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)=1 , 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

The function f(x) satisfying the equation f^2 (x) + 4 f'(x) f(x) + (f'(x))^2 = 0

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

A function f : R→R satisfies the equation f(x)f(y) - f(xy) = x + y ∀ x, y ∈ R and f (1)>0 , then

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

A function f (x) is defined for all x in R and satisfies, f(x + y) = f (x) + 2y^2 + kxy AA x, y in R , where k is a given constant. If f(1) = 2 and f(2) = 8 , find f(x) and show that f (x+y).f(1/(x+y))=k,x+y != 0 .

Let f(x) be a real valued function satisfying the relation f(x/y) = f(x) - f(y) and lim_(x rarr 0) f(1+x)/x = 3. The area bounded by the curve y = f(x), y-axis and the line y = 3 is equal to

Let f(x) be a function satisfying f(x+y)=f(x)+f(y) and f(x)=x g(x)"For all "x,y in R , where g(x) is continuous. Then,

A function f: R -> R satisfy the equation f (x)f(y) - f (xy)= x+y for all x, y in R and f(y) > 0 , then