Let $A=\{1,2,3, \ldots, 10\}$ and $f: A \rightarrow A$ be defined as $f(k)=\left\{\begin{array}{cc}k+1 & \text { if } k \text { is odd } \\ k & \text { if } k \text { is even }\end{array}\right.$ Then the number of possible functions $g: A \rightarrow A$ such that $g$ of $=f$ is :

(A) ${ }^{10} \mathrm{C}_{5}$

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Sets Relations and Functions

Let

A = {1, 2, 3, \dots, 10}

and

f : A \to A

be defined as

f (k) = {k + 1 k if k is odd if k is even

Then the number of possible functions

g : A \to A

such that

g

= f

is :

Answer: (C)

The condition

g \circ f = f

reads

g (f (k)) = f (k)

for every

k \in A

, so

g

must fix every element of the range of

f

. For odd

k

f (k) = k + 1

, and for even

k

f (k) = k

, so the range of

f

{2, 4, 6, 8, 10}

and

g (m) = m

is forced at these five values.
The other five elements

1, 3, 5, 7, 9

carry no constraint, and each may be sent to any of the

10

elements of

A

, independently. The number of possible functions

g

is therefore

1 0^{5}

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option