$$f,g:N - \{ 1\} \to N$$ defined as

$$f(a) = \alpha $$, where $$\alpha$$ is the maximum power of those primes p such that p^$$\alpha$$ divides a.

$$g(a) = a + 1$$,

Now,

$$\matrix{ {f(2) = 1,} & {g(2) = 3} & \Rightarrow & {(f + g)\,(2) = 4} \cr {f(3) = 1,} & {g(3) = 4} & \Rightarrow & {(f + g)\,(3) = 5} \cr {f(4) = 2,} & {g(4) = 5} & \Rightarrow & {(f + g)\,(4) = 7} \cr {f(5) = 1,} & {g(5) = 6} & \Rightarrow & {(f + g)\,(5) = 7} \cr } $$

$$\because$$ $$(f + g)\,(5) = (f + g)\,(4)$$

$$\therefore$$ $$f + g$$ is not one-one

Now, $$\because$$ $${f_{\min }} = 1,\,{g_{\min }} = 3$$

So, there does not exist any $$x \in N - \{ 1\} $$ such that $$(f + g)(x) = 1,2,3$$

$$\therefore$$ $$f + g$$ is not onto