The number of points where `f(x) =[sin x + cos x]` (where [.] denotes the greatest integer.function] `x in (0,2pi)` is discontinuous is:

Options

A. 3
B. 4
C. 5
D. 6

Answer

To find the number of points where the function \( f(x) = \lfloor \sin x + \cos x \rfloor \) is discontinuous in the interval \( (0, 2\pi) \), we will follow these steps:

### Step 1: Analyze the function \( \sin x + \cos x \)
We start by rewriting \( \sin x + \cos x \) using the sine addition formula. We can express it as:
\[ \sin x + \cos x = \sqrt{2} \left( \sin x \cdot \frac{1}{\sqrt{2}} + \cos x \cdot \frac{1}{\sqrt{2}} \right) = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) \]
This transformation helps us understand the behavior of the function.

### Step 2: Determine the range of \( \sin x + \cos x \)
The maximum value of \( \sin x + \cos x \) occurs when \( \sin \left( x + \frac{\pi}{4} \right) = 1 \), which gives:
\[ \max(\sin x + \cos x) = \sqrt{2} \]
The minimum value occurs when \( \sin \left( x + \frac{\pi}{4} \right) = -1 \), giving:
\[ \min(\sin x + \cos x) = -\sqrt{2} \]
Thus, the range of \( \sin x + \cos x \) is:
\[ [-\sqrt{2}, \sqrt{2}] \]

### Step 3: Apply the greatest integer function
The greatest integer function \( \lfloor x \rfloor \) is discontinuous at integer values. Therefore, we need to find the integers within the range of \( \sin x + \cos x \).

### Step 4: Identify the integers in the range
The approximate values of \( -\sqrt{2} \) and \( \sqrt{2} \) are:
\[ -\sqrt{2} \approx -1.414 \quad \text{and} \quad \sqrt{2} \approx 1.414 \]
Thus, the integers in the range \( [-\sqrt{2}, \sqrt{2}] \) are \( -1, 0, 1 \).

### Step 5: Find points of discontinuity
The function \( f(x) \) will be discontinuous at the points where \( \sin x + \cos x \) crosses these integer values. We need to find the points where:
1. \( \sin x + \cos x = -1 \)
2. \( \sin x + \cos x = 0 \)
3. \( \sin x + \cos x = 1 \)

### Step 6: Solve for each case
1. **For \( \sin x + \cos x = -1 \)**:
\[ \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) = -1 \implies \sin \left( x + \frac{\pi}{4} \right) = -\frac{1}{\sqrt{2}} \implies x + \frac{\pi}{4} = \frac{7\pi}{4} \text{ or } \frac{5\pi}{4} \]
This gives two solutions in \( (0, 2\pi) \).

2. **For \( \sin x + \cos x = 0 \)**:
\[ \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) = 0 \implies \sin \left( x + \frac{\pi}{4} \right) = 0 \implies x + \frac{\pi}{4} = n\pi \]
This gives two solutions in \( (0, 2\pi) \).

3. **For \( \sin x + \cos x = 1 \)**:
\[ \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) = 1 \implies \sin \left( x + \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \implies x + \frac{\pi}{4} = \frac{\pi}{4} \text{ or } \frac{3\pi}{4} \]
This gives two solutions in \( (0, 2\pi) \).

### Step 7: Count the points of discontinuity
Summing the points from all cases, we find:
- From \( \sin x + \cos x = -1 \): 2 points
- From \( \sin x + \cos x = 0 \): 2 points
- From \( \sin x + \cos x = 1 \): 2 points

Thus, the total number of points of discontinuity is:
\[ 2 + 2 + 2 = 6 \]

### Final Answer
The number of points where \( f(x) = \lfloor \sin x + \cos x \rfloor \) is discontinuous in the interval \( (0, 2\pi) \) is **6**.