A ral valued functin `f (x)` satisfies the functional equation `f (x-y)=f(x)f(y)- f(a-x) f(x+y)` where 'a' is a given constant and `f (0) =1, f(2a-x)` is equal to :

A ral valued functin f (x) satisfies the functional equation f (x-y)=f(x)f(y)- f(a-x) f(a+y) where 'a' is a given constant and f (0) =1, f(2a-x) is equal to :

A real valued function f(x) satisfies the functional equation f(x-y)=f(x)f(y)-f(a-x)f(a+y) where a is a given constant and f(0), f (2a-x)=

A real-valued functin f(x) satisfies the functional equation f(x-y)=f(x)f(y)-f(a-x)f(a+y), where a given constant and f(0)=1. Then prove that f(x) is symmetrical about point (a, 0).

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

A function f(x) satisfies the relation f(x+y) = f(x) + f(y) + xy(x+y), AA x, y in R . If f'(0) = - 1, then

If f(x) be a differentiable function such that f(x+y)=f(x)+f(y) and f(1)=2 then f'(2) is equal to

If f is a real valued function not identically zero,satisfying f(x+y)+f(x-y)=2f(x)*f(y) then f(x)