QJEEM21Mae387: JEE Main 2021 Mathematics - Class 11 JEEM21Mae387 Solution: Let and be defined as f ( k ) = { k + 1 i f k i s o d d k i f k …

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

$\therefore$ 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) $\Rightarrow$ g(2) = 2

when, x = 2, g(f(2)) = f(2) $\Rightarrow$ g(2) = 2

$\therefore$ 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.

$\therefore$ For 1, number of functions = 10

For 3, number of functions = 10

$$\begin{array}{rl}& .\\ & .\\ & .\\ & \end{array}$$
for 9, number of functions = 10

$\therefore$ Total number of functions = 10 $\times$ 10 $\times$ 10 $\times$ 10 $\times$ 10 = 10⁵