Let $f(x) = 2 + \cos x$ for all real $x$. \ STATEMENT I : For each real $t$, there exists a point $c$ in $[t, t + \pi]$ such that $f'(c) = 0$ because \ STATEMENT II : $f(t) = f(t + 2\pi)$ for each real $t$.

$f(x)=2+\cos x$ $\text{I: } \forall\, t\ \exists\, c\in[t,\,t+\pi]:\ f'(c)=0$ $\text{II: } f(t)=f(t+2\pi)$ $(1)\ \text{I True, II True; II explains I} \quad (2)\ \text{I True, II True; II not the explanation}$ $(3)\ \text{I True, II False} \qquad (4)\ \text{I False, II True}$ $\text{Plan: } f'\to \text{check I}\to \text{check II}\to \text{link}$ $f'(x)=-\sin x$ $f'(c)=0 \iff \sin c=0 \iff c=n\pi,\ n\in\mathbb{Z}$ $\text{zeros of } f' \text{ are spaced } \pi \text{ apart}$ $t\in[n\pi,(n+1)\pi] \Rightarrow (n+1)\pi\in[t,t+\pi]$ $\Rightarrow \text{Statement I is } \textbf{TRUE}$ $f(t+2\pi)=2+\cos(t+2\pi)=2+\cos t=f(t)$ $\Rightarrow \text{Statement II is } \textbf{TRUE}$ $\text{II says: period } 2\pi;\quad \text{I needs a critical point in length-}\pi \text{ window}$ $\text{Rolle with } f(t)=f(t+2\pi)\Rightarrow c\in(t,t+2\pi),\ \text{not } (t,t+\pi)$ $\Rightarrow \text{II does not explain I}$ $\text{actual reason: zeros of } f'=-\sin x \text{ are } \pi\text{-spaced}$ $\text{I True, II True; II not the correct explanation} \Rightarrow (2)$ $\boxed{(2)}$