Explanation:

Let’s break it down:

- Rolle's theorem says that if a function is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there’s at least one c in (a, b) with f'(c) = 0.

- Here, f(x) = e? sin x on [0, p].

- At x = 0: f(0) = e° sin 0 = 0

- At x = p: f(p) = e^p sin p = 0

- f(0) = f(p), and the function is continuous and differentiable everywhere. Rolle’s theorem applies.

Now, let’s find c where f'(c) = 0:

- f(x) = e? sin x

- f'(x) = e? sin x + e? cos x = e? (sin x + cos x)

- Set f'(x) = 0

? e? (sin x + cos x) = 0

? sin x + cos x = 0 (since e? ? 0)

- So, sin x = –cos x

? tan x = –1

Where does that happen in [0, p]?

- x = 3p/4, since tan(3p/4) = –1

Now the options:

1. p/6 — Nope, that’s not it.

2. p/4 — Not this one either.

3. p/2 — tan(p/2) is undefined.

4. 3p/4 — That’s the spot.

Correct option: Option 4, 3p/4.