Consider a function `f (x)` in `[0,2pi]` defined as :
` f(x)=[{:([sinx]+ [cos x],,, 0 le x le pi),( [sin x] -[cos x],,, pi lt x le 2pi):}`
where {.} denotes greatest integer function then.
Number of points where `f (x)` is non-derivable :

Options

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

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined in the intervals \( [0, \pi] \) and \( [\pi, 2\pi] \) using the greatest integer function (also known as the floor function). The function is defined as follows:

\[ f(x) = \begin{cases} \lfloor \sin x + \cos x \rfloor & \text{for } 0 \leq x \leq \pi \\ \lfloor \sin x - \cos x \rfloor & \text{for } \pi < x \leq 2\pi \end{cases} \]

### Step 1: Analyze \( f(x) \) on the interval \( [0, \pi] \)

In this interval, we need to evaluate \( \sin x + \cos x \):

1. **Find the maximum and minimum values of \( \sin x + \cos x \)**:
- The maximum occurs at \( x = \frac{\pi}{4} \), where \( \sin x + \cos x = \sqrt{2} \).
- The minimum occurs at \( x = \frac{5\pi}{4} \), but this is outside the interval \( [0, \pi] \). The minimum value in this interval is \( 0 \) (at \( x = 0 \) and \( x = \pi \)).

2. **Determine the range of \( \sin x + \cos x \)**:
- The function varies from \( 0 \) to \( \sqrt{2} \) in the interval \( [0, \pi] \).

3. **Apply the greatest integer function**:
- \( \lfloor \sin x + \cos x \rfloor \) will take values \( 0 \) or \( 1 \) in this interval.

### Step 2: Analyze \( f(x) \) on the interval \( [\pi, 2\pi] \)

Now, we evaluate \( \sin x - \cos x \):

1. **Find the maximum and minimum values of \( \sin x - \cos x \)**:
- The maximum occurs at \( x = \frac{3\pi}{4} \), where \( \sin x - \cos x = \sqrt{2} \).
- The minimum occurs at \( x = \frac{7\pi}{4} \), where \( \sin x - \cos x = -\sqrt{2} \).

2. **Determine the range of \( \sin x - \cos x \)**:
- The function varies from \( -\sqrt{2} \) to \( \sqrt{2} \) in the interval \( [\pi, 2\pi] \).

3. **Apply the greatest integer function**:
- \( \lfloor \sin x - \cos x \rfloor \) will take values \( -2, -1, 0 \) in this interval.

### Step 3: Identify points of non-differentiability

The function \( f(x) \) will be non-differentiable at points where it is discontinuous. We need to check the points where the greatest integer function changes its value:

1. **At \( x = 0 \)**:
- \( f(0) = \lfloor \sin(0) + \cos(0) \rfloor = \lfloor 1 \rfloor = 1 \).

2. **At \( x = \frac{\pi}{2} \)**:
- \( f(\frac{\pi}{2}) = \lfloor \sin(\frac{\pi}{2}) + \cos(\frac{\pi}{2}) \rfloor = \lfloor 1 + 0 \rfloor = 1 \).
- The left-hand limit approaches \( 1 \) and the right-hand limit approaches \( -1 \) (from the next interval), indicating a discontinuity.

3. **At \( x = \pi \)**:
- \( f(\pi) = \lfloor \sin(\pi) + \cos(\pi) \rfloor = \lfloor 0 - 1 \rfloor = -1 \).
- The left-hand limit approaches \( 1 \) and the right-hand limit approaches \( -1 \), indicating a discontinuity.

4. **At \( x = \frac{3\pi}{2} \)**:
- \( f(\frac{3\pi}{2}) = \lfloor \sin(\frac{3\pi}{2}) - \cos(\frac{3\pi}{2}) \rfloor = \lfloor -1 - 0 \rfloor = -1 \).
- The left-hand limit approaches \( 0 \) and the right-hand limit approaches \( 0 \), indicating a discontinuity.

### Conclusion

The points of non-differentiability are \( x = \frac{\pi}{2}, \pi, \frac{3\pi}{2} \). Therefore, the number of points where \( f(x) \) is non-derivable is:

\[ \text{Number of points where } f(x) \text{ is non-derivable} = 3 \]