If `f(x)=sin (sqrt([lambda])x)` is a function with period `pi` , [ ] where denotes the greatest integer less than or equal to `lambda" is "pi`, then

Options

A. `lambda in [4,5)`
B. `lambda in [4,5]`
C. `lambda=4,5`
D. None of these

Answer

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

### Step-by-Step Solution:

1. **Understanding the Period of the Sine Function**:
The standard sine function, \(\sin(x)\), has a period of \(2\pi\). If we modify the argument of the sine function, the period changes. Specifically, if \(f(x) = \sin(kx)\), the period of \(f\) is given by:
\[ \text{Period} = \frac{2\pi}{|k|} \]
In our case, \(k = \sqrt{[\lambda]}\).

2. **Setting the Period Equal to \(\pi\)**:
According to the problem, we want the period of \(f(x)\) to be \(\pi\). Therefore, we set up the equation:
\[ \frac{2\pi}{\sqrt{[\lambda]}} = \pi \]

3. **Solving for \(\sqrt{[\lambda]}\)**:
To solve for \(\sqrt{[\lambda]}\), we can simplify the equation:
\[ \frac{2\pi}{\sqrt{[\lambda]}} = \pi \implies \sqrt{[\lambda]} = 2 \]

4. **Squaring Both Sides**:
Next, we square both sides to eliminate the square root:
\[ [\lambda] = 4 \]

5. **Understanding the Greatest Integer Function**:
The greatest integer function \([\lambda]\) returns the largest integer less than or equal to \(\lambda\). Therefore, for \([\lambda] = 4\), we have:
\[ 4 \leq \lambda < 5 \]

6. **Conclusion**:
Thus, the values of \(\lambda\) that satisfy the condition are:
\[ \lambda \in [4, 5) \]

### Final Answer:
The values of \(\lambda\) such that \(f(x) = \sin(\sqrt{[\lambda]} x)\) has a period of \(\pi\) are:
\[ \lambda \in [4, 5) \]