Mathematics - Functions Question with Solution | TestHub

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

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

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

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

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

Now, again we can see $f_{\min }=1,g_{\min }=3$

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

$\therefore f+g$ is not onto

So function is neither one-one nor onto.