Given,
f,g:N−1→N defined as
f(a)=α, where α is the maximum power of those primes p such that pα divides a.
g(a)=a+1
Now, f(2)=1,g(2)=3⇒(f+g)(2)=4
f(3)=1,g(3)=4⇒(f+g)(3)=5
f(4)=2,g(4)=5⇒(f+g)(4)=7
f(5)=1,g(5)=6⇒(f+g)(5)=7
We can see (f+g)(5)=(f+g)(4), sof+g is not one-one function
Now, again we can see fmin=1,gmin=3
So, there does not exist any x∈N−1 such that (f+g)(x)=1,2,3
∴f+g is not onto
So function is neither one-one nor onto.