Let [x] greatest integer `le x`. Define `f : R to R` by `f(x) = cos(5x) cos [5x] + sin(5x) sin [5x]` then period of f is ………..

Answer

To find the period of the function \( f(x) = \cos(5x) \cos([5x]) + \sin(5x) \sin([5x]) \), we can use the cosine addition formula, which states that:

\[ \cos A \cos B + \sin A \sin B = \cos(A - B) \]

### Step 1: Rewrite the function using the cosine addition formula

We can rewrite \( f(x) \) as:

\[ f(x) = \cos(5x - [5x]) \]

### Step 2: Analyze the term \([5x]\)

The term \([5x]\) is the greatest integer less than or equal to \(5x\). This means that for any \(x\) in the interval \([n/5, (n+1)/5)\) for integer \(n\), \([5x] = n\).

### Step 3: Determine the intervals for \(f(x)\)

For \(x \in [0, \frac{1}{5})\):
- \(5x \in [0, 1)\)
- \([5x] = 0\)
- Thus, \(f(x) = \cos(5x - 0) = \cos(5x)\)

For \(x \in [\frac{1}{5}, \frac{2}{5})\):
- \(5x \in [1, 2)\)
- \([5x] = 1\)
- Thus, \(f(x) = \cos(5x - 1)\)

For \(x \in [\frac{2}{5}, \frac{3}{5})\):
- \(5x \in [2, 3)\)
- \([5x] = 2\)
- Thus, \(f(x) = \cos(5x - 2)\)

For \(x \in [\frac{3}{5}, \frac{4}{5})\):
- \(5x \in [3, 4)\)
- \([5x] = 3\)
- Thus, \(f(x) = \cos(5x - 3)\)

For \(x \in [\frac{4}{5}, 1)\):
- \(5x \in [4, 5)\)
- \([5x] = 4\)
- Thus, \(f(x) = \cos(5x - 4)\)

### Step 4: Identify the periodicity

Notice that the function \(f(x)\) is defined piecewise over intervals of length \(\frac{1}{5}\). The cosine function itself has a period of \(2\pi\), but since we are looking at \(5x\), the effective period of \(f(x)\) is determined by the intervals of \(\frac{1}{5}\).

### Step 5: Conclusion

Since \(f(x)\) repeats its values every \(\frac{1}{5}\) interval, the period of the function \(f\) is:

\[ \text{Period} = \frac{1}{5} = 0.2 \]