Solve this following

Question:

Let $\mathrm{N}$ be the set of natural numbers and two functions $f$ and $g$ be defined as $f, g: N \rightarrow N$

such that $: f(n)=\left(\begin{array}{ll}\frac{n+1}{2} & \text { if } n \text { is odd } \\ \frac{n}{2} & \text { if } n \text { is even }\end{array}\right.$

and $g(n)=n-(-1)^{n}$. The fog is :

 

  1. Both one-one and onto

  2. One-one but not onto

  3. Neither one-one nor onto

  4. onto but not one-one


Correct Option: , 4

Solution:

$f(x)= \begin{cases}\frac{n+1}{2} & n \text { is odd } \\ n / 2 & n \text { is even }\end{cases}$

$\mathrm{g}(\mathrm{x})=\mathrm{n}-(-1)^{\mathrm{n}} \begin{cases}\mathrm{n}+1 ; & \mathrm{n} \text { is odd } \\ \mathrm{n}-1 ; & \mathrm{n} \text { is even }\end{cases}$

$\mathrm{f}(\mathrm{g}(\mathrm{n}))= \begin{cases}\frac{\mathrm{n}}{2} ; & \mathrm{n} \text { is even } \\ \frac{\mathrm{n}+1}{2} ; & \mathrm{n} \text { is odd }\end{cases}$

$\therefore$ many one but onto

Option (4)

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