Let `f(x + y) = f(x).f(y) AA x, y in R and f(0) != 0`. Then the function `phi` defined by `phi(x)=(f(x))/(1+(f(x))^2)` is

Let f(x+y)=f(x)*f(y)AA x,y in R and f(0)!=0* Then the function phi defined by .Then the function phi(x)=(f(x))/(1+(f(x))^(2)) is

Let f(x+y)+f(x-y)=2f(x)f(y) AA x,y in R and f(0)=k , then

Let f(x+y)+f(x-y)=2f(x)f(y)AA x,y in R and f(0)=k, then

If f(x+y) = f(x) + f(y) + |x|y+xy^(2),AA x, y in R and f'(0) = 0 , then

If f(x+y) = f(x) + f(y) + |x|y+xy^(2),AA x, y in R and f'(0) = 0 , then

If f(x+y) = f(x) + f(y) + |x|y+xy^(2),AA x, y in R and f'(0) = 0 , then

Let f(x+y)=f(x)f(y)AA x, in IR, f(6)=5 and f'(0)=1 . Then the value of f' (6) is

f(x+y)=f(x).f(y)AA x,y in R and f(x) is a differentiable function and f'(0)=1,f(x)!=0 for any x.Findf(x)

Let a real valued function f satisfy f(x + y) = f(x)f(y)AA x, y in R and f(0)!=0 Then g(x)=f(x)/(1+[f(x)]^2) is

Let f(x+y)=f(x)f(y) for all x,y in R and f(0)!=0. Let phi(x)=(f(x))/(1+f(x)^(2)). Then prove that phi(x)-phi(-x)=0