If `f(x)=sin{(pi)/(3)[x]-x^(2)}" for "2ltxlt3` and [x] denotes the greatest integer less than or equal to x, then `f'"("sqrt(pi//3)")"` is equal to

Options

A. `sqrt(pi//3)`
B. `-sqrt(pi//3)`
C. `-sqrt(pi)`
D. None of these

Answer

To solve the problem, we need to find the second derivative of the function \( f(x) = \sin\left(\frac{\pi}{3} [x] - x^2\right) \) at \( x = \sqrt{\frac{\pi}{3}} \), where \( [x] \) denotes the greatest integer less than or equal to \( x \).

### Step-by-Step Solution:

1. **Identify the function**:
The function is given as:
\[ f(x) = \sin\left(\frac{\pi}{3} [x] - x^2\right) \]
For \( 2 < x < 3 \), we have \( [x] = 2 \). Hence, we can simplify the function:
\[ f(x) = \sin\left(\frac{\pi}{3} \cdot 2 - x^2\right) = \sin\left(\frac{2\pi}{3} - x^2\right) \]

2. **Differentiate \( f(x) \)**:
We will differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx} \left(\sin\left(\frac{2\pi}{3} - x^2\right)\right) \]
Using the chain rule:
\[ f'(x) = \cos\left(\frac{2\pi}{3} - x^2\right) \cdot \frac{d}{dx}\left(\frac{2\pi}{3} - x^2\right) \]
The derivative of \( \frac{2\pi}{3} - x^2 \) is \( -2x \):
\[ f'(x) = \cos\left(\frac{2\pi}{3} - x^2\right) \cdot (-2x) = -2x \cos\left(\frac{2\pi}{3} - x^2\right) \]

3. **Evaluate \( f'(\sqrt{\frac{\pi}{3}}) \)**:
Now we substitute \( x = \sqrt{\frac{\pi}{3}} \):
\[ f'\left(\sqrt{\frac{\pi}{3}}\right) = -2\sqrt{\frac{\pi}{3}} \cos\left(\frac{2\pi}{3} - \left(\sqrt{\frac{\pi}{3}}\right)^2\right) \]
Since \( \left(\sqrt{\frac{\pi}{3}}\right)^2 = \frac{\pi}{3} \):
\[ f'\left(\sqrt{\frac{\pi}{3}}\right) = -2\sqrt{\frac{\pi}{3}} \cos\left(\frac{2\pi}{3} - \frac{\pi}{3}\right) = -2\sqrt{\frac{\pi}{3}} \cos\left(\frac{\pi}{3}\right) \]
We know that \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ f'\left(\sqrt{\frac{\pi}{3}}\right) = -2\sqrt{\frac{\pi}{3}} \cdot \frac{1}{2} = -\sqrt{\frac{\pi}{3}} \]

4. **Conclusion**:
Therefore, the value of \( f'(\sqrt{\frac{\pi}{3}}) \) is:
\[ f'(\sqrt{\frac{\pi}{3}}) = -\sqrt{\frac{\pi}{3}} \]