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Question Number 123356 by ZiYangLee last updated on 25/Nov/20

If f(x)=((ax+1)/(4x−3)) , x∈R , x≠(3/4) and ff(x)=x, find the values of a

$$\mathrm{If}\:{f}\left({x}\right)=\frac{{ax}+\mathrm{1}}{\mathrm{4}{x}−\mathrm{3}}\:,\:{x}\in\mathbb{R}\:,\:{x}\neq\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\:\:\mathrm{and}\:{ff}\left({x}\right)={x},\:\mathrm{find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{a} \\ $$

Answered by liberty last updated on 25/Nov/20

→ { ((f(x)=((ax+1)/(4x−3)))),((f^(−1) (x)=((3x+1)/(4x−a)))) :} ⇒ f(f(x))=x ; f^(−1) (x)=f(x) ⇒((3x+1)/(4x−a)) = ((ax+1)/(4x−3)) ; (3x+1)(4x−3)=(ax+1)(4x−a) 12x^2 −5x−3=4ax^2 +(4−a^2 )x−a { ((12=4a; a=3)),((−5=4−a^2 ; a=3 , a=−3 (rejected))) :}

$$\rightarrow\begin{cases}{{f}\left({x}\right)=\frac{{ax}+\mathrm{1}}{\mathrm{4}{x}−\mathrm{3}}}\\{{f}\:^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{3}{x}+\mathrm{1}}{\mathrm{4}{x}−{a}}}\end{cases} \\ $$$$\Rightarrow\:{f}\left({f}\left({x}\right)\right)={x}\:;\:{f}^{−\mathrm{1}} \left({x}\right)={f}\left({x}\right) \\ $$$$\Rightarrow\frac{\mathrm{3}{x}+\mathrm{1}}{\mathrm{4}{x}−{a}}\:=\:\frac{{ax}+\mathrm{1}}{\mathrm{4}{x}−\mathrm{3}}\:;\:\left(\mathrm{3}{x}+\mathrm{1}\right)\left(\mathrm{4}{x}−\mathrm{3}\right)=\left({ax}+\mathrm{1}\right)\left(\mathrm{4}{x}−{a}\right) \\ $$$$\mathrm{12}{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{3}=\mathrm{4}{ax}^{\mathrm{2}} +\left(\mathrm{4}−{a}^{\mathrm{2}} \right){x}−{a} \\ $$$$\begin{cases}{\mathrm{12}=\mathrm{4}{a};\:{a}=\mathrm{3}}\\{−\mathrm{5}=\mathrm{4}−{a}^{\mathrm{2}} \:;\:{a}=\mathrm{3}\:,\:{a}=−\mathrm{3}\:\left({rejected}\right)}\end{cases} \\ $$

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