If `f(x)=sin(sqrt([a])x)` (where [.] denotes the greatest integer function) has `pi` as its fundamental period, then

Options

A. `a=1`
B. `a=9`
C. `aepsilon[1,2)`
D. `aepsilon[4,5)`

Explanation

To solve the problem, we need to determine the value of \( a \) such that the function \( f(x) = \sin(\sqrt{[a]} x) \) has a fundamental period of \( \pi \). Here, \( [a] \) denotes the greatest integer function.

### Step-by-Step Solution:

1. **Understand the Period of the Sine Function**:
The fundamental period of the sine function \( \sin(kx) \) is given by:
\[ T = \frac{2\pi}{k} \]
where \( k \) is the coefficient of \( x \).

2. **Set the Period Equal to \( \pi \)**:
For our function, we have:
\[ k = \sqrt{[a]} \]
Therefore, the period of \( f(x) \) can be expressed as:
\[ T = \frac{2\pi}{\sqrt{[a]}} \]
Given that the period is \( \pi \), we can set up the equation:
\[ \frac{2\pi}{\sqrt{[a]}} = \pi \]

3. **Simplify the Equation**:
Dividing both sides by \( \pi \) (assuming \( \pi \neq 0 \)):
\[ \frac{2}{\sqrt{[a]}} = 1 \]

4. **Solve for \( \sqrt{[a]} \)**:
Multiplying both sides by \( \sqrt{[a]} \):
\[ 2 = \sqrt{[a]} \]

5. **Square Both Sides**:
Squaring both sides gives:
\[ 4 = [a] \]

6. **Determine the Range of \( a \)**:
The greatest integer function \( [a] = 4 \) implies that \( a \) must be in the range:
\[ 4 \leq a < 5 \]

### Final Answer:
Thus, the value of \( a \) for which the function \( f(x) = \sin(\sqrt{[a]} x) \) has a fundamental period of \( \pi \) is:
\[ a \in [4, 5) \]