1. The problem asks us to describe the language accepted by the finite automaton M1. 2. From the description, M1 has three states: $q_1$ (start state), $q_2$ (accept state), and $q_3$. 3. Transitions: - From $q_1$ to $q_2$ on input $1$. - $q_2$ loops to itself on input $1$. - From $q_2$ to $q_3$ on input $0$. - From $q_3$ back to $q_2$ on input $0$ or $1$. 4. Since $q_2$ is the only accept state, strings must end in $q_2$ to be accepted. 5. To reach $q_2$ from $q_1$, the string must start with $1$. 6. From $q_2$, the machine can read any number of $1$s (due to the self-loop). 7. If a $0$ is read, the machine moves to $q_3$, and from $q_3$ it can return to $q_2$ on either $0$ or $1$. 8. This means after the initial $1$, the string can have any combination of $1$s and blocks of $0$s and $1$s that move between $q_2$ and $q_3$, but must end in $q_2$. 9. Formally, the language consists of strings starting with $1$, followed by any number (including zero) of substrings of the form $1^*$ or $(0(0|1)^*1^*)$ that keep the machine in $q_2$ or $q_3$, but ultimately end in $q_2$. 10. Simplifying, the language is all strings over $\{0,1\}$ that start with $1$ and do not end with a $0$ (since ending in $q_3$ is not accepting), allowing any combination of $0$s and $1$s after the initial $1$. Final answer: The language accepted by M1 is all strings over $\{0,1\}$ that start with $1$ and do not end with $0$.