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


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

$If f (x) = \frac{a x + 1}{4 x - 3}, x \in R, x \neq \frac{3}{4}$ $and f f (x) = x, find the values of a$
	Answered by liberty last updated on 25/Nov/20
	$→ { ((f(x)=((ax+1)/(4x−3)))),((f^(−1) (x)=((3x+1)/(4x−a)))) :} ⇒ f(f(x))=x ; f^(−1) (x)=f(x) ⇒((3x+1)/(4x−a)) = ((ax+1)/(4x−3)) ; (3x+1)(4x−3)=(ax+1)(4x−a) 12x^2 −5x−3=4ax^2 +(4−a^2 )x−a { ((12=4a; a=3)),((−5=4−a^2 ; a=3 , a=−3 (rejected))) :}$
	$\to {\begin{cases} f (x) = \frac{a x + 1}{4 x - 3} \\ f^{- 1} (x) = \frac{3 x + 1}{4 x - a} \end{cases}$ $\Rightarrow f (f (x)) = x; f^{- 1} (x) = f (x)$ $\Rightarrow \frac{3 x + 1}{4 x - a} = \frac{a x + 1}{4 x - 3}; (3 x + 1) (4 x - 3) = (a x + 1) (4 x - a)$ $12 x^{2} - 5 x - 3 = 4 {a x}^{2} + (4 - a^{2}) x - a$ ${\begin{cases} 12 = 4 a; a = 3 \\ - 5 = 4 - a^{2}; a = 3, a = - 3 (r e j e c t e d) \end{cases}$
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