Mathematics

Mathematics

Problem

Let cc and dd be fixed real numbers with cd≠0cd\neq 0. Define f:R∖{−dc}→Rf: \mathbb R\setminus\left\{-\dfrac d c\right\}\to\mathbb R by f(x)=ax+bcx+d,f(x)=\frac{ax+b}{cx+d}, where a,ba,b are real parameters and ad−bc≠0ad-bc\neq 0. Which of the following conditions on aa and dd ensures that f(f(x))=xf(f(x))=x holds for every permissible xx? a) ad=1ad=1 b) a+d=0a+d=0 c) d=ad=a d) b=−db=-d
Intuitive explanation

The key insight is that f(f(x))=xf(f(x)) = x means ff is an involution—it's its own inverse function. Rather than computing f(f(x))f(f(x)) blindly, recognize that for a rational function of this form, an involution has a special structure: the function must "swap" in a symmetric way. This happens when the linear fractional transformation has a particular relationship between its coefficients. Focus on what constraint the involution property places on aa and dd specifically, since those are the parameters you're asked about.

Intuitive explanation

The key insight is that f(f(x))=xf(f(x)) = x means ff is an involution—it's its own inverse function. Rather than computing f(f(x))f(f(x)) blindly, recognize that for a rational function of this form, an involution has a special structure: the function must "swap" in a symmetric way. This happens when the linear fractional transformation has a particular relationship between its coefficients. Focus on what constraint the involution property places on aa and dd specifically, since those are the parameters you're asked about.

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