f(x)=[sinx]+[cosx] ,`x epsilon [0,2pi]`, where[.] denotes the greatest integer function. Total number of point where f (x) is non-differentiable is equal to
i) 2
ii) 3
iii) 4
iv) 5

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to analyze the function \( f(x) = [\sin x] + [\cos x] \) over the interval \( x \in [0, 2\pi] \), where \([.]\) denotes the greatest integer function.

### Step 1: Determine the values of \([\sin x]\) over the interval \([0, 2\pi]\)

1. **For \( x \in [0, \frac{\pi}{2}) \)**:
- \( \sin x \) varies from 0 to 1.
- Therefore, \([\sin x] = 0\).

2. **At \( x = \frac{\pi}{2} \)**:
- \( \sin(\frac{\pi}{2}) = 1 \).
- Therefore, \([\sin(\frac{\pi}{2})] = 1\).

3. **For \( x \in (\frac{\pi}{2}, \pi) \)**:
- \( \sin x \) decreases from 1 to 0.
- Therefore, \([\sin x] = 0\).

4. **For \( x \in [\pi, \frac{3\pi}{2}) \)**:
- \( \sin x \) varies from 0 to -1.
- Therefore, \([\sin x] = -1\).

5. **For \( x = \frac{3\pi}{2} \)**:
- \( \sin(\frac{3\pi}{2}) = -1 \).
- Therefore, \([\sin(\frac{3\pi}{2})] = -1\).

6. **For \( x \in (\frac{3\pi}{2}, 2\pi) \)**:
- \( \sin x \) increases from -1 to 0.
- Therefore, \([\sin x] = -1\).

7. **At \( x = 2\pi \)**:
- \( \sin(2\pi) = 0 \).
- Therefore, \([\sin(2\pi)] = 0\).

### Step 2: Determine the values of \([\cos x]\) over the interval \([0, 2\pi]\)

1. **At \( x = 0 \)**:
- \( \cos(0) = 1 \).
- Therefore, \([\cos(0)] = 1\).

2. **For \( x \in (0, \frac{\pi}{2}) \)**:
- \( \cos x \) decreases from 1 to 0.
- Therefore, \([\cos x] = 0\).

3. **At \( x = \frac{\pi}{2} \)**:
- \( \cos(\frac{\pi}{2}) = 0 \).
- Therefore, \([\cos(\frac{\pi}{2})] = 0\).

4. **For \( x \in (\frac{\pi}{2}, \frac{3\pi}{2}) \)**:
- \( \cos x \) decreases from 0 to -1.
- Therefore, \([\cos x] = -1\).

5. **At \( x = \frac{3\pi}{2} \)**:
- \( \cos(\frac{3\pi}{2}) = 0 \).
- Therefore, \([\cos(\frac{3\pi}{2})] = 0\).

6. **For \( x \in (\frac{3\pi}{2}, 2\pi) \)**:
- \( \cos x \) increases from -1 to 1.
- Therefore, \([\cos x] = -1\).

7. **At \( x = 2\pi \)**:
- \( \cos(2\pi) = 1 \).
- Therefore, \([\cos(2\pi)] = 1\).

### Step 3: Calculate \( f(x) = [\sin x] + [\cos x] \)

Now, we can combine the values of \([\sin x]\) and \([\cos x]\):

- **At \( x = 0 \)**: \( f(0) = [\sin(0)] + [\cos(0)] = 0 + 1 = 1 \)
- **For \( x \in (0, \frac{\pi}{2}) \)**: \( f(x) = 0 + 0 = 0 \)
- **At \( x = \frac{\pi}{2} \)**: \( f(\frac{\pi}{2}) = 1 + 0 = 1 \)
- **For \( x \in (\frac{\pi}{2}, \pi) \)**: \( f(x) = 0 - 1 = -1 \)
- **At \( x = \pi \)**: \( f(\pi) = -1 - 1 = -2 \)
- **For \( x \in (\pi, \frac{3\pi}{2}) \)**: \( f(x) = -1 - 1 = -2 \)
- **At \( x = \frac{3\pi}{2} \)**: \( f(\frac{3\pi}{2}) = -1 + 0 = -1 \)
- **For \( x \in (\frac{3\pi}{2}, 2\pi) \)**: \( f(x) = -1 + 0 = -1 \)
- **At \( x = 2\pi \)**: \( f(2\pi) = 0 + 1 = 1 \)

### Step 4: Identify points of non-differentiability

The function \( f(x) \) is non-differentiable at points where there are jumps or discontinuities in the function. The points where \( f(x) \) changes its value abruptly are:

1. \( x = 0 \) (from 1 to 0)
2. \( x = \frac{\pi}{2} \) (from 0 to 1)
3. \( x = \pi \) (from 1 to -2)
4. \( x = \frac{3\pi}{2} \) (from -2 to -1)
5. \( x = 2\pi \) (from -1 to 1)

Thus, the total number of points where \( f(x) \) is non-differentiable is **5**.

### Final Answer

The total number of points where \( f(x) \) is non-differentiable is **5** (Option iv).

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