The function `f(x)= cos ""(x)/(2)+{x}`, where {x}= the fractional part of x , is a

Options

A. periodic function with period `4pi`
B. periodic function with period 1
C. periodic function with indeterminate period
D. none of these

Explanation

To determine the nature of the function \( f(x) = \cos\left(\frac{x}{2}\right) + \{x\} \), where \(\{x\}\) is the fractional part of \(x\), we need to analyze the periodicity of both components of the function.

### Step 1: Analyze the periodicity of \(\cos\left(\frac{x}{2}\right)\)

The cosine function \(\cos(x)\) has a standard period of \(2\pi\). For the function \(\cos\left(\frac{x}{2}\right)\), we can find its period by setting:

\[ \cos\left(\frac{x + T}{2}\right) = \cos\left(\frac{x}{2}\right) \]

This implies:

\[ \frac{x + T}{2} = \frac{x}{2} + 2k\pi \quad \text{for some integer } k \]

Solving for \(T\):

\[ \frac{T}{2} = 2k\pi \implies T = 4k\pi \]

Thus, the smallest positive period occurs when \(k = 1\):

\[ T = 4\pi \]

### Step 2: Analyze the periodicity of \(\{x\}\)

The fractional part function \(\{x\}\) is defined as:

\[ \{x\} = x - \lfloor x \rfloor \]

This function has a period of \(1\) because:

\[ \{x + 1\} = (x + 1) - \lfloor x + 1 \rfloor = x + 1 - (\lfloor x \rfloor + 1) = x - \lfloor x \rfloor = \{x\} \]

### Step 3: Determine the overall periodicity of \(f(x)\)

The overall periodicity of \(f(x)\) will be the least common multiple (LCM) of the individual periods of the components:

- Period of \(\cos\left(\frac{x}{2}\right)\) is \(4\pi\)
- Period of \(\{x\}\) is \(1\)

To find the LCM of \(4\pi\) and \(1\):

Since \(4\pi\) is an irrational number and \(1\) is a rational number, the LCM of an irrational number with a rational number is not defined in the standard sense. Therefore, the function \(f(x)\) does not have a period.

### Conclusion

The function \(f(x) = \cos\left(\frac{x}{2}\right) + \{x\}\) is not periodic.