If `f (x) = cos (x ^(2) -4[x]), 0 lt x lt 1,` (whre [.] denotes greatest integer function) then `f '((sqrtpi)/(2))` is equal to:

Options

A. `-sqrt((pi)/(2))`
B. `sqrt((pi)/(2))`
C. 0
D. `sqrt((pi)/(4))`

Answer

To find \( f'(\sqrt{\pi}/2) \) for the function \( f(x) = \cos(x^2 - 4[x]) \) where \( 0 < x < 1 \) and \([x]\) denotes the greatest integer function, we will follow these steps:

### Step 1: Understand the function
The function \( f(x) = \cos(x^2 - 4[x]) \) needs to be evaluated at \( x = \sqrt{\pi}/2 \). Since \( \sqrt{\pi}/2 \approx 0.886 \), which is between 0 and 1, we have \([x] = 0\) for this range.

### Step 2: Substitute into the function
Since \([x] = 0\) for \( x = \sqrt{\pi}/2 \):
\[ f(x) = \cos(x^2 - 4 \cdot 0) = \cos(x^2) \]
Thus, we need to compute:
\[ f(\sqrt{\pi}/2) = \cos\left(\left(\frac{\sqrt{\pi}}{2}\right)^2\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

### Step 3: Find the derivative
We will use the limit definition of the derivative:
\[ f'(\sqrt{\pi}/2) = \lim_{h \to 0} \frac{f(\sqrt{\pi}/2 + h) - f(\sqrt{\pi}/2)}{h} \]
Substituting \( f(x) \):
\[ f'(\sqrt{\pi}/2) = \lim_{h \to 0} \frac{\cos\left(\left(\frac{\sqrt{\pi}}{2} + h\right)^2\right) - \frac{1}{\sqrt{2}}}{h} \]

### Step 4: Expand the cosine term
Using the identity \( \cos(a + b) = \cos a \cos b - \sin a \sin b \), we can expand:
\[ \left(\frac{\sqrt{\pi}}{2} + h\right)^2 = \frac{\pi}{4} + \sqrt{\pi}h + h^2 \]
Thus,
\[ f\left(\frac{\sqrt{\pi}}{2} + h\right) = \cos\left(\frac{\pi}{4} + \sqrt{\pi}h + h^2\right) \]
Using the Taylor expansion for cosine around \( \frac{\pi}{4} \):
\[ \cos\left(\frac{\pi}{4} + x\right) \approx \cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right)x \]
where \( x = \sqrt{\pi}h + h^2 \).

### Step 5: Calculate the limit
Substituting back:
\[ f'(\sqrt{\pi}/2) = \lim_{h \to 0} \frac{\left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}(\sqrt{\pi}h + h^2)\right) - \frac{1}{\sqrt{2}}}{h} \]
This simplifies to:
\[ f'(\sqrt{\pi}/2) = \lim_{h \to 0} -\frac{1}{\sqrt{2}}(\sqrt{\pi} + 0) = -\frac{\sqrt{\pi}}{\sqrt{2}} \]
Thus, we have:
\[ f'(\sqrt{\pi}/2) = -\sqrt{\frac{\pi}{2}} \]

### Final Answer
The value of \( f'(\sqrt{\pi}/2) \) is:
\[ -\sqrt{\frac{\pi}{2}} \]