If `f(x)=[sqrt(2)sinx]`, where `[x]` represents the greatest integer function, then choose the incorrect option.

Options

A. f(x) is periodic
B. Maximum value of f(x) is 1 in the invertal `[-2,2]`
C. f(x) is discontinuous is `x=(npi)/2+(pi)/4,n epsilonI`
D. f(x) is differentiable at `n= n pi, n epsilon I`

Answer

To solve the problem, we need to analyze the function \( f(x) = \lfloor \sqrt{2} \sin x \rfloor \), where \( \lfloor x \rfloor \) denotes the greatest integer function. We will evaluate the given options to determine which one is incorrect.

### Step 1: Determine if \( f(x) \) is periodic.
The function \( \sin x \) is periodic with a period of \( 2\pi \). Since \( f(x) \) is derived from \( \sin x \) (specifically, \( \sqrt{2} \sin x \)), it will also be periodic with the same period. Therefore, \( f(x) \) is periodic.

**Hint:** A function is periodic if it repeats its values in regular intervals.

### Step 2: Find the maximum value of \( f(x) \) in the interval \((-2, 2)\).
The maximum value of \( \sqrt{2} \sin x \) occurs when \( \sin x = 1 \), which gives \( \sqrt{2} \). The greatest integer less than or equal to \( \sqrt{2} \) is 1. Thus, the maximum value of \( f(x) \) in the interval \((-2, 2)\) is indeed 1.

**Hint:** The maximum value of a function can often be found by evaluating it at critical points.

### Step 3: Check for discontinuity at \( x = n\pi + \frac{\pi}{4} \).
To check the continuity of \( f(x) \) at \( x = n\pi + \frac{\pi}{4} \), we need to evaluate the left-hand limit (LHL) and right-hand limit (RHL) at these points.

1. For \( n = 0 \), \( x = \frac{\pi}{4} \):
- LHL: \( \lfloor \sqrt{2} \sin(\frac{\pi}{4}^-) \rfloor = \lfloor \sqrt{2} \cdot \frac{1}{\sqrt{2}} \rfloor = \lfloor 1 \rfloor = 1 \)
- RHL: \( \lfloor \sqrt{2} \sin(\frac{\pi}{4}^+) \rfloor = \lfloor \sqrt{2} \cdot \frac{1}{\sqrt{2}} \rfloor = \lfloor 1 \rfloor = 1 \)

Thus, \( f(x) \) is continuous at \( x = \frac{\pi}{4} \).

2. For \( n = 1 \), \( x = \frac{3\pi}{4} \):
- LHL: \( \lfloor \sqrt{2} \sin(\frac{3\pi}{4}^-) \rfloor = \lfloor \sqrt{2} \cdot \frac{1}{\sqrt{2}} \rfloor = \lfloor 1 \rfloor = 1 \)
- RHL: \( \lfloor \sqrt{2} \sin(\frac{3\pi}{4}^+) \rfloor = \lfloor \sqrt{2} \cdot \frac{1}{\sqrt{2}} \rfloor = \lfloor 1 \rfloor = 1 \)

Thus, \( f(x) \) is continuous at \( x = \frac{3\pi}{4} \).

**Hint:** A function is discontinuous at a point if the left-hand limit does not equal the right-hand limit.

### Step 4: Check differentiability at \( x = n\pi \).
To check if \( f(x) \) is differentiable at \( x = n\pi \), we compute the derivative using the limit definition:

\[ f'(n\pi) = \lim_{h \to 0} \frac{f(n\pi + h) - f(n\pi)}{h} \]

Since \( \sin(n\pi) = 0 \), we have:

\[ f(n\pi) = \lfloor \sqrt{2} \cdot 0 \rfloor = 0 \]

For small \( h \), \( \sin(n\pi + h) \) will be approximately \( h \), thus:

\[ f(n\pi + h) = \lfloor \sqrt{2} h \rfloor \]

As \( h \) approaches 0, \( \lfloor \sqrt{2} h \rfloor \) will be 0 for small enough \( h \). Therefore, the limit becomes:

\[ \lim_{h \to 0} \frac{0 - 0}{h} = 0 \]

However, the function \( f(x) \) is not differentiable at \( n\pi \) because it jumps from 0 to 1 as \( h \) crosses 0.

**Hint:** A function is not differentiable at a point if it has a jump discontinuity there.

### Conclusion:
The incorrect option is that \( f(x) \) is differentiable at \( n\pi \), as we have shown that it is not differentiable at these points.

### Summary of Steps:
1. Check periodicity of \( f(x) \).
2. Determine the maximum value of \( f(x) \) in the given interval.
3. Analyze continuity at specific points.
4. Evaluate differentiability at \( n\pi \).