If f( x) = \(\frac{\alpha x}{x+1}\) , x ≠ – 1 , for what value o

1.	If f( x) = \(\frac{\alpha x}{x+1}\) , x ≠ – 1 , for what value of α is f (f (x)) = x?
Answer» f (f (x)) = \(f\big(\frac{\alpha x}{x+1}\big) = \frac{\alpha . \big( \frac{\alpha x}{x+1}\big)}{\big(\frac{\alpha x}{x+1}\big) +1}\) = \(\frac{\frac{\alpha^2 x}{x+1}}{\frac{\alpha x+ x +1}{x+1}} = \frac{\alpha^2 x}{\alpha x +x+1}\) Given, f (f (x)) = x ⇒ \(\frac{\alpha^2 x}{\alpha x+x+1}\) = x ⇒ α²x =αx² +x² +x ⇒ α² - 1 = (α+1) x ⇒ (α – 1) (α + 1) – (α + 1)x = 0 ⇒ (α + 1) (α – 1 – x) = 0 ⇒ α + 1 = 0 ⇒ α = – 1 [∵ α – 1 – x ≠ 0]

If f( x) = \(\frac{\alpha x}{x+1}\) , x ≠ – 1 , for what value of α is f (f (x)) = x?

Answer»

f (f (x)) = \(f\big(\frac{\alpha x}{x+1}\big) = \frac{\alpha . \big( \frac{\alpha x}{x+1}\big)}{\big(\frac{\alpha x}{x+1}\big) +1}\)

= \(\frac{\frac{\alpha^2 x}{x+1}}{\frac{\alpha x+ x +1}{x+1}} = \frac{\alpha^2 x}{\alpha x +x+1}\)

Given, f (f (x)) = x ⇒ \(\frac{\alpha^2 x}{\alpha x+x+1}\) = x

⇒ α²x =αx² +x² +x

⇒ α² - 1 = (α+1) x

⇒ (α – 1) (α + 1) – (α + 1)x = 0

⇒ (α + 1) (α – 1 – x) = 0

⇒ α + 1 = 0

⇒ α = – 1 [∵ α – 1 – x ≠ 0]