Let [x]= greatest integer ` le x` and `f(x) = cos([pi^(2)]x) + sin([e^(2)]x)` then `f(pi//4)` is equal to

Answer

To solve the problem, we need to evaluate the function \( f(x) = \cos(\pi^2 [x]) + \sin(e^2 [x]) \) at \( x = \frac{\pi}{4} \).

### Step-by-Step Solution:

1. **Identify the greatest integer function**:
The greatest integer function \( [x] \) gives the largest integer less than or equal to \( x \). For \( x = \frac{\pi}{4} \):
\[ \frac{\pi}{4} \approx \frac{3.14}{4} \approx 0.785 \]
Thus, \( [\frac{\pi}{4}] = 0 \).

2. **Calculate \( \pi^2 \) and \( e^2 \)**:
- \( \pi^2 \approx 9.87 \) (since \( \pi \approx 3.14 \))
- \( e^2 \approx 7.39 \) (since \( e \approx 2.71 \))

3. **Evaluate \( [\pi^2] \) and \( [e^2] \)**:
- \( [\pi^2] = 9 \) (the greatest integer less than or equal to \( 9.87 \))
- \( [e^2] = 7 \) (the greatest integer less than or equal to \( 7.39 \))

4. **Substitute into the function**:
Now substitute \( [\frac{\pi}{4}] = 0 \) into the function:
\[ f\left(\frac{\pi}{4}\right) = \cos(\pi^2 [\frac{\pi}{4}]) + \sin(e^2 [\frac{\pi}{4}]) \]
This simplifies to:
\[ f\left(\frac{\pi}{4}\right) = \cos(\pi^2 \cdot 0) + \sin(e^2 \cdot 0) \]

5. **Evaluate the trigonometric functions**:
- \( \cos(0) = 1 \)
- \( \sin(0) = 0 \)

6. **Combine the results**:
Therefore,
\[ f\left(\frac{\pi}{4}\right) = 1 + 0 = 1 \]

### Final Answer:
\[ f\left(\frac{\pi}{4}\right) = 1 \]