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1.Relation and Function
hard
અહી વિધેય $\mathrm{f}: N \rightarrow N$ આપેલ છે કે જેથી દરેક $\mathrm{m}, \mathrm{n} \in N$ માટે $\mathrm{f}(\mathrm{m}+\mathrm{n})=\mathrm{f}(\mathrm{m})+\mathrm{f}(\mathrm{n})$ થાય. જો $\mathrm{f}(6)=18$ હોય તો $\mathrm{f}(2) \cdot \mathrm{f}(3)$ ની કિમંત મેળવો.
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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