If `f(mn)=f(m+n)` for all m, `n epsilonR` and `f(2)=2006` then find `f(2006^(2006))`

Let f:N to N for which f(m+n)=f(m)+f(n) forall m,n in N .If f(6)=18 then the value of f(2)*f(3) is

If f(x) satisfies the relation, f(x+y)=f(x)+f(y) for all x,y in R and f(1)=5, then find sum_(n=1)^(m)f(n) . Also, prove that f(x) is odd.

Le the be a real valued functions satisfying f(x+1) + f(x-1) = 2 f(x) for all x, y in R and f(0) = 0 , then for any n in N , f(n) =

If f(x) satisfies the relation f(x+y)=f(x)+f(y) for all x,y in R and f(1)=5, then f(x) is an odd function f(x) is an even function sum_(n=1)^(m)f(r)=5^(m+1)C_(2)sum_(n=1)^(m)f(r)=(5m(m+2))/(3)

If f(x)=(a-x^n)^(1/n) then find f(f(x))

Let f : N rarr N be a function such that f(m + n) = f(m) + f(n) for every m, n in N . If f(6) = 18 , then f(2).f(3) is equal to :

f_(n)(x)=e^(f_(n-1)(x))" for all "n in N and f_(0)(x)=x," then "(d)/(dx){f_(n)(x)} is

If f:RtoR satisfies f(x+y)=f(x)+f(y), for all real x,y and f(1)=m. then: sum_(r=1)^(n)f(r)=

Let f be a one-to-one function from set of natural numbers to itself such that f(mn)-f(m)xx f(n) for all m,n in N. what is least possible value of f(999)?

Let f be real valued function from N to N satisfying. The relation f(m+n)=f(m)+f(n) for all m,n in N . The range of f contains all the even numbers, the value of f(1) is