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Let $f(x)=\frac{a x+b}{c x+d},$ then fof $(x)=x,$ provided that:
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The correct answer is:
$\mathrm{d}=-\mathrm{a}$
$f(x)=\frac{a x+b}{c x+d}$
fof $(x)=\frac{a\left\{\frac{a x+b}{c x+d}\right\}+b}{c\left\{\frac{a x+b}{c x+d}\right\}+d} \Rightarrow \frac{a^{2} x+a b+b c x+b d}{a c x+b c+c d x+d^{2}}=x$
$\Rightarrow(a c+d c) x^{2}+\left(b c+d^{2}-b c-a^{2}\right) x$
$-a b-b d=0, \quad \forall x \in R$
$\Rightarrow(a+d) c=0, d^{2}-a^{2}=0$ and $(a+d) b=0$
$\Rightarrow a+d=0$
fof $(x)=\frac{a\left\{\frac{a x+b}{c x+d}\right\}+b}{c\left\{\frac{a x+b}{c x+d}\right\}+d} \Rightarrow \frac{a^{2} x+a b+b c x+b d}{a c x+b c+c d x+d^{2}}=x$
$\Rightarrow(a c+d c) x^{2}+\left(b c+d^{2}-b c-a^{2}\right) x$
$-a b-b d=0, \quad \forall x \in R$
$\Rightarrow(a+d) c=0, d^{2}-a^{2}=0$ and $(a+d) b=0$
$\Rightarrow a+d=0$
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