If for all `x,y` the function `f` is defined by `f(x)+f(y)+f(x).f(y)=andf(x)gt0.` Then show `f'(x)=0` when `f'(x)=0` is differentiable.

A function f : R -> R^+ satisfies f(x+y)= f(x) f(y) AA x in R If f'(0)=2 then f'(x)=

Let f be differentiable function satisfying f((x)/(y))=f(x) - f(y)"for all" x, y gt 0 . If f'(1) = 1, then f(x) is

If the function / satisfies the relation f(x+y)+f(x-y)=2f(x),f(y)AAx , y in R and f(0)!=0 , then

Let f((x+y)/(2))=(f(x)+f(y))/(2) and f(0)=b. Find f''(x) (where y is independent of x), when f(x) is differentiable.

Let f:RtoR be a function given by f(x+y)=f(x)f(y) for all x,y in R .If f'(0)=2 then f(x) is equal to

Let f(x+y) + f(x-y) = 2f(x)f(y) for x, y in R and f(0) != 0 . Then f(x) must be

A function is defined as f(x) = {{:(e^(x)",",x le 0),(|x-1|",",x gt 0):} , then f(x) is

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

A function f : R to R Satisfies the following conditions (i) f (x) ne 0 AA x in R (ii) f(x +y)= f(x) f(y) AA x, y, in R (iii) f(x) is differentiable (iv ) f'(0) =2 The derivative of f(x) satisfies the equation

If a real valued function f(x) satisfies the equation f(x +y)=f(x)+f (y) for all x,y in R then f(x) is