f(n+1)−f(n)=f(1)
A.P. with common difference = f(1)
General term = Tn=f(1)+(n−1)f(1)=nf(1)
⇒f(n)=nf(1)⇒f is one-one
Now, Let f(g(x2))=f(g(x1))
⇒g(x2)=g(x1) (as f is one-one)
⇒x1=x2 (as fog is one-one)
⇒g is 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
Now, f(g(n))=g(n)f(1)
may be many-one if
g(n) is many-one
So “if g is onto, then fog is one-one” is incorrect.