Let mathrm{f}: N → N be a function such that mathrm{f}(mathrm{m}+mathrm{n})=mathrm{f}(mathrm{m})+m

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1.Relation and Function

hard

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

A

$6$

B

$54$

C

$18$

D

$36$

(JEE MAIN-2021)

Solution

$f(\mathrm{~m}+\mathrm{n})=f(\mathrm{~m})+f(\mathrm{n})$

$\text { Put } \mathrm{m}=1, \mathrm{n}=1$

$f(2)=2 f(1)$

$\text { Put } \mathrm{m}=2, \mathrm{n}=1$

$f(3)=f(2)+f(1)=3 f(1)$

$\text { Put } \mathrm{m}=3, \mathrm{n}=3$

$f(6)=2 f(3) \Rightarrow f(3)=9$

$\Rightarrow f(1)=3, f(2)=6$

$f(2) \cdot f(3)=6 \times 9=54$

Standard 12

Mathematics

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