Given functional equation is
f(m+n)=f(m)+f(n);m,n∈N....(1)
Put m=n=3
f(3+3)=f(3)+f(3)
⇒f(3)=9[∵f(6)=18]
Put m=2,n=1in 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
Let f:N→N be a function such that f(m+n)=f(m)+f(n) for every m,n∈N. If f(6)=18 then f(2)⋅f(3) is equal to :
Held on 31 Aug 2021 · Verified 6 Jul 2026.
54
6
36
18
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