Test Index

Functions Part 1

Section: Mathematics

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Question : 46 of 100

Marks: +1, -0

Let

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

f: A \rightarrow A

f(k)=\begin{cases} k+1, & \text{if } k \text{ is odd}; \\ k, & \text{if } k \text{ is even}. \end{cases}

Then, the number of possible functions

g: A \rightarrow A

g \circ f = f

[26 Feb 2021 Shift 2]

$10^{5}$
${}^{10}C_{5}$
$5^{5}$
$5!$

Solution:

f(x)=\begin{cases} x+1, & x \text{ is odd.}; \\ x, & x \text{ is even.} \end{cases}

Given,

g: A \rightarrow A

such that,

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

When

x

is even, then

g(x)=x

. . . (i)

When

x

is odd, then

g(x+1)=x+1

. . . (ii)

This implies,

g(x)=x, x

is even.

\Rightarrow

If

x

is odd, then

g(x)

can take any value in set

A

.

So, number of

g(x)=10^{5}

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