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Question Number 122128 by bemath last updated on 14/Nov/20

F o r w h i c h n u m b e r s a, b, c, d

w i l l t h e f u n c t i o n

ψ (x) = \frac{a x + b}{c x + d} s a t i s f y ψ (ψ (x)) = x

f o r a l l x .

Commented by liberty last updated on 14/Nov/20

from ψ (ψ (x)) = x, we get ψ (x) = ψ^{- 1} (x)

first, we find ψ^{- 1} (x) .

\Rightarrow ψ^{- 1} (x) = \frac{- dx + b}{cx - a} = \frac{ax + b}{cx + d}

\Rightarrow (- dx + b) (cx + d) = (ax + b) (cx - a)

\Rightarrow - cd x^{2} + (bc - d^{2}) x + bd = ac x^{2} + (bc - a^{2}) x - ba

\to {\begin{cases} - cd = ac, a = - d \\ bc - d^{2} = bc - a^{2}; a^{2} = d^{2} \\ bd = - ba; a = - d \end{cases}

therefore we get a = - d and {\begin{cases} b is arbitary constant \\ c is arbitary constant \end{cases}

Answered by TANMAY PANACEA last updated on 14/Nov/20

ψ (ψ (x)) = \frac{a (\frac{a x + b}{c x + d}) + b}{c (\frac{a x + b}{c x + d}) + d} = x

\frac{a^{2} x + a b + b c x + b d}{a c x + b c + c d x + d^{2}} = x

\frac{x (a^{2} + b c) + a b + b d}{x (a c + c d) + b c + d^{2}} =

a^{2} + b c = 1

a b + b d = 0 \to b (a + d) = 0

a c + c d = 0 \to c (a + d) = 0

b c + d^{2} = 1

n o w i f a + d \neq 0 b = 0 a n d c = 0

a^{2} = 1 a = 1 a n d d = 1

a = 1 b = 0 c = 0 d = 1