We have to recall couple of theorems before solving this question.
- Number of odd degree vertices in any connected component of a graph is even.
- The degree sum of all the vertices in any connected component of a graph is even.
Now in this question graph $G$ has $n-2$ vertices with degree $10,$ one vertex with degree $1.$
Because of theorem (1), vertex $n$ and vertex $1$ should be in the same connected component.
Proof: Lets suppose vertex $1$ and vertex $n$ are not in the same connected component. Also assume that vertex $1$ is in component $C_1.$ Now $C_1$ has only one vertex with odd degree which is vertex $1,$ because vertex $2$ to vertex $n-1$ has even degree (10) and this violates theorem (1). So, $C_1$ must have vertex $1$ as well as vertex $n$ and vertex $n$ should have odd degree.
Now, since vertex $1$ and vertex $n$ are in the same connected component and the given graph is undirected so, there must be a path from vertex $1$ to vertex $n.$ Hence option (a) is true.
Option (b) is false. Proof by counter example:
Lets take a graph $G$ where $n=13$ with two connected components $C_1$ and $C_2.$ $C_1$ has two vertices vertex $1$ and vertex $n$ and only one edge $(\text{vertex } 1,\text{vertex } n).$ Hence, degree of vertex $1$ as well as vertex $n$ is $1.$ $C_2$ has rest $11$ vertices (vertex $2$ to vertex $n-1)$ and $C_2$ is a complete component (there is exactly one edge between every pair of vertices of $C_2),$ hence degree of vertex $2,\ldots,$ vertex $n-1$ is $10.$ In this graph there is no path from vertex $1$ to vertex $2,\ldots,$ vertex $n-1.$.
Option (c) is False. Proof by counter example:
Lets take graph $G$ where $n=12$ and each pair of vertices from vertex $2$ to vertex $n$ are connected with each other by an edge and additionally there is an ede between vertex $n$ and vertex $1.$ Hence, vertex $1$ has degree $1$ and vertex $2,\ldots,$ vertex $n-1$ has degree $10$ and vertex $n$ has degree $11.$ In this graph vertex $n$ has degree $11,$ and hence option C is wrong.
Option (d) is false.
Proof: As we have seen an example as part of proving that option (b) is wrong where, graph $G$ can be disconnected and a disconnected graph has infinite diameter. Hence, option (d) is wrong.
Option (e) is false since options $b,c,d$ are false.