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JEE Advanced 2025Paper 1
Exam
JEE Advanced
Year
2025
Paper
Paper 1
Subject
Mathematics
Chapter
Functions

JEE Advanced 2025 · Paper 1 · Mathematics · Functions

Let ℕ denote the set of all natural numbers, and ℤ denote the set...

Let N\mathbb{N} denote the set of all natural numbers, and Z\mathbb{Z} denote the set of all integers. Consider the functions f:N→Zf: \mathbb{N} \rightarrow \mathbb{Z} and g:Z→Ng: \mathbb{Z} \rightarrow \mathbb{N} defined by f(n)=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 $g(n)=\left\{\begin{array}{cc}3+2 n & \text{\text{ if } } n \geq 0 \\ -2 n & \text{\text{ if } } n <0\end{array}$\right. Define (g∘f)(n)=g(f(n))(g \circ f)(n)=g(f(n)) for all n∈Nn \in \mathbb{N}, and (f∘g)(n)=f(g(n))(f \circ g)(n)=f(g(n)) for all n∈Zn \in \mathbb{Z}. Then which of the following statements is (are) TRUE?

Options

  1. Ag∘fg \circ f is NOT one-one and g∘fg \circ f is NOT onto
  2. Bf∘ gf \circ \mathrm{~g} is NOT one-one but f∘gf \circ g is onto
  3. Cgg is one-one and gg is onto
  4. Dff is NOT one-one but ff is onto

Correct answer

A. g∘fg \circ f is NOT one-one and g∘fg \circ f is NOT onto

Step-by-step solution

f(n)={(n+1)/2 if n is odd (4−n)/2 if n is even f(n)={(1,1),(2,1),(3,2),(4,0),(5,3),(6,−1),…}\begin{aligned} & 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\} \end{aligned}\therefore \mathrm{f}(\mathrm{n})ismanyoneandontofunctionis many one and onto functiong(n)={3+2n if  n≥0−2n if  n<0g(n)={(−3,6),(−2,4),(−1,2),(0,3),(1,5),(2,7),(3,9),(4,15),…}\begin{aligned} & \mathrm{g}(\mathrm{n})=\left\{\begin{array}{cc} 3+2 \mathrm{n} & \text{\text{ if } } \mathrm{n} \geq 0 \\ -2 \mathrm{n} & \text{\text{ if } } \mathrm{n} <0 \end{array}\right. \\ & \mathrm{g}(\mathrm{n})=\{(-3,6),(-2,4),(-1,2),(0,3),(1,5),(2,7),(3,9),(4,15), \ldots\} \end{aligned}∴g\therefore \mathrm{g}(n)\mathrm{n}) is one-one and into function f\mathrm{f}( g\mathrm{~g}(n\mathrm{n}))=2+n\mathrm{n}, n\mathrm{n} \in  N\mathrm{~N} fog is one-one and into g(f(n))=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∴g(f(2))=g(f(1))=5 \therefore gof is many one and into
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