Let mathbb N denote the set of all natural numbers, and mathbb Z denote the set of all integers. Consider the functions f: mathbb N rightarrow mathbb Z and g: mathbb Z rightarrow mathbb N defined by f(n)= begin cases (n+1) / 2 & if n is odd \\ (4-n) / 2 & if n is even end cases and g(n)= \ begin array cc 3+2 n & if n geq 0 \\ -2 n & if n<0 end array . Define (g ° f)(n)=g(f(n)) for all n in mathbb N , and (f ° g)(n)=f(g(n)) for all n in mathbb Z . Then which of the following statements is (are) TRUE ?

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

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

Mathematics · Functions

JEE Advanced 2025 — Paper 1 — Question 20

Let $\mathbb{N}$ denote the set of all natural numbers, and $\mathbb{Z}$ denote the set of all integers. Consider the functions $f: \mathbb{N} \rightarrow \mathbb{Z}$ and $g: \mathbb{Z} \rightarrow \mathbb{N}$ defined by $f(n)= \begin{cases}(n+1) / 2 & \text { if } n \text { is odd } \\ (4-n) / 2 & \text { if } n \text { is even }\end{cases} $$ and $$ g(n)=\left\{\begin{array}{cc} 3+2 n & \text { if } n \geq 0 \\ -2 n & \text { if } n<0 \end{array}\right.$

Define $(g \circ f)(n)=g(f(n))$ for all $n \in \mathbb{N}$ , and $(f \circ g)(n)=f(g(n))$ for all $n \in \mathbb{Z}$ . Then which of the following statements is (are) TRUE ?

Option A:
$g \circ f$ is NOT one-one and $g \circ f$ is NOT onto
Correct
Option B:
$f \circ g$ is NOT one-one but $f \circ g$ is onto
Option C:
$g$ is one-one and $g$ is onto
Option D:
$f$ is NOT one-one but $f$ is onto
Correct

Answer: A, D

Step-by-step solution

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

$f(n)=\{(1,1),(2,1),(3,2),(4,0),(5,3),(6,-1), \ldots\}$

$\therefore f(n)$ is many one and onto function

$g(n)=\left\{\begin{array}{cc}3+2 n & \text { if } n \geq 0 \\ -2 n & \text { if } n<0\end{array}\right.$

$\mathrm{g}(\mathrm{n})=\{(-3,6),(-2,4),(-1,2),(0,3),(1,5),(2,7),(3,9),(1,45), \ldots\}$

$\therefore \mathrm{g}(\mathrm{n})$ is one-one and into function

$\mathrm{f}(\mathrm{g}(\mathrm{n}))=2+\mathrm{n}, \mathrm{n} \in \mathrm{N}$

fog is one-one and into

$g(f(n))= \begin{cases}4+n & \text { if } n \text { is odd natural number } \\ 7-n & \text { if } n=2,4 \\ n-4 & \text { if } n \text { is even natural number and } n \geq 6\end{cases}$ $g(f(2))=g(f(1))=5$

$\therefore$ gof is many one and into

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2025
Paper: Paper 1
Subject: Mathematics
Chapter: Functions
Topic: One-One, many-one, onto, into, bijective functions

Answer: A, D

Step-by-step solution

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