Given-f-2x-3-2x-1-f-2x-3-1-2x-4x-f-x-

Question Number 150044 by bramlexs22 last updated on 09/Aug/21

Given f (\frac{2 x - 3}{2 x + 1}) + f (\frac{2 x + 3}{1 - 2 x}) = 4 x

f (x) = ?

Answered by liberty last updated on 09/Aug/21

f (\frac{2 x - 3}{2 x + 1}) + f (\frac{2 x + 3}{1 - 2 x}) = 4 x \dots (i)

let \frac{2 x - 3}{2 x + 1} = y;

x = \frac{- y - 3}{2 y - 2}

\Rightarrow f (y) + f (\frac{y - 3}{y + 1}) = \frac{- 2 y + 6}{y - 1} \dots (ii)

\Rightarrow f (x) + f (\frac{x - 3}{x + 1}) = \frac{- 2 x + 6}{x - 1}

let \frac{y - 3}{y + 1} = u; y = \frac{- u - 3}{u - 1}

\Rightarrow f (\frac{- u - 3}{u - 1}) + f (u) = - \frac{4 u}{u + 1} \dots (iii)

\Rightarrow f (\frac{- x - 3}{x - 1}) + f (x) = \frac{- 4 x}{x + 1}

let \frac{- x - 3}{x - 1} = t; x = \frac{t - 3}{t + 1}

\Rightarrow f (x) + f (\frac{x - 3}{x + 1}) = \frac{- 2 x + 6}{x - 1} \dots (iv)

Answered by Rasheed.Sindhi last updated on 09/Aug/21

f (\frac{2 x - 3}{2 x + 1}) + f (\frac{2 x + 3}{1 - 2 x}) = 4 x \dots (i); f (x) = ?

x \to - x

f (\frac{- 2 x - 3}{- 2 x + 1}) + f (\frac{- 2 x + 3}{1 + 2 x}) = - 4 x

f (- \frac{2 x + 3}{1 - 2 x}) + f (- \frac{2 x - 3}{1 + 2 x}) = - 4 x \dots . (i i)

L e t \frac{2 x - 3}{2 x + 1} = u, \frac{2 x + 3}{1 - 2 x} = v

{\begin{cases} (i) \Rightarrow f (u) + f (v) = 4 x \\ (ii) \Rightarrow f (- u) + f (- v) = - 4 x \end{cases}

\Rightarrow f (u) + f (- u) + f (v) + (- v) = 0

\frac{2 x - 3}{2 x + 1} = u \Rightarrow 2 x - 3 = 2 ux + u

2 ux - 2 x + u + 3 = 0

x = - \frac{u + 3}{2 (u - 1)}

v = \frac{2 x + 3}{1 - 2 x} = \frac{2 (- \frac{u + 3}{2 (u - 1)}) + 3}{1 - 2 (- \frac{u + 3}{2 (u - 1)})} = \frac{- u - 3 + 3 u - 3}{u - 1 + u + 3}

v = \frac{2 u - 6}{2 u + 2} = \frac{u - 3}{u + 1}

f (u) + f (- u) + f (v) + f (- v) = 0

f (u) + f (- u) + f (\frac{u - 3}{u + 1}) + f (- \frac{u - 3}{u + 1}) = 0

\dots . \underset{\dots . .}{\overset{\dots . .}{}} \dots .