A) \[\sqrt{2}\]
B) \[-\sqrt{2}\]
C) 1
D) 2
E) \[-1\]
Correct Answer: E
Solution :
\[\because \]\[f(x)=\frac{\alpha x}{x+1}\] \[\therefore \] \[f(f(x))=f\left( \frac{\alpha x}{x+1} \right)\] \[=\frac{\alpha \left( \frac{\alpha x}{x+1} \right)}{\left( \frac{\alpha \,x}{x+1} \right)+1}=\frac{{{\alpha }^{2}}x}{\alpha x+x+1}\] But \[f[f(x)]=x\] \[\therefore \] \[\frac{{{\alpha }^{2}}x}{\alpha x+x+1}=x\] \[\Rightarrow \] \[{{\alpha }^{2}}=\alpha x+x+1\] \[\Rightarrow \] \[{{\alpha }^{2}}-1=(\alpha +1)x\] \[\Rightarrow \] \[(\alpha -1)(\alpha +1)-(\alpha +1)x=0\] \[\Rightarrow \] \[(\alpha +1)(\alpha -1-x)=0\] \[\Rightarrow \] \[\alpha +1=0\] \[(\therefore \alpha -1-x\ne 0)\] \[\Rightarrow \] \[\alpha =-1\]
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