Correct Answer - A::B::C
`r_(n)propn^(2) "therefore"` graph between `r_(n)`,n will be parabola
`I_(n)propn^(2) Rightarrow r_(n)/r_(1)=2`log n hence the graph is a straigth line passing through origin
`r_(n)propn^(2)RightarrowA_(n)propn^(4)RightarrowA_(n)/A_(1)=n^(n)`
` log(A_(n)/A_(1))=4log` n: i.e the graph is a straight line passing through origin.
`fprop1/n^(3) Rightarrow f_(n)/f_(1)Rightarrow1^(3)/n^(3)Rightarrow(f_(n)/f_(1))=-3logn`brgt i.e the graph is a straight line passing origin with negative slope.