Which of the following function are not periodic (where [.] denotes greatest integer function:
`f(x)=sinsqrt(x)`

Options

A. `f(x)=1^([x])+(-1)^([x])`
B. `g(x)=1^([5x])+(-1)^([5x])`
C. `h(x)=2^([x])-(-2)^([x])`
D. `phi(x)=1^([x])-(-1)^([x])`

Answer

To determine whether the function \( f(x) = \sin(\sqrt{x}) \) is periodic or not, we will analyze the properties of the sine function and the square root function.

### Step-by-Step Solution:

1. **Understanding Periodicity**:
A function \( f(x) \) is periodic if there exists a positive number \( P \) such that \( f(x + P) = f(x) \) for all \( x \) in the domain of \( f \). The smallest such \( P \) is called the period of the function.

2. **Analyzing the Function**:
The given function is \( f(x) = \sin(\sqrt{x}) \). Here, \( \sin(x) \) is a periodic function with a period of \( 2\pi \). However, we need to consider how the argument of the sine function behaves.

3. **Behavior of \( \sqrt{x} \)**:
The function \( \sqrt{x} \) is defined for \( x \geq 0 \) and is an increasing function. As \( x \) increases, \( \sqrt{x} \) also increases without bound. This means that the argument of the sine function, \( \sqrt{x} \), does not repeat values in a regular interval.

4. **Checking for Periodicity**:
For \( f(x) = \sin(\sqrt{x}) \) to be periodic, there must be some \( P > 0 \) such that:
\[ \sin(\sqrt{x + P}) = \sin(\sqrt{x}) \]
This implies:
\[ \sqrt{x + P} = \sqrt{x} + 2k\pi \quad \text{for some integer } k \]
Squaring both sides gives:
\[ x + P = x + 2k\pi\sqrt{x} + 4k^2\pi^2 \]
Simplifying, we find:
\[ P = 2k\pi\sqrt{x} + 4k^2\pi^2 \]
This equation shows that \( P \) depends on \( x \), which means there is no single \( P \) that works for all \( x \).

5. **Conclusion**:
Since we cannot find a constant period \( P \) that satisfies the periodicity condition for all \( x \), we conclude that the function \( f(x) = \sin(\sqrt{x}) \) is not periodic.

### Final Answer:
The function \( f(x) = \sin(\sqrt{x}) \) is not periodic.