A) \[f(x)=\frac{1-x}{1+x}\]
B) \[f(x)={{5}^{\log x}}\]
C) \[f(x)={{2}^{x(x-1)}}\]
D) None of these
Correct Answer: A
Solution :
Since \[fof(x)=f\,(f(x)\,)=f\left( \frac{1-x}{1+x} \right)=\frac{1-\frac{1-x}{1+x}}{1+\frac{1-x}{1+x}}=x,\,\forall x\] \[\therefore \,\,\,fof=I\,\Rightarrow \,\,f\] is the inverse of itself.
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