Question

Let  \[f\left( x \right) = \frac{\alpha x}{x + 1}, x \neq - 1\] . Then write the value of α satisfying f(f(x)) = x for all x ≠ −1.

 

 

Answer 1

Given:

\[f\left( x \right) = \frac{\alpha x}{x + 1}, x \neq - 1\]

\[Since f(f(x)) = x, \]

\[\frac{\alpha\left( \frac{\alpha x}{x + 1} \right)}{\frac{\alpha x}{x + 1} + 1} = x\]

\[ \Rightarrow \frac{\alpha^2 x}{\alpha x + x + 1} = x\]

\[ \Rightarrow \alpha^2 x - \alpha x^2 - ( x^2 + x) = 0\]

\[\text{ Solving the quadratic equation in }  \alpha: \]

\[\alpha = \frac{x^2 \pm \sqrt{x^4 + 4x( x^2 + x)}}{2x} \]

\[ \Rightarrow \alpha = x + 1 \text{ or } - 1\]

\[\text{ Since } , \alpha \neq x + 1, \]

\[\alpha = - 1 . \]