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 :
Correct Option: 1
Solution:
$f(x+y)=f(x)+f(y)$
$\Rightarrow \mathrm{f}(\mathrm{n})=\mathrm{nf}(1)$
$\mathrm{f}(\mathrm{n})=2 \mathrm{n}$
$g(n)=\sum_{k=1}^{n-1} 2 n=2\left(\frac{(n-1) n}{2}\right)=n(n-1)$
$g(n)=20 \Rightarrow n(n-1)=20$
$\mathrm{n}=5$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.