Question

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

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

Answer 1

Answer: (A)

d = - a

Answer 2

$f\left(x\right) = \frac{ax+b}{cx+d}$ 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 $\Rightarrow \left(ac+dc\right)x^{2} +\left(bc +d^{2}-bc -a^{2}\right)x -ab-bd=0 , \forall x \in R$ $\Rightarrow \left(a+d\right)c = 0, d^{2}-a^{2} =0 \left(a + d\right)b = 0$ $\Rightarrow a +d = 0$