Question

If $f\left(x\right)$ satisfies the relation $f(x+y)=f\left(x\right)+f\left(y\right)$ for all $x,y\in R$ and $f\left(1\right)=5$, then

• A. $f\left(x\right)$ is an odd function
• B. $f\left(x\right)$ is an even function
• C. $\sum _{r=1}^{m}f\left(r\right)=5{}^{m+1}{C}_2$
• D. $\sum _{r=1}^{m}f\left(r\right)=\frac{5m(m+2)}{3}$

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(x\right)$ is an odd function
C $\sum _{r=1}^{m}f\left(r\right)=5{}^{m+1}{C}_2$

$f\left(2\right)=f(1+1)=2f\left(1\right)=10$
$f\left(3\right)=f(1+2)=f\left(1\right)+f\left(2\right)=5+10=15$
$\Rightarrow f\left(n\right)=5n$
or, $\sum _{r=1}^{m}f\left(r\right)=5\sum _{r=1}^{m}r=\frac{5m(m+1)}{2}$
Replacing $y$ by $-x$ in the given equation, we get
$f\left(0\right)=f\left(x\right)+f(-x)$
Also, putting $x=y=0$, we get
$f\left(0\right)=f\left(0\right)+f\left(0\right)\Rightarrow f\left(0\right)=0$
So, $f\left(x\right)+f(-x)=0$
Hence, $f\left(x\right)$ is an odd function.