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

$f(x)=x+\frac{1}{2}$ $f(x)=(\frac{1}{2})x+1$ $f(x)=(\frac{1}{2})x-1$ none of these

none of these

Taking x = y = 1, we get $f (1) f (1)− f (1) =1+1 ⇒ f(1)^2- f(1) -2 =0$ $⇒ ( f (1)− 2)( f (1)+1) = 0 ⇒ f (1) = 2\,\, (∵ f (1) > 0)$ Taking y = 1 we get $f (x) f (1)− f (x) = x +1 ⇒ 2 f (x)− f (x) = x +1 ⇒ f (x) = x +1$