Question

Let $f$ be a real valued function satisfying $f(x+y)=f\left(x\right)+f\left(y\right)$ for all $x,y.$ If $f\left(1\right)=\frac{1}{2}$, then the value of $f\left(16\right)$ is

• A. $16$
• B. $8$
• C. $4$
• D. $2$

Answer 1

The correct option is B $8$

Assuming $m$ be an integers.
$f(x+(m-1\left)x\right)=f\left(x\right)+f\left(\right(m-1\left)x\right)$
$=f\left(x\right)+f\left(x\right)+f\left(\right(m-2\left)x\right)$
$=f\left(x\right)+f\left(x\right)+f\left(x\right)+f\left(\right(m-3\left)x\right)$
$.$
$.$
$=mf\left(x\right)+f\left(0\right).....\left(i\right)$
Using $f(x+y)=f\left(x\right)+f\left(y\right)$
Putting $x=0,y=0\Rightarrow f\left(0\right)=2f\left(0\right)\Rightarrow f\left(0\right)=0$
$\therefore f\left(mx\right)=mf\left(x\right)\text{where}m\text{is an integers}\Rightarrow f(16\times 1)=16f\left(1\right)\Rightarrow f\left(16\right)=16\times \frac{1}{2}=8$

Alternate Solution:
$f\left(2\right)=f\left(1\right)+f\left(1\right)=1\Rightarrow f\left(4\right)=f\left(2\right)+f\left(2\right)=2\Rightarrow f\left(8\right)=f\left(4\right)+\left(4\right)=4\Rightarrow f\left(16\right)=f\left(8\right)+f\left(8\right)=8$