The fundamental period of function `f(x) = [x] + [x + (1)/(3)] + [x + (2)/(3)] - 3x + 15`

Options

A. `1//3`
B. `2//3`
C. `1`
D. Non- periodic

Answer

To find the fundamental period of the function \( f(x) = [x] + [x + \frac{1}{3}] + [x + \frac{2}{3}] - 3x + 15 \), we will follow these steps:

### Step 1: Understand the components of the function
The function consists of the greatest integer function (also known as the floor function) applied to \( x \), \( x + \frac{1}{3} \), and \( x + \frac{2}{3} \). The greatest integer function has a fundamental period of 1.

**Hint:** Recall that the greatest integer function \( [x] \) returns the largest integer less than or equal to \( x \), and it has a periodicity of 1.

### Step 2: Analyze the periodicity of the terms
Since the greatest integer function has a period of 1, we need to check how the terms \( [x + \frac{1}{3}] \) and \( [x + \frac{2}{3}] \) affect the overall periodicity.

**Hint:** Consider how shifting \( x \) by a constant affects the greatest integer function.

### Step 3: Substitute \( x \) with \( x + \frac{1}{3} \)
Now, let's substitute \( x \) with \( x + \frac{1}{3} \) in the function to check if \( f(x + \frac{1}{3}) = f(x) \):

\[ f\left(x + \frac{1}{3}\right) = \left[x + \frac{1}{3}\right] + \left[x + \frac{1}{3} + \frac{1}{3}\right] + \left[x + \frac{1}{3} + \frac{2}{3}\right] - 3\left(x + \frac{1}{3}\right) + 15 \]

This simplifies to:

\[ = \left[x + \frac{1}{3}\right] + \left[x + \frac{2}{3}\right] + \left[x + 1\right] - 3x - 1 + 15 \]

**Hint:** Remember that \( [x + 1] = [x] + 1 \) and apply this property.

### Step 4: Simplify the expression
Now, we can simplify the expression:

\[ = [x] + [x] + 1 + [x] + 1 - 3x - 1 + 15 \]
\[ = 3[x] + 1 - 3x + 15 \]
\[ = 3[x] - 3x + 15 + 1 \]
\[ = 3[x] - 3x + 16 \]

### Step 5: Compare with the original function
Now, we need to compare \( f\left(x + \frac{1}{3}\right) \) with \( f(x) \):

\[ f(x) = [x] + [x + \frac{1}{3}] + [x + \frac{2}{3}] - 3x + 15 \]

From our previous calculation, we see that:

\[ f\left(x + \frac{1}{3}\right) = f(x) \]

### Conclusion
Since \( f\left(x + \frac{1}{3}\right) = f(x) \), we conclude that the fundamental period of the function \( f(x) \) is \( \frac{1}{3} \).

**Final Answer:** The fundamental period of the function \( f(x) \) is \( \frac{1}{3} \).