Set the two functions equal to find intersection points:**
\[ \sin^{-1} |\sin x| = \cos^{-1} |\sin x| \]
Using the identity \( \cos^{-1} y = \frac{\pi}{2} - \sin^{-1} y \), we can rewrite the equation:
\[ \sin^{-1} |\sin x| = \frac{\pi}{2} - \sin^{-1} |\sin x| \]
Rearranging gives:
\[ 2\sin^{-1} |\sin x| = \frac{\pi}{2} \]
Thus,
\[ \sin^{-1} |\sin x| = \frac{\pi}{4} \]
Taking the sine of both sides:
\[ |\sin x| = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \]
This gives us:
\[ \sin x = \frac{\sqrt{2}}{2} \quad \text{or} \quad \sin x = -\frac{\sqrt{2}}{2} \]

2. **Find values of \( x \) for \( \sin x = \frac{\sqrt{2}}{2} \):**
\[ x = \frac{\pi}{4}, \quad \frac{3\pi}{4} \quad \text{(in the interval \( 0 < x < 2\pi \))} \]

3. **Find values of \( x \) for \( \sin x = -\frac{\sqrt{2}}{2} \):**
\[ x = \frac{5\pi}{4}, \quad \frac{7\pi}{4} \quad \text{(in the interval \( 0 < x < 2\pi \))} \]

### Step 4: Check for Points Where Functions are Not Differentiable
We also need to check where \( |\sin x| = 0 \):
- \( \sin x = 0 \) at \( x = 0, \pi, 2\pi \), but these points are not in the open interval \( (0, 2\pi) \).

### Step 5: Count the Points of Non-Differentiability
From the above analysis, we have identified the following points:
- \( x = \frac{\pi}{4} \)
- \( x = \frac{3\pi}{4} \)
- \( x = \frac{5\pi}{4} \)
- \( x = \frac{7\pi}{4} \)

Thus, there are **4 points** of non-differentiability.

### Final Answer
The number of points of non-differentiability of the function \( f(x) \) is **4**.

---