Question

Let $f(x)=\frac{ax+b}{cx+d}.$ Then $fof(x)=x$ provided that

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

Answer 1

Answer: (A)

Given, $f(x)=\frac{ax+b}{cx+d}$ and fof $(x)=x$
$\Rightarrow f\left(\frac{ax+b}{cx+d}\right)=x$
$\Rightarrow \frac{a\left(\frac{ax+b}{cx+d}\right)+b}{c\left(\frac{ax+b}{cx+d}\right)+d}=x$
$\Rightarrow \frac{x\left(a^2+bc\right)+ab+bd}{x(ac+cd)+bc+d^2}=x$
$\Rightarrow a^2+bc=bc+d^2,ab+bd=0$ and
$ac+cd=0$
$\Rightarrow d=-a$