Question:
Let $\mathrm{f}: \mathbf{N} \rightarrow \mathbf{N}$ be a function such that $\mathrm{f}(\mathrm{m}+\mathrm{n})=\mathrm{f}(\mathrm{m})+\mathrm{f}(\mathrm{n})$ for every $\mathrm{m}, \mathrm{n} \in \mathbf{N} .$ If $\mathrm{f}(6)=18$ then $f(2) \cdot f(3)$ is equal to :
Correct Option: , 2
Solution:
$f(\mathrm{~m}+\mathrm{n})=f(\mathrm{~m})+f(\mathrm{n})$
Put $\mathrm{m}=1, \mathrm{n}=1$
$f(2)=2 f(1)$
Put $\mathrm{m}=2, \mathrm{n}=1$
$f(3)=f(2)+f(1)=3 f(1)$
$f(6)=2 f(3) \Rightarrow f(3)=9$
$\Rightarrow f(1)=3, f(2)=6$
$f(2) \cdot f(3)=6 \times 9=54$
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