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 will follow these steps:

### Step 1: Identify the periods of the individual functions
The period of \( |\sin x| \) is \( \pi \) and the period of \( |\cos x| \) is also \( \pi \).

### Step 2: Adjust for the coefficient of \( x \)
Since we have \( \sin 3x \) and \( \cos 3x \), we need to adjust the periods by dividing by the coefficient of \( x \), which is \( 3 \).

- The period of \( |\sin 3x| \) is:
\[ \text{Period of } |\sin 3x| = \frac{\pi}{3} \]

- The period of \( |\cos 3x| \) is:
\[ \text{Period of } |\cos 3x| = \frac{\pi}{3} \]

### Step 3: Find the least common multiple (LCM)
Now, we need to find the least common multiple of the two periods we just calculated:
\[ \text{LCM}\left(\frac{\pi}{3}, \frac{\pi}{3}\right) = \frac{\pi}{3} \]

### Step 4: Check for periodicity with a phase shift
Next, we check if \( f(x) \) has a smaller period by evaluating \( f(x + \frac{\pi}{2}) \):
\[ f\left(x + \frac{\pi}{2}\right) = |\sin(3(x + \frac{\pi}{2}))| + |\cos(3(x + \frac{\pi}{2})| \]
This simplifies to:
\[ = |\sin(3x + \frac{3\pi}{2})| + |\cos(3x + \frac{3\pi}{2})| \]
Using the identities \( \sin(a + \frac{3\pi}{2}) = -\cos(a) \) and \( \cos(a + \frac{3\pi}{2}) = \sin(a) \):
\[ = |-\cos(3x)| + |\sin(3x)| = |\cos(3x)| + |\sin(3x)| = f(x) \]
This shows that \( f(x + \frac{\pi}{2}) = f(x) \), indicating that the function is periodic with a period of \( \frac{\pi}{2} \).

### Step 5: Adjust for the coefficient of \( x \)
Since the function is periodic with respect to \( x \), we need to divide the period \( \frac{\pi}{2} \) by the coefficient \( 3 \):
\[ \text{Final Period} = \frac{\frac{\pi}{2}}{3} = \frac{\pi}{6} \]

### Conclusion
Thus, the period of the function \( f(x) = |\sin 3x| + |\cos 3x| \) is:
\[ \boxed{\frac{\pi}{6}} \]