Question

If $f\left(x\right)=\frac{x+1}{x-1},x\ne 1$ then $f^{-1}\left(x\right)$ is

• A. $\frac{x+1}{x-1},x\ne 1$
• B. $\frac{x+1}{1-x},x\ne 1$
• C. $\frac{1}{1+x},x\ne 1$
• D. $\frac{1}{1-x},x\ne 1$

Answer 1

Answer: (A)

$\frac{x+1}{x-1},x\ne 1$

Given that:

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

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

Then

$y=\frac{x+1}{x-1}$

Cross multipication

$\Rightarrow y(x-1)=(x+1)$

$\Rightarrow yx-y=x+1$

$\Rightarrow yx-y-x-1=0$

$\to yx-x-y-1=0$

$\Rightarrow x(y-1)=y+1$

$\Rightarrow x=\frac{y+1}{y-1}$

Put $x=f^{-1}\left(y\right)$

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

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

Answer