JEE Main 2021MathematicsFunctionsEasyMCQ

JEE Main 2021 — Functions Question with Solution

JEE Main 2021 (31 Aug Shift 2)

Question

Let f:N→N be a function such that fm+n=fm+fn for every m, n∈N. If f6=18 then f2·f3 is equal to :

Choose an option

Show full solutionCorrect option: A
Correct answer
A54

Step-by-step explanation

Given functional equation is 

fm+n=fm+fn;m,n∈N ....1

Put m=n=3

f(3+3)=f(3)+f(3)

⇒f(3)=9∵f6=18

Put m=2,n=1 in equation 1

∴f(3)=f(2+1)=f(2)+f(1)

=f(1+1)+f(1)

=f(1)+f(1)+f(1)

9=3f(1)

⇒f(1)=3

∴f(2)=f(1+1)=f(1)+f(1)=6

Now,

f(2)·f(3)=(6)(9)=54

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About this question

This is a previous-year question from JEE Main 2021, covering the Functions chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.