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(B) Given the functional equation $f(m+n) = f(m) + f(n)$ for all $m, n \in N$.This is a Cauchy functional equation on natural numbers,which implies $f

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(B) Given the functional equation $f(m+n) = f(m) + f(n)$ for all $m, n \in N$.This is a Cauchy functional equation on natural numbers,which implies $f(n) = cn$ for some constant $c$.Given $f(6) = 18$,we substitute $n=6$ into $f(n) = cn$:$c \cdot 6 = 18 \Rightarrow c = 3$.Thus,the function is $f(n) = 3n$.Now,calculate $f(2)$ and $f(3)$:$f(2) = 3 \cdot 2 = 6$.$f(3) = 3 \cdot 3 = 9$.Finally,the product is $f(2) \cdot f(3) = 6 \cdot 9 = 54$.

Let $f: N \rightarrow N$ be a function such that $f(m+n)=f(m)+f(n)$ for every $m, n \in N$. If $f(6)=18$ then $f(2) \cdot f(3)$ is equal to :

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