Let f, g: ℕ - \ 1\ → ℕ be functions defined by f(a) = α , where α...

Given,

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

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

$g\left(a\right) = a + 1$

Now, $f\left(2\right) = 1, g\left(2\right) = 3 ⇒ \left(f + g\right)\left(2\right) = 4$

$f\left(3\right) = 1, g\left(3\right) = 4 ⇒ \left(f + g\right)\left(3\right) = 5$

$f\left(4\right) = 2, g\left(4\right) = 5 ⇒ \left(f + g\right)\left(4\right) = 7$

$f\left(5\right) = 1, g\left(5\right) = 6 ⇒ \left(f + g\right)\left(5\right) = 7$

We can see \left(f + g\right)\left(5\right) = \left(f + g\right)\left(4\right)$, so$f + g$is not one-one function</p><p>Now, again we can see${f}_\min = 1, {g}_\min = 3$</p><p>So, there does not exist any$x \in N - \left\{1\right\} such that \left(f + g\right)\left(x\right) = 1, 2, 3$</p><p>$∴ f + g is not onto

So function is neither one-one nor onto.