Understand the domain and codomain

We have

Interpret $f(a)$

Write the prime factorization of $a$ as $a=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}.$ Then $f(a)$ is the maximum exponent among $e_1,e_2,\dots,e_k$.

For example:

• $12=2^2\cdot 3 \Rightarrow f(12)=2$
• $72=2^3\cdot 3^2 \Rightarrow f(72)=3$
• if $a$ is square-free, then $f(a)=1$

Compute $(f+g)(a)$

Since $g(a)=a+1,$ we get $(f+g)(a)=f(a)+g(a)=f(a)+a+1.$

Check whether $f+g$ is one-one

To be one-one, different inputs must give different outputs.

Let us test small values:

• For $a=2$: since $2=2^1$, $f(2)=1,$ so $(f+g)(2)=1+(2+1)=4.$
• For $a=3$: since $3=3^1$, $f(3)=1,$ so $(f+g)(3)=1+(3+1)=5.$
• For $a=4$: since $4=2^2$, $f(4)=2,$ so $(f+g)(4)=2+(4+1)=7.$
• For $a=5$: since $5=5^1$, $f(5)=1,$ so $(f+g)(5)=1+(5+1)=7.$

Thus, $(f+g)(4)=(f+g)(5)=7,$ but $4\ne 5$.

Therefore, $f+g$ is not one-one.

Check whether $f+g$ is onto

We need to see whether every natural number is attained.

For any $a\ge 2$, we have $f(a)\ge 1$ because every integer $a>1$ has at least one prime factor with exponent at least $1$.

Hence, $(f+g)(a)=a+1+f(a)\ge a+1+1=a+2.$ Since $a\ge 2$, we get $(f+g)(a)\ge 4.$

So the values $1,2,3$ are never attained.

Therefore, $f+g$ is not onto as a function into $\mathbb N$.

Conclusion

The function $f+g$ is:

• not one-one
• not onto

Hence the correct option is $\boxed{\text{D: neither one-one nor onto}}.$