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

A function f: R -> R satisfies the equation f(x)f(y)-f(x y)=x+yAAx ,y in Ra n df(1)>0 , then a. f(x)f^(-1)(x)=x^2-4 b. f(x)f^(-1)(x)=x^2-6 c. f(x)f^(-1)(x)=x^2-1 d. none of these

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

Statement-1 : A function f : R rarr R satisfies the f(x) - f(y) = x-y, AA x, y in R, f(0), f(1) gt 0 , then f(x) = x+1. and Statement-2 : If function f : R rarr R satisfied the above relationship then f(1) + f(0) + f(3) is equal to 7.

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 : R rarr R satisfies the equation f(x+y) = f(x). f(y) for all x y in R, f(x) ne 0 . Suppose that the function is differentiable at x = 0 and f'(0) = 2 , then prove that f' = 2f(x) .

A function f: R->R satisfies that equation f(x+y)=f(x)f(y) for all x ,\ y in R , f(x)!=0 . Suppose that the function f(x) is differentiable at x=0 and f^(prime)(0)=2 . Prove that f^(prime)(x)=2\ f(x) .

A function f: R->R satisfies that equation f(x+y)=f(x)f(y) for all x ,\ y in R , f(x)!=0 . Suppose that the function f(x) is differentiable at x=0 and f^(prime)(0)=2 . Prove that f^(prime)(x)=2\ f(x) .

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 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 : R rarr R satisfies the equation f(x + y) = f(x) . f(y) for all, f(x) ne 0 . Suppose that the function is differentiable at x = 0 and f'(0) = 2. Then,