(a) Theorem: The language $B = \setbuilder{1^k y}{y \in \set{0, 1}^* \tand \#_1(y) \geq k \text{ for some } k \geq 1}$ is a regular language.

---

Proof: Notice that strings in $B$ must start with $1$, since $k \geq 1$. Also, any string of the form $1x$ will certainly be in $B$ as long as $x$ contains at least one 1. In other words, we have $B = \setbuilder{1x}{\#_1(x) \geq 1}$. So, the following DFA accepts $B$:

(b) Theorem: The language $C = \setbuilder{1^k y}{y \in \set{0, 1}^* \tand \#_1(y) \leq k \text{ for some } k \geq 1}$ is not a regular language.

---

Proof: Suppose $C$ were regular and let $p$ be the pumping length. Take $z = 1^p 0 1^p$, noting that $z \in C$ and $|z| \geq p$. Then let $z = uvw$ with $|uv| \leq p$ and $|v| \geq 1$.

Because $|uv| \leq p$, and $uv$ is a prefix of $z$, it must be the case that $u$ and $v$ consist entirely of $1$'s. In particular, $v = 1^k$ for some $k \geq 1$. But then $u v^0 w = 1^{(p-k)} 0 1^p$, which is not in $C$, contradicting the pumping lemma, so $C$ is not regular.