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JEE Main 202125 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
25 July
Shift
Morning Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2021 · 25 July · Morning Shift · Mathematics · Functions

Let g: N → N be defined as g ( 3n + 1 ) = 3n + 2 g ( 3n + 2 ) = 3n...

Let g:N→Ng: N \to N be defined as

g(3n+1)=3n+2g ( 3n + 1 ) = 3n + 2

g(3n+2)=3n+3g ( 3n + 2 ) = 3n + 3

g(3n+3)=3n+1g ( 3n + 3 ) = 3n + 1, for all n≥0n \ge 0

Then which of the following statements is true ?

Options

  1. AThere exists an onto function f:N→Nf: N \to N such that fog=ffog = f
  2. BThere exists a one-one function f:N→Nf: N \to N such
    that fog=ffog = f
  3. Cgogog=ggogog = g
  4. DThere exists a function f:N→Nf: N \to N such that gof=fgof = f

Correct answer

A. There exists an onto function f:N→Nf: N \to N such that fog=ffog = f

Step-by-step solution

g: N \to N</p><p></p><p>g ( 3n + 1 ) = 3n + 2</p><p></p><p>g ( 3n + 2 ) = 3n + 3</p><p></p><p>g ( 3n + 3 ) = 3n + 1</p><p></p><p>g\left(x\right) = \left\{x+1;x=3k+1x+1;x=3k+2x−2;x=3k+3\begin{array}{ll}x + 1; x = 3k + 1 \\ x + 1; & x = 3k + 2 \\ x - 2; & x = 3k + 3\end{array}\right)

g\left(g\left(x\right)\right) = \left\{x+2;x=3k+1x−1;x=3k+2x−1;x=3k+3\begin{array}{ll}x + 2; x = 3k + 1 \\ x - 1; & x = 3k + 2 \\ x - 1; & x = 3k + 3\end{array}\right)</p><p>g\left(g\left(g\left(x\right)\right)\right) = \left\{{x;x=3k+1x;x=3k+2x;x=3k+3\begin{cases}x; & x = 3k + 1 \\ x; & x = 3k + 2 \\ x; & x = 3k + 3\end{cases}\right)

If f:N→N,ff: N \to N, f is a one-one function such that f(g(x))=f(x)⇒g(x)=xf ( g ( x ) ) = f ( x ) ⇒ g ( x ) = x, which is not the case

If f:N→N,ff: N \to N, f is an onto function such that f(g(x))=f(x)f ( g ( x ) ) = f ( x ) one possibility is

f\left(x\right) = \left\{\begin{cases}n; & x = 3n + 1 \\ n; & x = 3n + 2 \\ n; & x = 3n + 3\end{cases}\right; n \in N

Here f(x)f\left(x\right) is onto, also f(g(x))=f(x)∀x∈Nf ( g ( x ) ) = f ( x ) \forall x \in N

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