Let $(R,\mathfrak m,k)$ be a complete local Cohen-Macaulay ring of dimension $1$. The type of $R$ is then given by $\dim_k \text{Ext}^1_R(k,R)=\mu(\omega)$, where $\omega$ is the canonical module of $R$.
My question is: If $R$ has minimal multiplicity (i.e. $\mathfrak m^2=x\mathfrak m$ for some $x\in\mathfrak m$ i.e. $e(R)=\mu(\mathfrak m)$ ) and the type of $R$ equals $\mu(\mathfrak m)-1$, then is it true that $R$ must be Gorenstein?
No, it is not true. It is possible to construct several examples using numerical semigroups. A numerical semigroup $S$ is simply a submonoid of $\mathbb{N}$ such that $\mathbb{N}\setminus S$ is finite. It is easy to see that every numerical semigroup has a unique minimal system of generators, which is finite. In this case I write $S=\langle s_1, \dots, s_\mu\rangle=\{\lambda_1 s_1 + \dots +\lambda_\mu s_\mu \mid \lambda_1, \dots, \lambda_\mu \in \mathbb N\}$.
Consider the ring $k[[S]]=k[[t^{s_1}, \dots, t^{s_\mu}]] \subseteq k[[t]]$, where $k$ is a field and $t$ is an indeterminate. This is a complete local Cohen-Macaulay ring of dimension one (it is indeed a domain). It is well known that in these rings the multiplicity equals the smallest non-zero element of $S$, while the embedding dimension is equal to the number of minimal generators, which is $\mu$ for us. Moreover, if $k[[S]]$ has minimal multiplicity, then the type is always $\mu-1$. Hence, every $S$ with at least three minimal generators and such that $k[[S]]$ has minimal multiplicity is a counterxample to your question.
For instance, for every $n \geq 3$ the ring $k[[t^n, t^{n+1}, \dots, t^{2n-1}]]$ has multiplicity and embedding dimension equal to $n$, whereas its type is $n-1 > 1$, and then it is not Gorenstein.
