Period of `f(x) = sin 3x cos[3x]-cos 3x sin [3x]` (where[] denotes the greatest integer function), is

Options

A. `1//6`
B. `2//3`
C. `5//6`
D. `1//3`

Explanation

To find the period of the function \( f(x) = \sin(3x) \cos[\mod(3x)] - \cos(3x) \sin[\mod(3x)] \), we can follow these steps:

### Step 1: Rewrite the Function
We can use the sine subtraction formula:
\[ \sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b) \]
In our case, let \( a = 3x \) and \( b = \mod(3x) \). Thus, we can rewrite the function as:
\[ f(x) = \sin(3x - \mod(3x)) \]

### Step 2: Understand the Greatest Integer Function
The term \(\mod(3x)\) represents the greatest integer less than or equal to \(3x\). Therefore, we can express \(3x\) as:
\[ 3x = n + \{3x\} \]
where \(n = \lfloor 3x \rfloor\) (the greatest integer less than or equal to \(3x\)) and \(\{3x\}\) is the fractional part of \(3x\).

### Step 3: Simplify the Argument of the Sine Function
Thus, we can rewrite the argument of the sine function:
\[ 3x - \mod(3x) = 3x - n = \{3x\} \]
So we have:
\[ f(x) = \sin(\{3x\}) \]

### Step 4: Determine the Period of \(\{3x\}\)
The function \(\{3x\}\) has a period of \(\frac{1}{3}\) because it resets every time \(3x\) increases by 1 (i.e., when \(x\) increases by \(\frac{1}{3}\)).

### Step 5: Conclusion
Thus, the period of \(f(x) = \sin(3x - \mod(3x))\) is:
\[ \text{Period} = \frac{1}{3} \]

### Final Answer
The period of the function \( f(x) = \sin(3x) \cos[\mod(3x)] - \cos(3x) \sin[\mod(3x)] \) is \(\frac{1}{3}\).

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