Question

Which of the following functions is inverse of itself?

• A. $f\left(x\right)=\frac{1-x}{1+x}$
• B. $f\left(x\right)=3logx$
• C. $f\left(x\right)=3x(x+1)$
• D. None of these

Answer 1

The correct option is A

$f\left(x\right)=\frac{1-x}{1+x}$

Explanation for the correct option:

Finding the inverse of the function

Given $f\left(x\right)=\frac{1-x}{1+x}$

$$\begin{gathered} y=f\left(x\right) \\ {f}^{-1}\left(y\right)=x \end{gathered}$$

Therefore,

$$\begin{gathered} y=\frac{1-x}{1+x} \\ \Rightarrow y(1+x)=1-x \\ \Rightarrow y+yx=1-x \\ \Rightarrow yx+x=1-y \\ \Rightarrow x(y+1)=1-y \\ \Rightarrow x=\frac{1-y}{1+y} \end{gathered}$$

Hence, functions $\frac{1-x}{1+x}$ is inverse of itself.

Explanation for the incorrect options:

Finding the inverse of the function of $\mathbf{3}\mathit{l}\mathit{o}\mathit{g}\mathit{x}$

For option (b)

Given $f\left(x\right)=3logx$

$$\begin{gathered} y=f\left(x\right) \\ {f}^{-1}\left(y\right)=x \end{gathered}$$

Therefore,

$$\begin{gathered} \Rightarrow y=3logx \\ \Rightarrow logx=\frac{y}{3} \\ \Rightarrow x=e^{\frac{y}{3}} \end{gathered}$$

Hence, function $3\log x$ is not the inverse of itself.

Finding the inverse of the function of $\mathbf{3}\mathit{x}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{)}$

For option (c)

Given $f\left(x\right)=3x(x+1)$

$y=f\left(x\right)$

$y=3x\left(x+1\right)$

$\Rightarrow$$3x^2+3x-y=0$

$\Rightarrow$ $x=\frac{-3\pm \sqrt{3^2-4\times 3\times \left(-y\right)}}{2\times 3}$

$\Rightarrow$ $x=\frac{-3\pm \sqrt{9+12y}}{6}$

Replace $x$ by $y$ and $y$ by $x$

$y=\frac{-3\pm \sqrt{9+12x}}{6}$

Therefore $y=\frac{-3\pm \sqrt{9+12x}}{6}$

$f^{-1}\left(x\right)=\frac{-3\pm \sqrt{9+12x}}{6}$

Hence, function $3x(x+1)$ is not the inverse of itself.

Hence, functions $\frac{1-x}{1+x}$ is inverse of itself.

Therefore, the correct option is option (A).