f:R->R is continuous such that f(x+y)=f(x)+f(y), for x,y belongs to R.

If f(1)=2,then f(100)=?

a)50

b)100

c)150

d)200

Plz explain the procedure...

put x=y .. we get f(2x)=2f(x) .. put x=1 .. we get f(2)=2f(1)=4 .. f(4)=2f(2)=8 and so on f(8)=16.. f(16)=32..f(32)=64..f(64)=128.... now f ( 32+64)=f(96)=f(32)+f(64)=64+128=192 so f(96)=192... now f(100)= f(96+4)=f(96)+ f(4)=192+8 = 200 ... so f(100) = 200 ... please approve my asnwer now as i have explained you the procedure