Question

Let $f:(0,1)\to R$ be defined by $f\left(x\right)=\frac{b-x}{1-bx}$ where $b$ is a constant such that $0<b<1$. Then

• A. $f^{–1}$ is differentiable on $(0,1)$
• B. $f\ne f^{-1}$ on $(0,1)$ and -----
• C. $f=f^{-1}$ on $(0,1)$ and $f^{\prime }\left(b\right)=\frac{1}{f^{\prime }\left(0\right)}$
• D. $f$ is not invertible on $(0,1)$

Answer 1

The correct option is D $f$ is not invertible on $(0,1)$

Let $f:(0,1)\to R$ defined by
$f\left(x\right)=\frac{b-x}{1-bx}$, where $0<b<1$
We observe that
$f^{\prime }\left(x\right)=\frac{1+b^2}{(1-bx{)}^2}>0$
$\Rightarrow f\left(x\right)$ is strictly increasing $\forall x\in (0,1)$
It is obvious that $f\left(x\right)$ does not take all real values for $0<b<1$
$\Rightarrow f:(0,1)\to R$ is into function, and hence its increase does not exist.