A) \[f(x)=\frac{1-x}{1+x}\]
B) \[f(x)={{3}^{\log x}}\]
C) \[f(x)={{3}^{x(x+1)}}\]
D) none of these
Correct Answer: A
Solution :
Let us consider\[f(x)=\frac{1-x}{1+x}\] \[\therefore \] \[fof=f[f(x)]\] \[=\frac{1-f(x)}{1+f(x)}\] \[=\frac{1-\frac{1-x}{1+x}}{1+\frac{1-x}{1+x}}\] \[=\frac{1+x-1+x}{1+x+1-x}\] \[=\frac{2x}{2}\] \[\Rightarrow \] \[f[f(x)]=x\] \[\Rightarrow \] \[f(x)={{f}^{-1}}(x)\] Hence, \[f(x)\] is inverse of itself. \[\therefore \]Option (a) is correct.
You need to login to perform this action.
You will be redirected in
3 sec
