Which one of the following functions are periodic

Options

A. f(x)=x-[x], where[x] denotes GIF
B. f(x)= x sin (1/ x) for x `ne` 0,f (0)=0 .
C. f (x) = X COS X
D. none of these

Answer

To determine which of the given functions are periodic, we will analyze each function step by step.

### Step 1: Analyze the first function \( f(x) = x - \lfloor x \rfloor \)

1. **Understanding the function**: The function \( f(x) = x - \lfloor x \rfloor \) represents the fractional part of \( x \). It takes any real number \( x \) and subtracts the greatest integer less than or equal to \( x \) from \( x \).

2. **Finding periodicity**: To check if this function is periodic, we need to find a positive number \( T \) such that:
\[ f(x + T) = f(x) \quad \text{for all } x \]
If we let \( T = 1 \):
\[ f(x + 1) = (x + 1) - \lfloor x + 1 \rfloor = (x + 1) - (\lfloor x \rfloor + 1) = x - \lfloor x \rfloor = f(x) \]
This shows that \( f(x) \) is periodic with a period of \( T = 1 \).

### Step 2: Analyze the second function \( f(x) = \begin{cases} 0 & \text{if } x = 0 \\ x \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \end{cases} \)

1. **Understanding the function**: The function is defined piecewise. For \( x = 0 \), it is 0, and for \( x \neq 0 \), it oscillates based on the sine function multiplied by \( x \).

2. **Finding periodicity**: To check if this function is periodic, we need to see if there exists a \( T > 0 \) such that:
\[ f(x + T) = f(x) \quad \text{for all } x \]
As \( x \) approaches 0, \( \sin\left(\frac{1}{x}\right) \) oscillates infinitely between -1 and 1, and thus \( x \sin\left(\frac{1}{x}\right) \) does not settle into a repeating pattern. Therefore, this function is not periodic.

### Step 3: Analyze the third function \( f(x) = x \cos x \)

1. **Understanding the function**: This function is the product of \( x \) and \( \cos x \).

2. **Finding periodicity**: To check if this function is periodic, we need to find \( T > 0 \) such that:
\[ f(x + T) = f(x) \quad \text{for all } x \]
We can express this as:
\[ f(x + T) = (x + T) \cos(x + T) \]
Expanding this using the cosine addition formula gives:
\[ f(x + T) = (x + T)(\cos x \cos T - \sin x \sin T) \]
This expression cannot equal \( x \cos x \) for all \( x \) unless \( T = 0 \), which means it does not repeat. Hence, this function is also not periodic.

### Conclusion

From the analysis:
- **Option 1**: \( f(x) = x - \lfloor x \rfloor \) is periodic with a period of 1.
- **Option 2**: \( f(x) = x \sin\left(\frac{1}{x}\right) \) is not periodic.
- **Option 3**: \( f(x) = x \cos x \) is not periodic.

Thus, the only periodic function among the options is **Option 1**.

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