Question

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

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

Answer 1

The correct option is A $d=-a$

$f\left(x\right)=\frac{ax+b}{cx+d}$

$fof\left(x\right)=\frac{a\left\{\frac{ax+b}{cx+d}\right\}+b}{c\left\{\frac{ax+b}{cx+d}\right\}+d}\Rightarrow \frac{a^{2}x+ab+bcx+bd}{acx+bc+cdx+d^2}=x$

$\Rightarrow (ac+dc)x^2+(bc+d^2-bc-a^2)x-ab-bd=0,\forall xϵR$

$\Rightarrow (a+d)c=0,d^2-a^2=0$ and $(a+d)b=0$

$\Rightarrow a+d=0$

$\Rightarrow d=-a$