Question 14

1. From start state $s$:
   • Reading $0$ transitions to state $r$, which is a dead (trap) state with self-loops on $0, 1$.
   • Reading $1$ transitions to state $p$, which is the only final (accepting) state.
2. From accepting state $p$:
   • Reading $0$ stays in $p$ (self-loop).
   • Reading $1$ transitions to state $q$.
3. From state $q$:
   • Reading $0$ transitions to dead state $r$.
   • Reading $1$ transitions back to state $p$.
4. Therefore, every accepted string must start with $1$ (to reach $p$). Once at $p$, any sequence composed of single $0$'s (staying at $p$) or blocks of $11$ ($p \xrightarrow{1} q \xrightarrow{1} p$) returns to $p$. Any other symbol sequence leads irreversibly to dead state $r$.

Thus, the language accepted is $1(0 + 11)^*$.