Question

Let $f\left(x\right)=\frac{\alpha x}{x+1},x\ne -1$. Then, for what value of α is $f\left(f\left(x\right)\right)=x?$
(a) $\sqrt{2}$
(b) $-\sqrt{2}$
(c) 1
(d) −1

Answer 1

(d) −1

$$\begin{gathered} f\left(f\left(x\right)\right)=x \\ \Rightarrow f\left(\frac{\alpha x}{x+1}\right)=x \\ \Rightarrow \frac{\alpha \left(\frac{\alpha x}{x+1}\right)}{\left(\frac{\alpha x}{x+1}\right)+1}=x \\ \Rightarrow \frac{{\alpha }^{2}x}{\alpha x+x+1}=x \\ \Rightarrow {\alpha }^{2}x=\alpha x^2+x^2+x \\ \Rightarrow {\alpha }^{2}x-\alpha x^2-x^2-x=0 \\ \Rightarrow {\alpha }^{2}x-\alpha x^2-\left(x^2+x\right)=0 \\ \mathrm{Solving}\mathrm{for}\mathrm{the}\mathrm{\alpha }\mathrm{we}\mathrm{get}, \\ \alpha =\frac{-\left(-x^2\right)\pm \sqrt{{\left(-x^2\right)}^2-4\times x\times \left[-\left(x^2+x\right)\right]}}{2x} \\ =\frac{x^2\pm \sqrt{x^4+4x^3+4x^2}}{2x} \\ =x+1,-1 \\ \mathrm{Here},-1\mathrm{is}\mathrm{independent}\mathrm{of}x, \\ \therefore \mathrm{for},\alpha =-1,f\left(f\left(x\right)\right)=x \end{gathered}$$