option D should be wrong, and Option C should be right !
@Smishra95
Option C if graph is disconnected then there is no path ib between 1 And N .
graph may be disconnected., But the n^{th} vertex should be connected to the component which is the 1^{st} vertex is connected !
WHY ?
let there are r components.... let the 1^{st} vertex connected component is C_{1}.
We all know that each component is a sub-graph, means it is also a graph.
therefore C_{1} is also a graph... 1st vertex is odd degree vertex, then there should be exist one more odd degree ( in worst case ) ===> all vertices have 4 as degree ===> n^{th} vertex should be connected to C1 only.
Therefore a path should be exist between them.
if you didn't get this try to draw a graph without n^{th} vertex connecting to C1.