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

The Correct Option is 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 \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$