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

Options

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

Answer

To find the value of \( c \) in Rolle's theorem for the function \( f(x) = \cos(2x) \) on the interval \([0, \pi]\), 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 \([a, b]\) and 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 \( f(x) = \cos(2x) \) is a trigonometric function, which is continuous everywhere. Hence, it is continuous on \([0, \pi]\).

2. **Differentiability**: The function \( f(x) = \cos(2x) \) is also differentiable everywhere, thus it is differentiable on \((0, \pi)\).

3. **Equal values at endpoints**: We need to check if \( f(0) = f(\pi) \):
\[ f(0) = \cos(2 \cdot 0) = \cos(0) = 1 \]
\[ f(\pi) = \cos(2 \cdot \pi) = \cos(2\pi) = 1 \]
Since \( f(0) = f(\pi) \), the conditions of Rolle's Theorem are satisfied.

### Step 2: Find the derivative of \( f(x) \)
Next, we calculate the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\cos(2x)) = -2\sin(2x) \]

### Step 3: Set the derivative equal to zero
Now, we need to find \( c \) such that \( f'(c) = 0 \):
\[ -2\sin(2c) = 0 \]
This simplifies to:
\[ \sin(2c) = 0 \]

### Step 4: Solve for \( c \)
The solutions to \( \sin(2c) = 0 \) occur when:
\[ 2c = n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus,
\[ c = \frac{n\pi}{2} \]

### Step 5: Determine the valid values of \( c \)
Since we are looking for \( c \) in the interval \( (0, \pi) \):
- For \( n = 0 \): \( c = 0 \) (not in the open interval)
- For \( n = 1 \): \( c = \frac{\pi}{2} \) (valid)
- For \( n = 2 \): \( c = \pi \) (not in the open interval)

Thus, the only valid solution for \( c \) in the interval \( (0, \pi) \) is:
\[ c = \frac{\pi}{2} \]

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