Question

If \[f(x)=\dfrac{ax+d}{cx+b}\] and $f\left( f(x) \right)=x$ for all $x$, then:
A. $a=b$
B. $c=d$
C. \[a+b=0\]
D. \[c+d=0\]

Answer 1

Hint: To solve this question, we need to determine the value of the composite function $f\left( f(x) \right)$. To determine value of $f\left( f(x) \right)$, we will substitute the value of term $x$ in $f(x)$ with \[f(x)=\dfrac{ax+d}{cx+b}\]. We will then equate the desired result with the given value of $f(x)$ which is $f\left( f(x) \right)=x$ and then we will find the relation.

Complete step by step solution: We are given function \[f(x)=\dfrac{ax+d}{cx+b}\] and $f\left( f(x) \right)=x$ and we have to determine the correct relationship from the options.
As we know that $f\left( f(x) \right)$ means function $f$ will have $f(x)$ as its value by replacing the term $x$ in $f(x)$ with \[f(x)=\dfrac{ax+d}{cx+b}\].
We will use the composite function $f\left( f(x) \right)=x$ by substituting the value of $f(x)$ in it.
$\begin{align}
  & f\left( f(x) \right)=f\left( \dfrac{ax+d}{cx+b} \right) \\
 & x=\dfrac{a\left( \dfrac{ax+d}{cx+b} \right)+d}{c\left( \dfrac{ax+d}{cx+b} \right)+b}
\end{align}$
We will now simplify the equation.
$\begin{align}
  & x=\dfrac{\dfrac{{{a}^{2}}x+ad+cdx+bd}{cx+b}}{\dfrac{acx+cd+bcx+{{b}^{2}}}{cx+b}} \\
 & x=\dfrac{{{a}^{2}}x+ad+cdx+bd}{acx+cd+bcx+{{b}^{2}}} \\
 & ac{{x}^{2}}+cdx+bc{{x}^{2}}+{{b}^{2}}x={{a}^{2}}x+ad+cdx+bd
\end{align}$
We will now put all the terms on one side and pair all the like terms with each other.
$\begin{align}
  & ac{{x}^{2}}+bc{{x}^{2}}-{{a}^{2}}x+{{b}^{2}}x+cdx-cdx-ad-bd=0 \\
 & (a+b)c{{x}^{2}}-({{a}^{2}}-{{b}^{2}})x-(a+b)d=0
\end{align}$
We will now use the formula of expansion $({{a}^{2}}-{{b}^{2}})=(a+b)(a-b)$.
$\begin{align}
  & (a+b)c{{x}^{2}}-(a-b)(a+b)x-(a+b)d=0 \\
 & (a+b)\left[ c{{x}^{2}}-(a-b)x-d \right]=0
\end{align}$
We will now equate both of the factors to zero.
$c{{x}^{2}}-(a-b)x-d$ will not be equal to zero for all the values of $x$, so we will not consider this.
Now equating other factor,
$a+b=0$
We can see that this relation is present in the options hence this will be the correct answer.

The correct relation is $a+b=0$ when \[f(x)=\dfrac{ax+d}{cx+b}\] and $f\left( f(x) \right)=x$ for all $x$. Hence the correct option is (C).

Note: The evaluation of a function with the value of another function is called composite functions such as $fof(x)=f(f(x))$ which was present in the question. It simply means that $f(x)$ is input of the function $f$and not the multiplication of $f$ and $x$.