Answer is: option4

\( \frac{1}{2} \) and \( \frac{3}{2} \)

Solution:

Rolle’s Theorem states that if a function \( f(x) \) satisfies the following three conditions on an interval \([a, b]\):

1. Continuity on \([a, b]\): The function \( f(x) = \sin(\pi x) \) is continuous everywhere, including on \([0,2]\).
2. Differentiability on \((a, b)\): The derivative of \( f(x) \), given by \( f'(x) = \pi \cos(\pi x) \), exists for all \( x \), so it is differentiable on \((0,2)\).
3. Equal Function Values at Endpoints:
   • \( f(0) = \sin(0) = 0 \)
   • \( f(2) = \sin(2\pi) = 0 \)

Since \( f(0) = f(2) \), all three conditions are satisfied.

Finding \( c \) where \( f'(c) = 0 \)

The derivative of \( f(x) \) is:

\[ f'(x) = \pi \cos(\pi x) \]

Setting \( f'(c) = 0 \):

\[ \pi \cos(\pi c) = 0 \]

Dividing by \( \pi \):

\[ \cos(\pi c) = 0 \]

The cosine function is zero at:

\[ \pi c = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \]

Solving for \( c \):

\[ c = \frac{1}{2}, \frac{3}{2} \]

Since \( c \) must be in the open interval \( (0,2) \), both \( c = \frac{1}{2} \) and \( c = \frac{3}{2} \) are valid.