Let $\mathrm{g}: \mathrm{N} \rightarrow \mathrm{N}$ be defined as \[ \begin{aligned} & g(3 n+1)=3 n+2 \\ & g(3 n+2)=3 n+3 \\ & g(3 n+3)=3 n+1, \text { for all } n \geq 0 \end{aligned} \] Then which of the following statements is true?

(D) There exists on onto function $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$ such that $\mathrm{fog}=\mathrm{f}$

(A) There exists a function $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$ such that gof $=\mathrm{f}$

(C) There exists a one-one function $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$ such that $\mathrm{fog}=\mathrm{f}$

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Sets Relations and Functions

Let

g : N \to 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 \geq 0

Then which of the following statements is true?

Answer: (D)

The map

g

permutes each block

{3 n + 1, 3 n + 2, 3 n + 3}

cyclically, so

g

is a bijection of

N

with no fixed point, and

g \circ g \circ g

is the identity rather than

g

.
A function with

g \circ f = f

would need

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

, making

f (x)

a fixed point of

g

; since

g

has none, no such

f

exists.
A function with

f \circ g = f

must satisfy

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

, because

g

carries each member of a block to the next. Such an

f

is constant on every block and so cannot be one-one, but it can be onto: setting

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

for all

n \geq 0

gives a surjection from

N

N

satisfying

f \circ g = f

.
So there exists an onto function

f : N \to N

with

f \circ g = f

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option