JEE Main 2021 — Functions Question with Solution
From: JEE Main 2021 (Online) 26th February Evening Shift
Question
Let and be defined as
f(k) = \left\{ {\matrix{ {k + 1} & {if\,k\,is\,odd} \cr k & {if\,k\,is\,even} \cr } } \right.
Then the number of possible functions such that is :
f(k) = \left\{ {\matrix{ {k + 1} & {if\,k\,is\,odd} \cr k & {if\,k\,is\,even} \cr } } \right.
Then the number of possible functions such that is :
Choose an option
Show full solutionCorrect option: B
Correct answer
B105
Step-by-step explanation

f(1) = 2
f(2) = 2
f(3) = 4
f(4) = 4
f(5) = 6
f(6) = 6
f(7) = 8
f(8) = 8
f(9) = 10
f(10) = 10
f(1) = f(2) = 2
f(3) = f(4) = 4
f(5) = f(6) = 6
f(7) = f(8) = 8
f(9) = f(10) = 10
Given, g(f(x)) = f(x)
when x = 1, g(f(1)) = f(1) g(2) = 2
when, x = 2, g(f(2)) = f(2) g(2) = 2
x = 1, 2, g(2) = 2
Similarly, at x = 3, 4, g(4) = 4
at x = 5, 6, g(6) = 6
at x = 7, 8, g(8) = 8
at x = 9, 10, g(10) = 10

Here, you can see for even terms mapping is fixed. But far odd terms 1, 3, 5, 7, 9 we can map to any one of the 10 elements.
For 1, number of functions = 10
For 3, number of functions = 10
\eqalign{ & . \cr & . \cr & . \cr & \cr}
for 9, number of functions = 10
Total number of functions = 10 10 10 10 10 = 105
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