Question

If $f\left(x\right)=\frac{ax+b}{cx+d}$, then $fof\left(x\right)=x$ provided that:

• A. $d=-a$
• B. $d=a$
• C. $a=b=c=d=1$
• D. $a=b=1$

Answer 1

The correct option is A

$d=-a$

Explanation for the correct option.

Find the relation:

$\begin{array}{rcl}f\left(x\right)& =& \frac{ax+b}{cx+d}\\ & \Rightarrow & f\left(f\left(x\right)\right)=\frac{a\left(f\left(x\right)\right)+b}{c\left(f\left(x\right)\right)+d}\\ & =& \frac{a\left({\displaystyle \frac{ax+b}{cx+d}}\right)+b}{c\left({\displaystyle \frac{ax+b}{cx+d}}\right)+d}\\ & =& \frac{a^{2}x+ab+bcx+bd}{acx+bc+cdx+d^2}\end{array}$

It is given that $fof\left(x\right)=x$.

$\begin{array}{rcl}\Rightarrow \frac{a^{2}x+ab+bcx+bd}{acx+bc+cdx+d^2}& =& x\\ \Rightarrow a^{2}x+ab+bcx+bd& =& ac{x}^2+bcx+cd{x}^2+d^{2}x\\ & \Rightarrow & \left(ac+cd\right)x^2+\left(d^2+bc-a^2-bc\right)x-\left(ab+bd\right)=0\end{array}$

$\forall x\in R$, so

$$\begin{gathered} \left(ac+cd\right)x^2=0,\left(d^2+bc-a^2-bc\right)x=0,-\left(ab+bd\right)=0 \\ \Rightarrow \left(a+d\right)c=0,\left(d+a\right)\left(d-a\right)=0,-\left(a+d\right)b=0 \\ \Rightarrow d=-a \end{gathered}$$

Hence, option A is correct.