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Question Number 167840 by mathlove last updated on 27/Mar/22

Commented by cortano1 last updated on 27/Mar/22

$$\{\begin{array}{l}f\left(x\right)=2+(a-1)x\\ g\left(x\right)=2-(a+1)x\end{array}$$$$f\left(g\left(x\right)\right)-g\left(f\left(x\right)\right)=f(a-1)+g(a+1)$$$$\{\begin{array}{l}f\left(g\left(x\right)\right)=2+(a-1)\{2-(a+1)x\}\\ g\left(f\left(x\right)\right)=2-(a+1)\{2+(a-1)x\}\end{array}$$$$\{\begin{array}{l}f\left(g\left(x\right)\right)=2a-(a^2-1)x\\ g\left(f\left(x\right)\right)=-2a-(a^2-1)x\end{array}$$$$f\left(g\left(x\right)\right)-g\left(f\left(x\right)\right)=4a$$$$f(a-1)=2+{(a-1)}^2$$$$g(a+1)=2-{(a+1)}^2$$$$f(a-1)+g(a+1)=4$$$$$$$$\therefore 4a=4\Rightarrow a=1$$

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