JEE Main 2021 — Functions Question with Solution
From: JEE Main 2021 (Online) 25th July Morning Shift
Question
Let g : N N be defined as
g(3n + 1) = 3n + 2,
g(3n + 2) = 3n + 3,
g(3n + 3) = 3n + 1, for all n 0.
Then which of the following statements is true?
g(3n + 1) = 3n + 2,
g(3n + 2) = 3n + 3,
g(3n + 3) = 3n + 1, for all n 0.
Then which of the following statements is true?
Choose an option
Show full solutionCorrect option: A
Correct answer
AThere exists an onto function f : N N such that fog = f
Step-by-step explanation
g : N N
g(3n + 1) = 3n + 2,
g(3n + 2) = 3n + 3,
g(3n + 3) = 3n + 1
g(x) = \left[ {\matrix{ {x + 1} & {x = 3k + 1} \cr {x + 1} & {x = 3k + 2} \cr {x - 2} & {x = 3k + 3} \cr } } \right.
g\left( {g(x)} \right) = \left[ {\matrix{ {x + 2} & {x = 3k + 1} \cr {x - 1} & {x = 3k + 2} \cr {x - 1} & {x = 3k + 3} \cr } } \right.
g\left( {g\left( {g\left( x \right)} \right)} \right) = \left[ {\matrix{ x & {x = 3k + 1} \cr x & {x = 3k + 2} \cr x & {x = 3k + 3} \cr } } \right.
If f : N N, if is a one-one function such that f(g(x)) = f(x) g(x) = x, which is not the case
If f : N N f is an onto function
such that f(g(x)) = f(x),
one possibility is
f(x) = \left[ {\matrix{ x & {x = 3n + 1} \cr x & {x = 3n + 2} \cr x & {x = 3n + 3} \cr } } \right. nN0
Here f(x) is onto, also f(g(x)) = f(x) xN
g(3n + 1) = 3n + 2,
g(3n + 2) = 3n + 3,
g(3n + 3) = 3n + 1
g(x) = \left[ {\matrix{ {x + 1} & {x = 3k + 1} \cr {x + 1} & {x = 3k + 2} \cr {x - 2} & {x = 3k + 3} \cr } } \right.
g\left( {g(x)} \right) = \left[ {\matrix{ {x + 2} & {x = 3k + 1} \cr {x - 1} & {x = 3k + 2} \cr {x - 1} & {x = 3k + 3} \cr } } \right.
g\left( {g\left( {g\left( x \right)} \right)} \right) = \left[ {\matrix{ x & {x = 3k + 1} \cr x & {x = 3k + 2} \cr x & {x = 3k + 3} \cr } } \right.
If f : N N, if is a one-one function such that f(g(x)) = f(x) g(x) = x, which is not the case
If f : N N f is an onto function
such that f(g(x)) = f(x),
one possibility is
f(x) = \left[ {\matrix{ x & {x = 3n + 1} \cr x & {x = 3n + 2} \cr x & {x = 3n + 3} \cr } } \right. nN0
Here f(x) is onto, also f(g(x)) = f(x) xN
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Functions chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2021, covering the Functions chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.