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1.Relation and Function
hard
વિધેય $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ માટે $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) \forall \mathrm{x}, \mathrm{y} \in \mathrm{R}$ થાય જો $\mathrm{f}(1)=2$ અને $g(n)=\sum \limits_{k=1}^{(n-1)} f(k), n \in N$ હોય તો $n$ કિમત મેળવો જ્યાં $\mathrm{g}(\mathrm{n})=20$ થાય
A
$5$
B
$9$
C
$20$
D
$4$
(JEE MAIN-2020)
Solution
$f(x+y)=f(x)+f(y)$
$\Rightarrow f(n)=n f(1)$
$f(n)=2 n$
$g(n)=\sum_{k=1}^{n-1} 2 n=2\left(\frac{(n-1) n}{2}\right)=n(n-1)$
$g(n)=20 \Rightarrow n(n-1)=20$
$n=5$
Standard 12
Mathematics
