Question

A function \[f:R\to R\]satisfies the equation, \[f\left( x \right)f\left( y \right)-f\left( xy \right)=x+y\]\[\forall x,y\in R\]and \[f\left( 1 \right)>0\]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).
\[\begin{align}
  & f\left( x \right)f\left( y \right)-f\left( xy \right)=x+y-(1) \\
 & \therefore f\left( 1 \right)f\left( 1 \right)-f\left( 1 \right)=1+1 \\
 & f\left( 1 \right)f\left( 1 \right)-f\left( 1 \right)=2 \\
 & {{f}^{2}}\left( 1 \right)-f\left( 1 \right)=2 \\
 & \Rightarrow {{f}^{2}}\left( 1 \right)-f\left( 1 \right)-2=0 \\
\end{align}\]

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.
\[\begin{align}
  & {{f}^{2}}\left( 1 \right)-f\left( 1 \right)-2-f\left( 1 \right)+f\left( 1 \right)=0 \\
 & {{f}^{2}}\left( 1 \right)-2f\left( 1 \right)+f\left( 1 \right)-2=0 \\
\end{align}\]

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.
\[\begin{align}
  & f\left( x \right)f\left( y \right)-f\left( xy \right)=x+y \\
 & f\left( x \right)f\left( 1 \right)-f\left( x \right)=x+1 \\
 & f\left( x \right)\times 2-f\left( x \right)=x+1 \\
 & 2f\left( x \right)-f\left( x \right)=x+1 \\
 & \therefore f\left( x \right)=x+1 \\
\end{align}\]

Let us put, \[y=f\left( x \right)\].
\[\begin{align}
  & \therefore y=x+1 \\
 & \therefore x=y-1 \\
\end{align}\]

Thus we can say that, \[{{f}^{-1}}\left( x \right)=x-1\].
\[\therefore f\left( x \right){{f}^{-1}}\left( x \right)=\left( x+1 \right)\left( x-1 \right)\]
                             \[\begin{align}
  & ={{x}^{2}}-x+x-1 \\
 & ={{x}^{2}}-1 \\
\end{align}\]
\[\therefore f\left( x \right){{f}^{-1}}\left( x \right)={{x}^{2}}-1\].

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 \[f\left( x \right){{f}^{-1}}\left( x \right)\].