Question

Which of the following functions is inverse to itself?

• A. $f\left(x\right)=\frac{1-x}{1+x}$
• B. $f\left(x\right)=5^{\log x}$
• C. $f\left(x\right)=\sin x$
• D. $f\left(x\right)=2^{x(x-1)}$

Answer 1

The correct option is A $f\left(x\right)=\frac{1-x}{1+x}$

We know,
$f\left(x\right)=f^{-1}\left(x\right)\Rightarrow f\left(f\right(x\left)\right)=x$
Now,
$f\left(x\right)=\frac{1-x}{1+x}$
Let $y=\frac{1-x}{1+x}$
$\Rightarrow xy+y=1-x\Rightarrow x=\frac{1-y}{1+y}$
Clearly, $f\left(x\right)=f^{-1}\left(x\right)$

For $f\left(x\right)=5^{\log x}$
$f\left(f\right(x\left)\right)=5^{\log \left(5^{\log x}\right)}=5^{\log x\cdot \log 5}$
Clearly $f\left(f\right(x\left)\right)\ne x$

Similarly, we find for the other $2$ functions that, $f\left(f\right(x\left)\right)\ne x$
Hence $f\left(x\right)=\frac{1-x}{1+x}$ is the only function which is inverse of itself.