Let g: N arrow N be defined as g(3 n+1)=3 n+2 g(3 n+2)=3 n+3 g(3 n+3)=3 n+1, for all n ≥ 0 Then which of the following statements is true? [25 Jul 2021 Shift 1]

Correct Answer is : There exists an onto function f:N→Nf: N arrow Nf:N→N such that fog =f=f=f

There exists an onto function f: N arrow N such that fog =f

There exists a one-one function f: N arrow N such that fog =f

There exists a function f: N arrow N such that gof =f

Correct Answer is : There exists an onto function f:N→Nf: N arrow Nf:N→N such that fog =f=f=f

Test Index

Functions Part 1

Section: Mathematics

Question : 36 of 100

Marks: +1, -0

Let

g: N \rightarrow N

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

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

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

n \geq 0

Solution:

g: N \rightarrow N

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

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

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

g(x)=\left[\begin{array}{lc} x+1 & x = 3k+1 \\ x+1 & x=3k+2 \\ x-2 & x=3k+3 \end{array}\right.

g(g(x))=\left[\begin{array}{ll} x+2 & x=3k+1 \\ x-1 & x=3k+2 \\ x-1 & x=3k+3 \end{array}\right.

g(g(g(x))) = \left[ \begin{array}{ll} x & x=3k+1 \\ x & x=3k+2 \\ x & x=3k+3 \end{array} \right]

f: N \rightarrow N, f

is a one-one function such that

f(g(x))=f(x) r g(x)=x

, which is not the caseIf

f f: N \rightarrow N f

is an onto function such that

f(g(x))=f(x)

one possibility is

f(x)=\left[\begin{array}{ll} n & x=3n+1 \\ n & x=3n+2 \\ n & x=3n+3 \end{array}\right.

n \in N_0.

Here

f(x)

is onto, also

f(g(x))=f(x) \forall x \in N

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