Q49 / 90 · JEE Main 2022 (27 Jun Shift 2) · Mathematics

Let S=1,2,3,4,5,6,7,8,9,10. Define f:S→S as fn=2n, if n=1,2,3,4,52n-11 if n=6,7,8,9,10
Let g:S≥S be a function such that fogn=n+1, if n is odd n-1, if n is even , then g10g1+g2+g3+g4+g5 is equal to

Correct answer: 190

Solution

Given, fn=2n, if n=1,2,3,4,52n-11 if n=6,7,8,9,10

So, f1=2, f2=4, .....f5=10

And f6=1, f7=3, f8=5 ......f10=9

fgn=n+1;n∈ odd n-1;n∈ even 

So, fg10=9⇒g10=10

fg1=2⇒g1=1

fg2=1⇒g2=6

fg3=4⇒g3=2

fg4=3⇒g4=7

fg5=6⇒g5=3

So, g10·g1+g2+g3+g4+g5

=10·1+6+2+7+3=190

Hence, the correct answer is 190.

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