Let f: R rightarrow R be a function which satisfies
Question:

Let $f: R \rightarrow R$ be a function which satisfies $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) \forall \mathrm{x}, \mathrm{y} \in \mathrm{R}$. If $\mathrm{f}(1)=2$ and

$g(n)=\sum_{k=1}^{(n-1)} f(k), n \in N$ then the value of $n$, for which $g(n)=20$, is :

  1. 5

  2. 9

  3. 20

  4. 4


Correct Option: 1

Solution:

$f(x+y)=f(x)+f(y)$

$\Rightarrow \mathrm{f}(\mathrm{n})=\mathrm{nf}(1)$

$\mathrm{f}(\mathrm{n})=2 \mathrm{n}$

$g(n)=\sum_{k=1}^{n-1} 2 n=2\left(\frac{(n-1) n}{2}\right)=n(n-1)$

$g(n)=20 \Rightarrow n(n-1)=20$

$\mathrm{n}=5$

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