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If f(x)=x−1x+1f(x) = \frac{x - 1}{x + 1} then f(f(a x)) is equal to

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Let x−1x+1=f(x)=y\frac{x - 1}{x + 1} = f(x) = y
∴x−1=yx+y⇒x(y−1)=−1−y⇒x=y+11−y\therefore x - 1 = yx + y \Rightarrow x(y - 1) = - 1 - y \Rightarrow x = \frac{y + 1}{1 - y}
Now,
f(f(x))=f(x−1x+1)=x−1x+1−1x−1x+1+1=(x−1)−(x+1)(x−1)+(x+1)=−22x=−1xf(f(x)) = f(\frac{x - 1}{x + 1}) = \frac{\frac{x - 1}{x + 1} - 1}{\frac{x - 1}{x + 1} + 1} = \frac{(x - 1) - (x + 1)}{(x - 1) + (x + 1)} = \frac{- 2}{2x} = - \frac{1}{x}
f(f(ax))=−1ax=−1a(y+11−y)=y−1a(y+1)=f(x)−1a(f(x)+1)f(f(ax)) = - \frac{1}{ax} = - \frac{1}{a(\frac{y + 1}{1 - y})} = \frac{y - 1}{a(y + 1)} = \frac{f(x) - 1}{a(f(x) + 1)}

Source: Grass-G / M-17 Mathematics Paper 26 Nov Second.docx

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