If ` f(x) = cos[pi^2] x + cos [-pi^2]x ` where [x] stands for the greatest integer function then

Options

A. `f(pi/2) =-1`
B. ` f(pi )=1`
C. `f (-pi)=0`
D. `f(pi/4)=(1)/(sqrt(2))`

Answer

To solve the problem, we need to analyze the function \( f(x) = \cos[\pi^2 x] + \cos[-\pi^2 x] \), where \([x]\) denotes the greatest integer function (also known as the floor function).

### Step-by-Step Solution:

1. **Evaluate \(\pi^2\)**:
\[ \pi \approx 3.14 \implies \pi^2 \approx 9.86 \]
The greatest integer less than or equal to \(9.86\) is \(9\). Thus, \([\pi^2] = 9\).

2. **Evaluate \(-\pi^2\)**:
\[ -\pi^2 \approx -9.86 \]
The greatest integer less than or equal to \(-9.86\) is \(-10\). Thus, \([- \pi^2] = -10\).

3. **Substitute into the function**:
Now we can rewrite \(f(x)\):
\[ f(x) = \cos[9x] + \cos[-10x] \]
Since \(\cos(-\theta) = \cos(\theta)\), we have:
\[ f(x) = \cos(9x) + \cos(10x) \]

4. **Evaluate \(f\left(\frac{\pi}{2}\right)\)**:
\[ f\left(\frac{\pi}{2}\right) = \cos\left(9 \cdot \frac{\pi}{2}\right) + \cos\left(10 \cdot \frac{\pi}{2}\right) \]
Simplifying:
\[ = \cos\left(\frac{9\pi}{2}\right) + \cos(5\pi) = \cos\left(\frac{9\pi}{2}\right) + (-1) \]
The angle \(\frac{9\pi}{2}\) can be simplified:
\[ \frac{9\pi}{2} = 4\pi + \frac{\pi}{2} \implies \cos\left(\frac{9\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]
Thus:
\[ f\left(\frac{\pi}{2}\right) = 0 - 1 = -1 \]

5. **Evaluate \(f(\pi)\)**:
\[ f(\pi) = \cos(9\pi) + \cos(10\pi) \]
Since \(\cos(9\pi) = -1\) and \(\cos(10\pi) = 1\):
\[ f(\pi) = -1 + 1 = 0 \]

6. **Evaluate \(f(-\pi)\)**:
\[ f(-\pi) = \cos(-9\pi) + \cos(-10\pi) = \cos(9\pi) + \cos(10\pi) \]
This gives:
\[ f(-\pi) = -1 + 1 = 0 \]

7. **Evaluate \(f\left(\frac{\pi}{4}\right)\)**:
\[ f\left(\frac{\pi}{4}\right) = \cos\left(9 \cdot \frac{\pi}{4}\right) + \cos\left(10 \cdot \frac{\pi}{4}\right) \]
This simplifies to:
\[ = \cos\left(\frac{9\pi}{4}\right) + \cos\left(\frac{10\pi}{4}\right) = \cos\left(2\pi + \frac{\pi}{4}\right) + \cos\left(2\pi + \frac{3\pi}{2}\right) \]
Thus:
\[ = \cos\left(\frac{\pi}{4}\right) + \cos\left(\frac{3\pi}{2}\right) = \frac{1}{\sqrt{2}} + 0 = \frac{1}{\sqrt{2}} \]

### Summary of Results:
- \( f\left(\frac{\pi}{2}\right) = -1 \)
- \( f(\pi) = 0 \)
- \( f(-\pi) = 0 \)
- \( f\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)