Question

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

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

Answer 1

The correct option is A d =- a

$f\left(f\right(x\left)\right)=\frac{af\left(x\right)+b}{cf\left(x\right)+d}$

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

$f\left(f\right(x\left)\right)=\frac{a^{2}x+ab+bcx+bd}{acx+bc+d^2+dcx}=x$

$\Rightarrow a^{2}x+ab+bcx+bd=ac{x}^2+bcx+d^{2}x+dc{x}^2$

Given $a=-d$, the above equation can also be verified