Functions Ques 15

  1. Let $N$ be the set of natural numbers and two functions $f$ and $g$ be defined as $f, g: N \longrightarrow N$ such that $f(n)= \begin{cases}\frac{n+1}{2} ; & \text { if } n \text { is odd } \ \frac{n}{2} ; & \text { if } n \text { is even }\end{cases}$
    and $g(n)=n-(-1)^n$. Then, fog is

(2019 Main, 10 Jan II)

(a) one-one but not onto

(b) onto but not one-one

(c) both one-one and onto

(d) neither one-one nor onto

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Answer:

Correct Answer: 15.(b)

Solution: (b) Given, $f(n)= \begin{cases}\frac{n+1}{2}, & \text { if } n \text { is odd } \\ \frac{n}{2}, & \text { if } n \text { is even, }\end{cases}$

and $\quad g(n)=n-(-1)^n=\left\{\begin{array}{l}n+1, \text { if } n \text { is odd } \\ n-1, \text { if } n \text { is even }\end{array}\right.$

Now, $f(g(n))=\left\{\begin{array}{l}f(n+1), \text { if } n \text { is odd } \\ f(n-1), \text { if } n \text { is even }\end{array}\right.$

$=\left\{\begin{array}{l}\frac{n+1}{2}, \text { if } n \text { is odd } \\ \frac{n-1+1}{2}=\frac{n}{2}, \text { if } n \text { is even }\end{array}\right.$

$=f(x)$

$[\because \quad $ if $n$ is odd, then $(n+1)$ is even and if $n$ is even, then $(n-1)$ is odd]

Clearly, function is not one-one as $f(2)=f(1)=1$

But it is onto function.

$[\because \quad $ If $m \in N$ (codomain) is odd, then $2 m \in N$ (domain) such that $f(2 m)=m$ and

if $m \in N$ codomain is even, then

$2 m-1 \in N$ (domain) such that $f(2 m-1)=m]$

$\therefore \quad $ Function is onto but not one-one



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