Take a tree for example
- False. Every vertex of tree (other than leaves) is a cut vertex.
- False. There is no cut-edge (an edge whose removal increases the number of connected components in graph) in $G_1 \cup G_2.$
- False. $G_1 \cup G_2$, $G_1$ and $G_2-$ all three graphs have same the chromatic number of $2$.
Now, we have given counter examples for options A,C and D. So, option B is the only possible answer and its proof is given at end.
Correct Answer: Option B
We are given that $G_1$ and $G_2$ are connected. So, if we take any two vertices say $v_i$ and $v_j$ there must be path between them in both $G_1$ and $G_2.$ Now, it is given that $G_1\cap G_2$ is disconnected. That is, we have at least two vertices $v_1$ and $v_2$ such that there is no path between them in $G_1 \cap G_2.$
This means the path between $v_1$ and $v_2$ in $G_1$ and $G_2$ are distinct.
When we have two distinct paths between a pair of vertices in a graph, it forms a cycle.