Prove that f(x) given by `f(x+y)=f(x) +f(y) AA x in R` is an odd function.

Prove that the function f is given by f(x)=|x-1| , x in R is not differentiable at x=1

f is a real-valued function not identically zero, satisfying f(x + y) + f(x - y) = 2f(x) ·f(y) AA x, y in R. f(x) is definitely a)odd b)even c)neither even nor odd d)none of these

Let f (x) = | x - 2| where x is real number. Which one of the following is true ? a) f is periodic b) f (x + y ) = f (x) + f (y) c)f is not one-one function d)f is an even function

The function f:NtoN given by f(x)=2x

Consider a real-valued function f(x) satisfying 2f(xy) =(f(x))^(y)+ (f(y))^(x) AA x, y in R and f(1) = a where a ne 1 then (a-1)sum_(i=1)^(n)f(i)= a) a^(n)-1 b) a^(n+1)+1 c) a^(n)+1 d) a^(n+1) -a

A function f: RrarrR satisfies the equation f(x) f(y) - f(xy) = x+ y, AAx,y in R and f(1) gt 0 , then

Let f: X rarr Y , be a function. Define a relation R in X given by: R={(a, b): f(a)=f(b)} Examine if R is an equivalence relation.

Prove that the function f: R rarr R given by f(x)=2 x is one-one and onto.