Given, f(n+1)=f(n)+f(1),∀n∈N ⇒f(n+1)−f(n)=f(1) It is an AP with common difference =f(1) Also, general term ll=Tn=f(1)+(n−)f(1)=nf(1) ⇒f(n)=nf(1) Clearly, f(n) is one-one. For fog to be one-one, g must be one-one. For f to be onto, f(n) should take all the values of natural numbers. As, f(x) is increasing, f(1)=1 ⇒f(n)=n If g is many-one, then fog is many one. So, if g is onto, then fog is one-one.