Question

A function f : R$\to$R satisfies the equation f(x)f(y) - f(xy) = x + y $\forall$ x, y $\in$ R and f (1)>0, then

• A. $f\left(x\right)f^{-1}\left(x\right)=x^2-4$
• B. $f\left(x\right)f^{-1}\left(x\right)=x^2-6$
• C. $f\left(x\right)f^{-1}\left(x\right)=x^2-1$
• D. none of these

Answer 1

The correct option is A $f\left(x\right)f^{-1}\left(x\right)=x^2-4$

$f:R\to Rf\left(x\right)f\left(y\right)-f\left(xy\right)=x+y\forall x,yϵR\because f\left(1\right)>0Put,x=1\&y=1\Rightarrow f\left(1\right)f\left(1\right)-f\left(1\right)=2\Rightarrow f^2\left(1\right)-f\left(1\right)=0f^2\left(1\right)-2f\left(1\right)+f\left(1\right)-2=0f\left(1\right)(f\left(1\right)-2)+1(f\left(1\right)-2)=0(f\left(1\right)+1)(f\left(1\right)-2)=0f\left(1\right)=-1\&f\left(1\right)=2\therefore f\left(1\right)=2True.Put,y=12f\left(x\right)-f\left(x\right)=2+x\Rightarrow f\left(x\right)=2+x\Rightarrow y=x+2x=y-2f^{-1}\left(x\right)=x-2\Rightarrow f\left(x\right)f^{-1}\left(x\right)=(x+2)(x-2)f\left(x\right)f^{-1}\left(x\right)=x^2-4$