Let $\mathrm{f}(x)=\frac{\mathrm{ax}}{x+1}, x \neq-1$, then for $\alpha=$ $\qquad$ ,…

Let $\mathrm{f}(x)=\frac{\mathrm{ax}}{x+1}, x \neq-1$, then for $\alpha=$ $\qquad$ , $\mathrm{f}(\mathrm{f}(\mathrm{x}) \mathrm{)}=x$.
  1. $\sqrt{2}$
  2. $-\sqrt{2}$
  3. 1
  4. -1

Solution

$\begin{array}{ll} & \mathrm{f}(x)=\frac{\alpha x}{x+1} \\ & \mathrm{f}(\mathrm{f}(x))=\mathrm{f}\left(\frac{\alpha x}{x+1}\right)=\frac{\alpha\left(\frac{\alpha x}{x+1}\right)}{\frac{\alpha x}{x+1}+1} \\ & \text { But } \mathrm{f}(\mathrm{f}(x))=x \\ \therefore \quad & \frac{\alpha^2 x}{\alpha x+x+1}=x \\ & \quad \text { In L.H.S., Put } \alpha=-1 \\ \therefore \quad & \frac{(-1)^2 x}{(-1) x+x+1}=\frac{x}{-x+x+1}=x \\ \therefore \quad & \alpha=-1\end{array}$

Asked in: MHT CET 2024 (10 May Shift 2)

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