Question

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

Hint: Put x = 1 and y = 1 in the equation which satisfies which satisfies, $f:R\to R$. Solve the equation formed by substituting this value and get the value for $f\left( 1 \right)$. Now put y = 1 in the equation. Then solve it to get the value of $f\left( x \right){{f}^{-1}}\left( x \right)$.

Complete step-by-step answer:
Given the function, $f:R\to R$, satisfies the equation,
$f\left( x \right)f\left( y \right)-f\left( xy \right)=x+y$, where x, y belongs to R, where R is a real number.
It is also given to us that $f\left( 1 \right)>0$.

Let us put x = 1 and y = 1 in equation (1).

& f\left( x \right)f\left( y \right)-f\left( xy \right)=x+y-(1) \\\ & \therefore f\left( 1 \right)f\left( 1 \right)-f\left( 1 \right)=1+1 \\\ & f\left( 1 \right)f\left( 1 \right)-f\left( 1 \right)=2 \\\ & {{f}^{2}}\left( 1 \right)-f\left( 1 \right)=2 \\\ & \Rightarrow {{f}^{2}}\left( 1 \right)-f\left( 1 \right)-2=0 \\\ \end{aligned}$$ In order to make the above equation in the form of a quadratic expression let us add and subtract $$f\left( 1 \right)$$to it. $$\begin{aligned} & {{f}^{2}}\left( 1 \right)-f\left( 1 \right)-2-f\left( 1 \right)+f\left( 1 \right)=0 \\\ & {{f}^{2}}\left( 1 \right)-2f\left( 1 \right)+f\left( 1 \right)-2=0 \\\ \end{aligned}$$ Let us take $$f\left( 1 \right)$$common from these terms. $$f\left( 1 \right)\left[ f\left( 1 \right)-2 \right]+2\left[ f\left( 1 \right)-2 \right]=0$$ Take $$\left[ f\left( 1 \right)-2 \right]$$common from LHS, we get, $$\left[ f\left( 1 \right)+1 \right]\left[ f\left( 1 \right)-2 \right]=0$$ Thus the above equation can be equated to zero. $$f\left( 1 \right)+1=0$$and $$f\left( 1 \right)-2=0$$ $$\therefore f\left( 1 \right)=-1$$and $$f\left( 1 \right)=2$$. We have been given the condition that, $$f\left( 1 \right)>0$$. $$\therefore f\left( 1 \right)=-1$$, can be neglected. So for $$f\left( 1 \right)=2$$, the condition is true. Now in equation (1) let us put the value of y as 1. i.e. y = 1. $$\begin{aligned} & f\left( x \right)f\left( y \right)-f\left( xy \right)=x+y \\\ & f\left( x \right)f\left( 1 \right)-f\left( x \right)=x+1 \\\ & f\left( x \right)\times 2-f\left( x \right)=x+1 \\\ & 2f\left( x \right)-f\left( x \right)=x+1 \\\ & \therefore f\left( x \right)=x+1 \\\ \end{aligned}$$ Let us put, $$y=f\left( x \right)$$. $$\begin{aligned} & \therefore y=x+1 \\\ & \therefore x=y-1 \\\ \end{aligned}$$ Thus we can say that, $${{f}^{-1}}\left( x \right)=x-1$$. $$\therefore f\left( x \right){{f}^{-1}}\left( x \right)=\left( x+1 \right)\left( x-1 \right)$$ $$\begin{aligned} & ={{x}^{2}}-x+x-1 \\\ & ={{x}^{2}}-1 \\\ \end{aligned}$$ $$\therefore f\left( x \right){{f}^{-1}}\left( x \right)={{x}^{2}}-1$$. Hence, we got that function $$f:R\to R$$satisfies the given equation and we got $$f\left( x \right){{f}^{-1}}\left( x \right)={{x}^{2}}-1$$. $$\therefore $$Option (c) is the correct answer. Note: As it is given, $$f\left( 1 \right)>0$$, it is wise to assume the value of x and y as 1. When finding the value of $$f\left( 1 \right)$$, check that it fulfills the condition of $$f\left( 1 \right)>0$$. As we need to find $$f\left( x \right){{f}^{-1}}\left( x \right)$$where number $$f\left( y \right)$$is mentioned, take again y = 1. Simplify it and you get $$f\left( x \right){{f}^{-1}}\left( x \right)$$.