The period of the function `f(x)=|sin 3x|+| cos 3x|` , is

Options

A. `(pi)/(2)`
B. `(pi)/(6)`
C. `(3pi)/(2)`
D. `pi`

Answer

To find the period of the function \( f(x) = |\sin(3x)| + |\cos(3x)| \), we need to analyze the periodicity of each component of the function.

### Step 1: Determine the period of \( |\sin(3x)| \)
The function \( \sin(x) \) has a period of \( 2\pi \). When we have \( \sin(3x) \), the period is reduced by a factor of \( 3 \). Therefore, the period of \( \sin(3x) \) is:
\[ \text{Period of } \sin(3x) = \frac{2\pi}{3} \]
Since we are taking the absolute value, \( |\sin(3x)| \) will have the same period as \( \sin(3x) \), which is \( \frac{2\pi}{3} \).

### Step 2: Determine the period of \( |\cos(3x)| \)
Similarly, the function \( \cos(x) \) also has a period of \( 2\pi \). For \( \cos(3x) \), the period is again reduced by a factor of \( 3 \):
\[ \text{Period of } \cos(3x) = \frac{2\pi}{3} \]
Taking the absolute value \( |\cos(3x)| \) does not change the period, so it remains \( \frac{2\pi}{3} \).

### Step 3: Find the combined period of \( f(x) \)
Now, we have both components of the function \( f(x) \):
- The period of \( |\sin(3x)| \) is \( \frac{2\pi}{3} \).
- The period of \( |\cos(3x)| \) is \( \frac{2\pi}{3} \).

Since both components have the same period, the overall period of the function \( f(x) \) is also:
\[ \text{Period of } f(x) = \frac{2\pi}{3} \]

### Conclusion
Thus, the period of the function \( f(x) = |\sin(3x)| + |\cos(3x)| \) is \( \frac{2\pi}{3} \).