If f(x+y) = f(x) . f(y) , then
f(x) = akx
f(x) = 0
Since this is a more than one correct answer question, we will go through each of the options and check if it satisfies the given conditions
f(x)=akx
⇒ f(x+y)=ak(x+y)=akx.aky
⇒ f(x) . f(y)
So, one of the functions which satisfies the given condition is akx
B. f(x) = 0 also satisfies the given conditions
If we go through other options, we will find that they don’t satisfy the given condition