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

Options

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

Answer

C

Explanation

Taking x = y = 1, we get
`f(1)f(1)-f(1)=2`
`rArr" "f^(2)(1)-f(1)-2=0`
`rArr" "(f(1)-2)(f(1)+1)=0`
`rArr" "f(1)=2`
Taking y = 1, we get
`f(x).f(1)-f(x)=x+1`
`rArr" "f(x)=x+1`
`rArr" "f^(-1)(x)=x-1`
`therefore" "f(x).f^(-1)(x)=x^(2)-1`