Let {N} denote the set of all natural numbers, and {Z} denote the set of all integers. Consider the functions f: {N} {Z} and g: {Z} {N} defined by f(n)={cases} n+1/2 & {if } n { is odd} \\ 4-n/2 & {if } n { is even} {cases} and g(n)={cases} 3+2n & {if } n ≥ 0 \\ -2n & {if } n Define (g f)(n)=g(f(n)) for all n {N}, and (f g)(n)=f(g(n)) for all n {Z}. Then which of the following statements is (are) TRUE? [JEE Adv 2025 P1]

g f is NOT one-one and g f is NOT onto

f g is NOT one-one but f g is onto

Test Index

Functions

Section: Mathematics

Question : 3 of 30

Marks: +1, -0

Let

\mathbb{N}

\mathbb{Z}

f: \mathbb{N} \longrightarrow \mathbb{Z}

g: \mathbb{Z} \longrightarrow \mathbb{N}

f(n)=\begin{cases} \frac{n+1}{2} & \text{if } n \text{ is odd} \\ \frac{4-n}{2} & \text{if } n \text{ is even} \end{cases}

g(n)=\begin{cases} 3+2n & \text{if } n \geq 0 \\ -2n & \text{if } n < 0 \end{cases}

(g \circ f)(n)=g(f(n))

n \in \mathbb{N}

(f \circ g)(n)=f(g(n))

n \in \mathbb{Z}

Solution:

f(n)= \begin{cases} \frac{n+1}{2}, & n=2k+1 \\ \frac{4-n}{2}, & n=2k \end{cases}

= \begin{cases} k+1, & k \in \mathbb{N} \cup \{0\} \\ 2-k, & k \in \mathbb{N} \end{cases}

k=0, f(1)=1=f(2)=1 \Rightarrow f

is not one-one

2-k

covers all integers

\{1,0,-1,\ldots\}

k+1

covers

\{1,2,3,\ldots\}

\Rightarrow f(n)

covers all integers

\Rightarrow f(n)

is onto but not one-one

g(n)= \begin{cases} 3+2n, & n \geq 0 \\ -2n, & n < 0 \end{cases}

Notice that

1 \notin

range of

g(x)

3+2n \neq 1, n \geq 0

and

-2n \neq 1, n < 0

\Rightarrow g(n)

is not onto

g(f(n))= \begin{cases} 3+2f(n), & f(n) \geq 0 \\ -2f(n), & f(n) < 0 \end{cases}

= \begin{cases} 3+2\left(\frac{n+1}{2}\right), & n=2k+1 \\ 3+2\left(\frac{4-n}{2}\right), & n=2k \end{cases}

= \begin{cases} n+4, & n=2k+1 \\ 7-n, & n=2k \end{cases}

g(f(n))

is always odd

\Rightarrow

not onto

g(f(1))=5=g(f(2)) \Rightarrow 0

not one-one

f(g(n))= \begin{cases} \frac{g(n)+1}{2}, & g(n) \text{ is odd natural} \\ \frac{4-g(n)}{2}, & g(n) \text{ is even natural} \end{cases}

= \begin{cases} n+2, & n \geq 0 \\ n+2, & n < 0 \end{cases}

\Rightarrow \;\; (n+2) \forall n \Rightarrow f(g(n)

is one-one

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