Let `[x]` denotes the greatest integer less than or equal to `x` . If the function `f(x)="tan"(sqrt([n])x)` has period `pi/3dot` then find the value of `ndot`

Explanation

To solve the problem, we need to find the value of \( n \) such that the function \( f(x) = \tan(\sqrt{[n]} x) \) has a period of \( \frac{\pi}{3} \).

### Step-by-step Solution:

1. **Understanding the Period of the Function**:
The function \( \tan(kx) \) has a period of \( \frac{\pi}{k} \). In our case, \( k = \sqrt{[n]} \). Therefore, the period of \( f(x) = \tan(\sqrt{[n]} x) \) is given by:
\[ \text{Period} = \frac{\pi}{\sqrt{[n]}} \]

2. **Setting the Period Equal to Given Value**:
We are given that the period of the function is \( \frac{\pi}{3} \). Thus, we can set up the equation:
\[ \frac{\pi}{\sqrt{[n]}} = \frac{\pi}{3} \]

3. **Cancelling \( \pi \)**:
Since \( \pi \) is a common factor on both sides, we can cancel it out:
\[ \frac{1}{\sqrt{[n]}} = \frac{1}{3} \]

4. **Cross-Multiplying**:
Cross-multiplying gives us:
\[ \sqrt{[n]} = 3 \]

5. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ [n] = 9 \]

6. **Finding the Range of \( n \)**:
The greatest integer function \( [n] = 9 \) means that \( n \) can take any value from 9 (inclusive) to 10 (exclusive). Therefore, we can write:
\[ n \in [9, 10) \]

### Final Answer:
The value of \( n \) is in the range:
\[ n \in [9, 10) \]