Question

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

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

Answer 1

The correct option is A $d=-a$

Given,

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

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

$=\frac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)}=x$

$a(ax+b)+b(cx+d)=x\left[c\right(ax+b)+d(cx+d\left)\right]$

$(a^2+bc)x+(ab+bd)=(ac+cd)x^2+(bc+d^2)x$

matching the coefficients of $x$, we get,

$a^2+bc=bc+d^2$

$a^2=d^2$

$d=\pm a$

matching the coefficients of $x^2$, we get,

$ac+cd=0$

$\therefore d=-a$

matching constant terms, we get,

$ab+bd=0$

$d=-a$