Question

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

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

Answer 1

The correct option is C

$f\left(x\right)=x+1$

Explanation for the correct answer:

Step 1: Finding $f\left(1\right)$

The function is $f\left(x\right)f\left(y\right)-f(xy)=x+y$ and

$x,y\in R$is always true.

Now take $x=1,y=1$

$$\begin{gathered} \Rightarrow f\left(1\right)f\left(1\right)-f\left(1\right)=2 \\ \Rightarrow f^2\left(1\right)-f\left(1\right)=2 \\ \Rightarrow f^2\left(1\right)-2f\left(1\right)+f\left(1\right)-2=0\left[\because -f\left(1\right)=2f\left(1\right)+f\left(1\right)\right] \\ \Rightarrow f\left(1\right)\left[f\left(1\right)-2\right]+1\left[f\left(1\right)-2\right]=0 \\ \Rightarrow \left[f\left(1\right)-2\right]\left[f\left(1\right)+1\right]=0 \\ f\left(1\right)=2or \\ f\left(1\right)=-1 \end{gathered}$$

We know that $f\left(1\right)>0$, so $f\left(1\right)=2$

Step 2: Finding $f\left(x\right)$

Now take, $y=1,$

$$\begin{gathered} \Rightarrow f\left(x\right)f\left(1\right)-f\left(\right(x\left)\right(1\left)\right)=x+1 \\ \Rightarrow f\left(x\right)\left(2\right)-f\left(x\right)=x+1 \\ \Rightarrow 2f\left(x\right)-f\left(x\right)=x+1 \\ \Rightarrow f\left(x\right)=x+1 \end{gathered}$$

Hence, option (C) is the correct answer.