The value of c in Rolle's theorem for the function
`f (x) = cos 2 (x - (pi)/(4)) ` in `[0, (pi)/(2)]` is

Options

A. `pi/3`
B. `pi/6`
C. `pi/2`
D. `pi/4`

Answer

To find the value of \( c \) in Rolle's theorem for the function \( f(x) = \cos(2(x - \frac{\pi}{4})) \) in the interval \([0, \frac{\pi}{2}]\), we will follow these steps:

### Step 1: Verify the conditions of Rolle's Theorem
Rolle's theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

1. **Continuity**: The function \( f(x) = \cos(2(x - \frac{\pi}{4})) \) is a composition of continuous functions (cosine and linear functions), hence it is continuous on \([0, \frac{\pi}{2}]\).

2. **Differentiability**: The function is differentiable on \((0, \frac{\pi}{2})\) since cosine is differentiable everywhere.

3. **Equal values at endpoints**:
- Calculate \( f(0) \):
\[ f(0) = \cos(2(0 - \frac{\pi}{4})) = \cos(-\frac{\pi}{2}) = 0 \]
- Calculate \( f\left(\frac{\pi}{2}\right) \):
\[ f\left(\frac{\pi}{2}\right) = \cos\left(2\left(\frac{\pi}{2} - \frac{\pi}{4}\right)\right) = \cos\left(2 \cdot \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]
- Since \( f(0) = f\left(\frac{\pi}{2}\right) = 0 \), the third condition is satisfied.

### Step 2: Find \( f'(x) \)
Now we differentiate \( f(x) \):
\[ f(x) = \cos(2(x - \frac{\pi}{4})) = \cos(2x - \frac{\pi}{2}) \]
Using the derivative of cosine:
\[ f'(x) = -\sin(2x - \frac{\pi}{2}) \cdot \frac{d}{dx}(2x - \frac{\pi}{2}) = -\sin(2x - \frac{\pi}{2}) \cdot 2 \]
Using the identity \( \sin\left(x - \frac{\pi}{2}\right) = -\cos(x) \):
\[ f'(x) = -2(-\cos(2x)) = 2\cos(2x) \]

### Step 3: Set \( f'(c) = 0 \)
To find \( c \):
\[ f'(c) = 2\cos(2c) = 0 \]
This implies:
\[ \cos(2c) = 0 \]
The cosine function is zero at odd multiples of \( \frac{\pi}{2} \):
\[ 2c = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
We will consider \( n = 0 \) for the first solution within the interval:
\[ 2c = \frac{\pi}{2} \implies c = \frac{\pi}{4} \]

### Conclusion
The value of \( c \) in Rolle's theorem for the function \( f(x) = \cos(2(x - \frac{\pi}{4})) \) in the interval \([0, \frac{\pi}{2}]\) is:
\[ \boxed{\frac{\pi}{4}} \]