Question

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

• A. $f\left(x\right)=x+\frac{1}{2}$
• B. $f\left(x\right)=\frac{x}{2}+1$
• C. $f\left(x\right)=-\frac{x}{2}+1$
• D. $f\left(x\right)=x+1$

Answer 1

The correct option is D $f\left(x\right)=x+1$

Taking $x=y=1$, we get
$f\left(1\right)f\left(1\right)-f\left(1\right)=1+1\Rightarrow {\left(f\left(1\right)\right)}^2-f\left(1\right)-2=0\Rightarrow (f\left(1\right)-2)(f\left(1\right)+1)=0\Rightarrow f\left(1\right)=2(\because f\left(1\right)>0)$
Taking $y=1$, we get
$f\left(x\right)f\left(1\right)-f\left(x\right)=x+1\Rightarrow 2f\left(x\right)-f\left(x\right)=x+1\Rightarrow f\left(x\right)=x+1$