Question

If $f:(0,1)\to R$is defined by $f\left(x\right)=[b-x]/[1-bx]$, where $b$ is a constant such that $0<b<1$. Then,the value of $f\text{'}\left(b\right)$ is

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

Answer 1

The correct option is A

$f^{(-1)}$ is not invertible on $(0,1)$

Explanation for the correct option:

Find the value of $f\text{'}\left(b\right)$:

Given that $f:(0,1)\to R$ and $f\left(x\right)=[b-x]/[1-bx]$

$$\begin{gathered} f:(0,1)\to R \\ {f}^{-1}:R\to (0,1) \end{gathered}$$

$\begin{array}{rcl}y& =& f\left(x\right)\\ & =& [b–x]/[1–bx]\end{array}$

$\Rightarrow$$y–byx=b–x$

$\Rightarrow$$x–byx=b–y$

$\Rightarrow$ $x=[b–y]/[1–by]$

$\because 0<x<1$

$\Rightarrow$$0<[b–y]/[1–by]<1$

$\Rightarrow$ $[b–y]/[1–by]>0$

$\Rightarrow$$y>bory>(1/b)$

$\Rightarrow$ $[b–y]/[1–by]<1$

$\Rightarrow$ $[b–y]/[1–by]–1<0$

$\Rightarrow$$[b–y–1+by]/[1–by]<0$

$\Rightarrow$ $\left[b\right(y+1)–1(y+1\left)\right]/[1–by]<0$

$\Rightarrow$ $(b–1)(y+1)/[1–by]<0$

$\Rightarrow$$-1<y<(1/b)$

$b\in (0,1)$

$f\left(x\right)=[b–x]/[1–bx]$

$\Rightarrow$ $\begin{array}{rcl}{f}^{\text{'}}\left(x\right)& =& [-1+bx+b^2–bx]/{[1–bx]}^2\\ & =& [b^2–1]/{[1–bx]}^2\end{array}$

$\Rightarrow$ $f^{\text{'}}\left(0\right)=b^2–1$

$\Rightarrow$ $f^{\text{'}}\left(b\right)=[b^2–1]/{[1–b^2]}^2\ne 1/f^{\text{'}}\left(0\right)$

$\Rightarrow$ ${\mathit{f}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{/}\mathbf{[}{\mathit{b}}^{\mathbf{2}}\mathbf{–}\mathbf{1}\mathbf{]}$

Hence option $A$is correct