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 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 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

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

Suppose the function f satisfies the conditions f(x+y)=f(x)f(y)AA x,y and f(x)=1+xg(x) where Lt_(x)rarr0(x)=1. show that f'(x) exists and f(x)=f(x)AA x

Solve: (dy)/(dx)+y*f'(x)=f(x)*f'(x), where f(x) is a given function.

A real valued function f(x) is satisfying f(x+y)=yf(x)+xf(y)+xy-2 for all x,y in R and f(1)=3, then f(11) is