Question

If $f:R\to R$ is a continuous function such that $f(x+y)=f\left(x\right)+f\left(y\right)\forall x,y\in R$ and, $f\left(1\right)=2,$ then $f\left(200\right)$ is

• A. $0$
• B. $100$
• C. $200$
• D. $400$

Answer 1

The correct option is D $400$

$f(x+y)=f\left(x\right)+f\left(y\right)$
Put $x=y=1$
$f\left(2\right)=f\left(1\right)+f\left(1\right)=2f\left(1\right)$
Put $x=2,y=1$
$f\left(3\right)=f\left(2\right)+f\left(1\right)=2f\left(1\right)+f\left(1\right)=3f\left(1\right)$
Similarly $f\left(4\right)=4f\left(1\right)$
$\therefore f\left(200\right)=200f\left(1\right)=200\times 2=400$