The total number of points of non-differentiability of `f(x)=max{sin^2 x,cos^2 x,3/4}` in `[0,10 pi],` is

Options

A. 40
B. 30
C. 20
D. 10

Answer

To find the total number of points of non-differentiability of the function \( f(x) = \max\{\sin^2 x, \cos^2 x, \frac{3}{4}\} \) in the interval \([0, 10\pi]\), we can follow these steps:

### Step 1: Identify the functions involved
The function \( f(x) \) is defined as the maximum of three functions: \( \sin^2 x \), \( \cos^2 x \), and the constant \( \frac{3}{4} \).

### Step 2: Find the points of intersection
To find the points of non-differentiability, we need to determine where the functions intersect:
1. Set \( \sin^2 x = \cos^2 x \).
- This occurs when \( \tan^2 x = 1 \), or \( x = \frac{\pi}{4} + n\frac{\pi}{2} \) for \( n \in \mathbb{Z} \).
2. Set \( \sin^2 x = \frac{3}{4} \).
- This gives \( \sin x = \pm \frac{\sqrt{3}}{2} \), leading to \( x = \frac{\pi}{3} + 2n\pi \) and \( x = \frac{2\pi}{3} + 2n\pi \).
3. Set \( \cos^2 x = \frac{3}{4} \).
- This gives \( \cos x = \pm \frac{\sqrt{3}}{2} \), leading to \( x = \frac{\pi}{6} + 2n\pi \) and \( x = \frac{5\pi}{6} + 2n\pi \).

### Step 3: Calculate the number of intersection points in \([0, 10\pi]\)
1. **For \( \sin^2 x = \cos^2 x \)**:
- The points are \( x = \frac{\pi}{4} + n\frac{\pi}{2} \).
- In the interval \([0, 10\pi]\), the values of \( n \) can be \( 0, 1, 2, \ldots, 19 \) (total of 20 points).

2. **For \( \sin^2 x = \frac{3}{4} \)**:
- The points are \( x = \frac{\pi}{3} + 2n\pi \) and \( x = \frac{2\pi}{3} + 2n\pi \).
- For \( n = 0, 1, 2, \ldots, 4 \), we get 5 points for each case, totaling \( 10 \) points.

3. **For \( \cos^2 x = \frac{3}{4} \)**:
- The points are \( x = \frac{\pi}{6} + 2n\pi \) and \( x = \frac{5\pi}{6} + 2n\pi \).
- Again, for \( n = 0, 1, 2, \ldots, 4 \), we get 5 points for each case, totaling \( 10 \) points.

### Step 4: Combine the counts
Now we need to combine the counts of intersection points:
- From \( \sin^2 x = \cos^2 x \): 20 points
- From \( \sin^2 x = \frac{3}{4} \): 10 points
- From \( \cos^2 x = \frac{3}{4} \): 10 points

### Step 5: Remove duplicates
Since some points may overlap, we need to check for duplicates. However, in this case, the points of intersection are distinct due to the nature of the sine and cosine functions.

### Total Points of Non-Differentiability
Thus, the total number of distinct points of non-differentiability in the interval \([0, 10\pi]\) is:
\[ 20 + 10 + 10 = 40 \]

### Final Answer
The total number of points of non-differentiability of \( f(x) \) in the interval \([0, 10\pi]\) is **40**.

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